So the last thing we proved was the path lifting lemma and the homotopy lifting lemma. So here's the corollary. So what does this do this is? So what is the situation? We have a covering, P, the ERB, a covering. And we fix some points, P of E0 equal B0 as before. So let F and G be pathless in B from B0 to B1, which are path homotopic. So they have the same end, which are path homotopic. Let F and G be pathless in B from B0 to B1, same initial point, same endpoint, which are path homotopic, and what? Sorry, I don't understand. Yes. Let F and G be pathless in B from B0 to B1, which are path homotopic. And let F tilde and G tilde be their liftings to E with initial point E0. So this we can lift, we know, OK? We fix the initial point and then we can lift it's even unique. Then also, F tilde and G tilde are path homotopic. And in particular, it's the same endpoint. Otherwise, we don't, and in particular, have the same endpoint. The same initial point is E0, have the same endpoint. In other words, if you lift homotopic passes with the same initial point, then they have also the same endpoint, the liftings, OK? They are also path homotopic, so the proof. So we have a path homotopic, OK? Between F and G, so there's a homotopic, H, homotopic. And the path lifting lemma, the homotopic lifting lemma, H lifts by the homotopic lifting lemma. H lifts to a path homotopic. Also, the lifting of a path homotopic is a path homotopic. To path homotopic. H tilde. So the situation is this. We have H, and it's a homotopic between F and G. Here's constant, which point, B0. And here it's constant, B1. So this is H, OK? A homotopic from F to G, a path homotopic from F to G. This means constant. So H tilde. Now we have lift H tilde. So the constant path, this is also a path homotopic, OK? The initial point here is E0, so this is E0. This is constant also here. It's constant also here, OK? It's also a path homotopic. I don't know the point, some point E1, OK? But these two paths, this is a path and this is a path. They are liftings of F. This is a lifting of F, and this is a lifting of G, OK? And they have the initial point, E0. Lost path homotopic H tilde. So by uniqueness, OK? So I say that by uniqueness of the lifting of F with this initial point, with the right initial point, this lifting can only be F tilde. There's only F tilde. So this here is F tilde, and this here is G tilde. So by the uniqueness of path lifting with a given initial point, in this point, E0 in this case, no? E0, we have that F tilde is homotopic to G tilde. H tilde is a path homotopic between F tilde and G tilde. This can be only F tilde, and this can be only V tilde. That's the uniqueness, OK? Because it's a lifting of F and G. OK, that's all. So this is a path homotopy, and so they have also the same endpoint, E1 in this case, OK? So this is the corollary of path lifting and homotopy lifting, this corollary. OK, for theorem, pi1 of the circle is base point 1, well complex number 1, OK? The circle we are in C is isomorphic to Z, the integers. So proof. So we define a mapping phi from pi1 of S1, pi1 of S1 to the integers. We want to define a mapping, OK? So what do we do? We take an element here, so that's the path. The homotopy class of a path, OK? So W is from I to S1 is a path with initial point and end point 1, this one here, with W tilde of 0 equal W tilde of 1 equal 1 in S1. So we do want to define the integer. This is a rotation number. How to get this rotation number? This can be interpreted as a rotation number. So what is a rotation number? How can we define this formally here? So we just take, let W tilde be the lifting of W. So we consider the P from R to S1, the covering, our covering. The P of t is equal to E2 pi I t, which we discussed yesterday. We take this covering. Let W tilde from I to E be the lifting of W with initial point, W tilde of 0 equal. So we have to choose one point which projects to 1 in S1. And that 0, the best choice is 0, is 0 in the reals. The lifting of W, this is not W1, W with initial point W tilde of 0, which is 0 in the reals. Then, of course, I mean W is a closed path, OK? So then W tilde of 1, the end point of 1, OK? What? Yes, that's E is R in this case, OK? OK, yes. This initial point, then W1, of course, is in the fiber pi minus 1 of 1, OK? But under this mapping, which goes to 1, the integers, exactly, OK, which is z. So that means W tilde of 1 is an integer, OK? This is a rotation number, OK? The rotation number can be, you can call this the rotation number of W. How to get this, the rotation number? You take the lifting with initial point 0. The end point is an integer, and this is the rotation number, OK? If you go downstairs, n times around, OK, in one direction, in S1, n times, OK?  And you get the rotation number in this way, at least in this special case. It's just a name, OK? You call it rotation number, and it doesn't matter. So we define phi of W equal to, well, at this point, W tilde of 1, which is an integer. So this is the mapping phi. Of course, there's a big problem here, you know? That's well-defined, OK? First, we have to lift. That's a path lifting lemma. And now, it has to be well-defined, OK? So if you have homotopic paths, if, so phi is well-defined, this is, if W is W prime, or if you have another representative, that means W is homotopic to W prime, OK? H, homotopy, some path homotopy. Then, by the homotopy lifting lemma. So let W tilde and W tilde prime be the liftings of W and W prime with initial point 0, OK? Our initial point is 0 in our, in the way. This initial point 0, in the way. By the corollary, by the path lifting lemma, by the corollary, W tilde of 1 is equal to, they have also the same endpoint, OK? W prime tilde, they have also, because downstairs they are homotopic, we lift the homotopy and upstairs it's also a path homotopy, so they have also the same endpoint. And this means, of course, that phi is well-defined, OK? So it doesn't depend on the representative of this cloud. So phi is well-defined. OK, this is the main part of the proof, OK? The rest is true, almost. Now we have to prove something still, but the rest is very easy. Here, this is an important part, OK? We use the path lifting lemma and we use the homotopy lifting lemma or the corollary, OK? So this is the main part of the proof. Now we have to prove something. What we have to prove, so after the definition, phi is a homomorphism of groups, OK? Phi is a group homomorphism. That's the first one we have to prove. So what does it mean? It means phi of w times v should be equal to phi of w plus in z now, plus phi of v, OK? That would be a homomorphism. So what is phi of w? This notation as before, this is w tilde of 1. So w tilde is a lifting with initial point 0 and then we look at the endpoint with the rotation number. So this is w tilde of 1. What is this? It takes the same for v. We lift it, v tilde with initial point 0 and takes the endpoint, so v tilde of 1. v tilde is the lift of v, obviously, in the same way, OK? Now we have to look at this. So this is phi of w times v. So this is defined as we have to lift now the product and look at the initial point 0, or this, and then look at the endpoint. So this is the lift w times v tilde, the lifting of this product and the endpoint. That's the definition, three times the definition of phi, three times the definition. Now we have to look, who is this pass? So we want to understand this lifting. Well, this means first we lift w, this initial point 0, that we have already, now that's w tilde. This lifts w with the right initial point, OK? And then we have some endpoint, which we don't know. And then we have to lift v. But not with 0 initial point, because maybe the endpoint here is not 0, it's the rotation number. But the initial point is the endpoint of this, of course, OK? The endpoint of this is w tilde of 1. So I cannot write v tilde here. This is the wrong initial point, OK? That's not defined in general, that's not defined, OK? But what I have to do is I have to translate v tilde, which starts from 0, so that it starts at the endpoint of this one, right? It must start in the endpoint of this for the product, OK? So what has been, I don't take w tilde, but I translate v tilde, OK? I translate it with which number? With the w tilde of 1, with the integer, OK? So w tilde 1, so this is this integer, which is some number in that, which I don't know, OK? So this means a translated path v tilde, OK? A translation, OK? A translation of v tilde. And now this is defined, OK? Now this starts, this w, v tilde of 0 is 0, OK? So it starts here, and it ends somewhere. So w times v tilde of 1 is, of course, this is first, this ends this. It's the endpoint of this, OK? The endpoint of this is v tilde plus w tilde of 1. This is just a constant, OK? This is just a constant, we add, we add the reals. Add 1, right? This is w tilde of 1. And then we have this translation plus w tilde, v tilde of 1 plus w tilde of 1, OK? So this is the endpoint. And this is, of course, exactly what we want, OK? This is w tilde of 1 plus v tilde of 1, OK? So this is what we need. So they are equal here. And that's the homomorphism. So this is just looking at the situation. There's nothing difficult here, OK? Just definitions. So this is a homomorphism of groups. And then we want an isomorphism, right? So second, injective. So we prove that the kernel is true here, OK? So we take something in the kernel, w in the kernel of 5. What does it mean? It's in the kernel of 5. This implies that w tilde of 0 is 0 anyway, no? What is w tilde of 1? So in the kernel, that's everything which finishes in 0, right? So it's 0 also. That means exactly this, OK? If and only if, OK? So this means w tilde is a closed path. It starts in 0, and it finishes in 0 because it's in the kernel. So it's a closed path. So it's an element where w tilde is an element in, it's a closed path in the reals, no? It's an element in the fundamental group of the reals with base 0, OK? That's our initial point, and also end point here. But what is this group? The fundamental group of Rn is trivial. So this is a trivial group. That's a straight line homotopy. This is a trivial group. So what means? That means this is a trivial element. So this implies that w tilde is equal to the trivial element. What is a trivial element? That's a constant path in 0. Or if you want, w tilde is homotopic to the constant path by some homotopy f, maybe h. And then we want to prove that w, what is w? w is w tilde and then the projection, no? It's homotopic to what? To c0 and then the projection. What is this? c0 and then the projection. This is c0 to 1 in s1, c1, constant path. This is constant path in s1. This is in R. This is projected. And what is the homotopy here? I mean, look at this. It's clear what the homotopy is. You project h. You project everything. h and the projection, OK? This is very trivial, this direction, OK? What is difficult is lifting in the other direction, OK? Here we just project everything, and it's OK. So this means that w is equal to the constant path, OK? And this means that the kernel is trivial, OK? This implies the kernel of phi is trivial. And of course, this means that phi is injective, because we know already that it is a homomorphism of groups, OK? So this is injectivity. And the last thing is surjectivity, which is even easier, 3 phi is surjective. So let n be in n. Let n be in zeta. No, not r. Sorry, that makes no sense. We take any integer, OK? And then, of course, now we want to find a path, OK? Which gives this n. So we take let w let u be any path in the reals from 0 to n. Let u from i to the reals be any path. u of 0 is 0, the initial point. And u of the n point, n. This is the rotation number. Now we project, OK? Now we project. Let w be the projection of u, u and then the projection. Then w is a closed path, because the integers go all to 1, OK? So let w, this is a closed path, because integer goes to 0. It goes to 1 in S1. And that means that this w is an element of pi 1 of S1, 1. It's a closed path. And now what is phi of this w? Then phi of w. So how do we find this? We take the lifting of w with initial point 0, OK? And then we take the n point, so w tilde of 1. But who is w tilde? I mean w is a projection of u, OK? So this is u of 1, because w tilde is u, OK? The lifting is because we project it. And u of 1 is what is this? n. So in other words, it's easy to find path with rotation number n in S1. You just go n times around, OK? And if you lift this, you have the linear path from 0 to n. That's trivial, OK? And this means phi is projective. And that's the end of the proof. That's an isomorphism of groups, OK? And this is a rotation number, no? You have any path in S1 which goes very complicated, no? I mean there are easy paths which just go n times with constant velocity around n times, no? Then the lift of this goes from 0 to n. That's clear, OK? That is trivial. But if you have a very complicated path, no, which there's a lot of, then it's not so clear. Now first we have to lift, and then we look at the end point, and that gives us a rotation number, OK? And that's all. And then it has to be well defined. These are the two main points, the lifting of any path, not just of a trivial path, which goes n times around, OK? But the path which may be very complicated, we have to lift this, and then it doesn't depend on the choice of representation. And that's the homotopy lifting lemma, OK? So these are the two main points of the proof. The path lifting lemma and the homotopy lifting lemma, or it's color. OK, so this is a proof. That's the fundamental group of the circle is the integers, and the isomorphism in a sense is given by the rotation number of a path, which we can define in this way, OK? We just define it in this way. That's we call, we give the name, oops, rotation number. OK, so I give two applications now. The first one is a fixed point theorem of Brouwer.  This is an important definition anyway.  R from X to A to the subspace is a retraction. That's a definition of retraction. That's a continuous map from X to the subspace, such that the restriction of R to the subspace A, OK, is the identity on A. On A, it doesn't move points. So it retracts X to A, OK, and A doesn't move. So this is the identity of A. That's the definition of, that's the definition of retraction. A continuous map to a subspace, which is the identity on A, which doesn't move points in the subspace, OK? This is equivalent to the fact that this can be written in a different way, that first the inclusion and then the retraction is the identity of A, where I from A to X is the inclusion. This is exactly the same as this. This or this is the same, OK? Here you have the inclusion and here the restriction, but it's exactly the same. So that's the definition of retraction. For example, I define a retraction from R minus 0, Rn minus 0 to Sn minus 1. What is Sn minus 1? These are all points in X in Rn such that the norm is 1, OK? This is a unit sphere. The norm of X is 1. This is a unit sphere. Unit sphere in Rn, that's Sn minus 1, OK? S1 is the unit sphere in R2, OK, on C, if you want. So what is a good definition of such a, what is a retraction, a retraction of this? For example, if you take any point X here, no? And what should be the image? Yes, exactly. X by the value of the norm of X. This is a retraction. That's an example. So small lemma, if R is a retraction, by the way, then if there exists a retraction, A is called a retract of X, OK? If R, this R, not V, the R, from X to A is a retraction, then the induced map R star from pi 1 of X, with some base point, which I choose here, this is a subspace, no? A0 to pi 1 of A, A0, is subjective. And the inclusion induced map, I star, which this is inclusion as before, from pi 1 of R, A, A0, to pi 1 of X, X0, A0, is injective. So here A0 is a point of A, which is a subset of X, a subspace of X. So our star of a retraction is subjective, and I star. This is very trivial because, to the proof, what means a retraction? That means that first the inclusion, and then the retraction, is the identity map of A, no? That's our definition, the second version of our definition. That by the factorial properties, OK? That implies, of course, that I and then R, the induced map, is the identity of A, induced map. And these are equal, the induced maps are equal, of course. And now we have the factorial properties, OK? This is I star and then R star. And this is the identity of the fundamental loop, A, A0. So they are equal. And this is by the factorial properties. There are two, no? By the two factorial properties. So we have this, and what does it mean? This means that I star is injective, and R star is subjective, no? As usual. So this implies I star, R star is subjective, and the inclusion star is injective. Because this is subjective and injective, OK? If this is subjective, injective, then this has to be injective and this is subjective, no? Because this map is both. OK, so for a retraction, we have the induced map is subjective. You mean this might be an isomorphism? This is an isomorphism. But that doesn't mean that these are isomorphisms, no? This composition is the identity. That's an isomorphism, of course. Yeah, but that doesn't mean you need more identity. OK, it's clear, right? The proof, yes. So now we give you an implication. So these are general stuff. And all the first application, proposition, oh, this is easy. There is no retraction. No, well, there is no retraction R from the unit ball. So what is the best name in the book, I forget? B or D? Disk, B. So these are all points. Z in x in R2, such that x norm is smaller equal to 1 now, OK? That's a unit disk. What? Well, yes, imagine 2 is disk, OK? And then it's unit ball in all dimensions. So R, B to S1. S1, by the way, which is the boundary, OK? So this is the boundary of B2, S1 B2, OK? So B2, this is the boundary of B2. There's no retraction from here to here, OK? There's no retraction. Proof is easy. I mean, it follows from what we proved. So proof, if R would be a retraction, would be either a retraction, by contradiction, if you want. If you have a retraction, suppose you have a retraction. If it's a retraction, then R star is subjective. Then R star from pi 1 of B2, base point, I take 1 also. And the same base point as here, no? So this then is Z in C, Z smaller equal to 1, the complex version, no? So pi 1 of S1, 1, the same base point, OK? 1 as a complex number. Then it's subjective. But what is this group? Now we have this group is isomorphic to the integers, OK? What is the group of B2, the fundamental group of B? What? Trivial, yes. So this is the trivial, not trivial, trivial, trivial, trivial, trivial group, right? This is by straight line homotopy, like Rn, OK? Here we have straight line homotopy. The same proof as for Rn. So this is a trivial group, OK? And of course, there's no surjection from the trivial group to the integers, right? So this is a contradiction, so contradiction. That's not possible. The important fact here is that this is not trivial, OK? That's the only, this is not trivial. That's the important fact. That this is trivial is trivial, almost, OK? This is easy. The important fact is that this is not trivial. So there's no retraction. By the way, there is a retraction, of course, remark. There is a retraction from R from, that's almost what we did before. If you take b2 minus 0, then we have a retraction 2s1, OK? And that's x goes to x over the norm of x. That's a retraction. How does it look like this retraction? So we have b2, and here's 0, right? Which is not there. So you take any point, and then you project it, OK? So if you have this x, then this is R of x, OK? That you see this retraction, this is this retraction, OK? But if you have 0, there's no retraction. You don't see any retraction. You don't know how to move 0, no? 0 cannot go to everything, OK? But you cannot move. So this is the fact that there's no retraction if we, OK, this is easy. And this uses the fact that it's a fundamental rule. So theorem, and this has a name. It's a famous theorem. It's a Brouwer fixed point theorem in dimension 2 law, OK? It's in all dimensions, but we have dimension 2. Brouwer fixed point theorem in dimension 2. So what does it say? Every continuous map, continuous map, f from b2 to b2 has a fixed point. Very easy, no? Formulation. Every continuous map from b2 has a fixed point. That's a famous Brouwer fixed point theorem for 2. This holds for bn, OK, for each n. Let's prove this first. So prove. Suppose f has no fixed point, OK? And you want the contradiction. So I make a picture now. This is b2. The unit disk, and we have f. So any each point, x, goes to another point. f of x, which is different from x. It has no fixed points, OK? So what you take is you take the segment from f of, well, to x, from f of x. So these two points define a segment, a line. And then you go in this direction on this line, from f of x to x, OK? And this meets the boundary, which is s1, in two points. And you choose this one where you go in this direction. And you call this r of x, OK? That's the definition of r. So this defines r from b2 to s1. So if you take a point x here on the boundary, f of x is different from this point, no? I don't know where it is. It's here, f of x, somewhere, but not the same point. Then you get the same point back. So then rx is equal to x, OK? So this r restricted to s1 is the identity of s1, OK? And r is continuous. Why is it continuous? Well, I don't want the formula, because since f is continuous, OK, since f is continuous. Well, very, I mean, you can compute, if you want, you have x coordinates, two coordinates, no? You have f of x, which depends in a continuous way of these two coordinates, OK? f is continuous. And then x and f on x, you can compute the line and then you compute this point. And it depends on the continuous way of the coordinates, OK, of the coordinates of x. Because f of x depends on the continuous way, and then also there's some square root of some expression, OK? In other words, if you vary x very little, OK, then f of x varies very little, because f is continuous. And then also this varies very little, OK? But you should see the formula, OK? You have x two coordinates in r2, x1, x2, xy, no, xy is not good, x1, x2, OK? And then you have f of x, which depends in a continuous way, f1 of x of these coordinates. And then you can compute this point, OK? And it depends in a continuous way on these coordinates, OK? That's clear. So this means that r is a retraction. But there is no retraction, OK? So this is a contradiction. There is no retraction from b2 to r, to s1. No retraction. There's an exercise which I gave or not, no, I don't remember even. The same for b1. What is b1? That's minus 1, 1, OK? This is in r, r1. That's a ball in r1, OK, the unit in some sense, no? So every continuous map, f from b1 to b1 has a fixed point. That's the problem of fixed point in dimension one. This is much easier, OK? Was it an exercise? No? No, it's an exercise. I forgot. Maybe you forgot also. 0, 1, 0, 1, it's the same. f from 0, 1 to 0, 1. Now I have minus 1, 1, OK? Yeah, it was 0, 1, 0, 1, maybe in the book, OK? So how do you prove it? I give the same proof as this, OK? What? No, no, no, that was a homework. Every continuous map from 0, 1 to 0, 1 has a fixed point. Yeah, and now it's minus 1, 1. Long time ago. Yes, that's the same exercise. Except that now I take minus 1, 1, b1, OK? But no difference. That's homeomorphic to 0, 1. All closed intervals are homeomorphic, OK? So this was an exercise. Yeah, I know I'm doing the exercise in this line, OK, along these lines. So what? So I define the retraction here in the same way, OK? I define r. Suppose f has no fixed point. So prove. It's the same proof. Suppose f has no fixed point. And then I define a retraction, OK? Now I have b1. This is b1, minus 1, 1. This is b1, no? And I define it exactly in the same way. What was the way? I take a point x. I look at f of x, which is different from x. And then I take the retraction, which goes from f of x to x, and goes to this point, OK? So you go from f of x to x, as here.  So db1, what's that called? That's s0, no? There's two points. s0 is minus 1, and minus 1, OK? And so this is r of x, in this case. So now we can write a formula. Here it's easier to write a formula, OK? No? What is r of x? Let me try. Don't write that immediately. f of x minus x, which is not 0. So I can divide by the absolute value, no? f of x minus x, absolute value. It's not the, wait a second, I have to check, OK? If this is this mapping, no? So where goes 1? I'm interested in 1. No, in minus 1 first, OK? In my, well, doesn't matter. f of minus 1 is on this side, it's not minus 1, OK?  This is the definition, of course. And r of 1 is 1, right? That's what we need. So is it true here? So I have to write minus, yeah? Let's see, if it's minus 1, if it's 1, x is 1. This is smaller, so this is negative. So it becomes minus 1, yeah? I need the other one. x minus f of x, x minus f of x. So that's OK now, OK? This implies r of 1 is equal 1, r of minus 1 is equal minus 1. Or if you want, r, restricted to s0, is the identity of s0. So this is a retraction. r from b2, from b1 to s0 is a retraction. As before, in particular, surjective, no? Retraction is surjective. Where's the contradiction? We need the contradiction, as in the first case. In the first case, we use fundamental rule. No, where's the contradiction? So r goes from minus 1, 1 to 1, minus 1, no? Yeah, exactly. This is connected. The image of a connected is connected, OK? This is not connected. Not connected. Obviously, it's not connected. No, you get a separation then. What is a separation? r minus 1 of 1. So you write minus 1, 1. You have a separation, which is r minus 1 of 1, union r minus 1 of minus 1. If they're both non-empty, there's a separation, no? But there is no separation. So we have a contradiction again. So it's the same proof, right? Formally trivial, straight line homotopy. Pi 1 of r n is trivial, pi 1 of b n is trivial. They are all trivial groups. You always have this straight line homotopy. Like b2, OK? So in general, yes. Now, fundamental group P is not very useful, no? Because here we have just two points, OK? Pi 1 of r n, 0 is trivial. Pi 1 of b n, 0 is trivial. Pi 1 of whatever you want, OK? Of any convex subset of r n is trivial, OK? Pi 1 of a convex subset of r n with some base point is trivial. Because in a convex subset, you don't need convex subset. What you need is star-shaped, OK? But let's say convex, that's OK, OK? That means with any two points, you have a straight line, OK? So you have a straight line. It's always straight line homotopy. Pi 1 of any convex subset of r n is a trivial group by straight line homotopy. And this is a special case of this, of course. And this is also a special case of this. Now, a fundamental group here, this is trivial, fundamental group, OK? That was your question, no? This is trivial group. And this is trivial also, OK? So everything, fundamental group doesn't help here. But it's easier. It's connectivity, no? So now I'll prove another application. I'll give some easy lemma first, one easy lemma set. If a continuous map, f from s1 to x extends to a continuous map, well, whatever, f, capital f, from b2, that's a unique disk, to x. So this means f restricted to s1 is f, no? That means this extends this, OK? Extends a continuous map. Then f star, the induced map, from pi 1 of s1, some base point, 1, to pi 1 of x. Well, I don't know, f of 1 I have to write. You have to preserve the base points, OK? f of 1, I don't know which one. Then this is a trivial homomorphism, OK? This is an exercise, or a lemma, whatever. Typical lemma, typical exercise also. So proof, very easy. So what you have, we have that with the inclusion that I, as before, from s1 to b2, b is inclusion, as before. Look at the inclusion. And then this means that f is nothing else than the inclusion and then the extended. That means exactly this again, OK? First inclusion, then f is f. And now we have functorial properties, OK?  And now I look at this, OK? Where does this inclusion go from? This goes from pi 1 of s1, 1 to inclusion, OK? Goes from this to pi 1 of b2, the same point, 1. And here we have f star, which returns to pi 1 of where f goes from there to there. No, x, f goes from b2 to x, yes. x and f of 1, whatever that is, OK? And this, what must it be, this mapping, this is f star, the composition. First i star, then f star. This is f star, right? That's exactly what it says here, no? This is a functorial property, OK? But what is this group here? This is z. What is this group? Trivial, no? Straight line homotopy. Trivial. And this we don't know, of course, no? That's x is any space. We don't know anything about it. However, so this is this, and this is this, no? And this is f star. So it goes from this to the trivial group, and then it goes on. So this means it says trivial, no? f star is trivial, because it factors over the trivial group, OK? So it can only be trivial, f star is trivial, OK? So this is easy. The second exercise, or lemma, small. So this is the first one, and now the second one, lemma. If f is homotopic to g by a homotopy h, and f and g are continuous maps from s1 to x, OK? We have two maps from s1 to x, which are homotopic. Then if f extends to b2 as before, if we can extend this f from s1 to b2 to x, OK? If f extends to b2, then also g extends to b2, OK? If one of these extends, then the other one also. And then by the lemma, f star extends as trivial, no? And then also g star is trivial, OK?  So also g star is trivial, and also g star is trivial. That's what we used the lemma before. So proof. So I make a picture. So what means h? h is a homotopy from s1 cross 0, 1 to x, OK? And now one time, instead of 0, 1, I take 1, 2. You will see why. It doesn't matter, no? Very much. If this is 0, 1, or 1, 2, no, 2x, b homotopy between what? f and g. So s is written here. So now I make a picture. So this is not a unit. This is radius 2, OK? Ball of radius 2, not 1. But here's also b, no? This is 1. So this is b2, radius 1, OK? But this is twice the radius, no? There's a bigger ball. So now I want to define this mapping. If x extends to, let me call it f from, sorry, f from b2 to x, right? Then also g extends to b2, OK? This is a bigger b2 here, but it doesn't matter. So I want to define a mapping from this to x, the extension. So here I have, on this, I have f on this, OK? And this extends to b2. So what I do here, f, OK? So on this part, it extends, OK? This is f on this circle, s1. And this extends to f from b2 to x, OK? What I want is g here outside. Extend to a continuous map, no? This part I have. I have to define it here. But this part is homomorphic. It's equal to what? What is this part? This part here. That's an annulus. It's homomorphic to what? Or it is equal to homomorphic to s1 cross 1, 2, OK? That's s1 times an interval, not clear? This is b2. And this part between these two, that's s1 times the interval, right? Well, you can write homomorphism, no? Of course. I mean, you see here the intervals, and this is s1, OK? So this is s1 times the interval. But I have a map from s1 times interval 2x, which is h, our homotopy, OK? So here you take h on this part. h goes from f to g, the homotopy. So by the pasting lemma, this mapping g is continuous, OK? And extends g. By the pasting lemma, g is continuous and extends, well, it extends to what? Not to b2 exactly, no? Because this is a larger b2, OK? But it doesn't matter. It's the same, it's the aromimorphic, OK? This is radius 2 now, but we don't care. It's a close ball. They are all aromimorphic, independent of the radius. So this is the picture. If f extends, so we have this f, OK, to b2, then g extends. And in this part, I take the homotopy between f and g, OK? So this is all. Now this next application, the second application, well, theorem, even more famous, which will prove in complex analysis, fundamental theorem of algebra. What does it say? I forgot the notation. Maybe not the fundamental theorem of algebra. I need some notation. So what does it say? Every polynomial, complex polynomial, non-constant complex polynomial has a root, has a 0, OK? Has a root. So every non-constant complex polynomial, so x to the n plus a n minus 1, x to the n minus 1 plus a 1x plus a 0, has a root. How do you say root or 0? What would you say? Root in C. This means the coefficients in C, a n minus 1, a 0, it's a complex polynomial. And n bigger than 1, OK, of course. This means non-constant, OK? n bigger equal to 1. The degree has to be at least n1. For n equal 1, it's trivial. It was just 0x minus a 0 divided by a 1. So this is trivial. For a 2, it's trivial, OK? You have the formula. For 3, there's some formula. For 4, there's some formula. For 5, there's no formula. What? Sorry? I don't understand the English. Then for 5, it's not, yes, right? Maybe in algebra. It's not resolvable, no? Because the alternating group A5 is non-solvable. Yeah, exactly. That's 5, starts with 5, OK? Exactly. So proof. And you understand why this is a case, OK? Hopefully. So there are two cases. The first case is an interesting case. The other case is easily reduced to the first case. So suppose we will see why this is strange. But in the proof, n minus 1 plus a 0 is smaller than 1. The absolute values of the coefficients, the sum is. So they have small coefficients, OK? Small coefficients. But the other case is you just reparameterize.  You divide by something or something. Then you go to this case anyway. Small coefficients. Suppose, so these are called f of x, sorry, f of x, or p of x. Whatever. p of x, that's a polynomial, OK? p of x, OK? That's my polynomial. p of x, complex polynomial. Suppose p x has no root. And we need a contradiction. So I define a mapping G from B2. So this is in C, the unit disk in C, in the complex numbers, to R2 minus 0. So this is our polynomial, but on B2. So it's just a polynomial. And it has no root, so no 0. So G of x is just p of x. That's our polynomial, OK? But we consider it on B2 only, not on the whole complex, OK? And even we restrict it to S1, OK? And so this is G and f from S1 also, if you want. This is not so nice, no R2 here. It should write C, of course, OK? Sorry. What? No, what I say is I should write C here, not R2, OK? Because we're in the complex setting. So it's R2, but it's better C minus 0, OK? So f from S1 to C minus 0, and it's again the polynomial. f of x is p of x. So f is what? f is the restriction of G to S1, no? Yes. It's obviously a polynomial, OK? One time considered on B2. First on C, then on B2, and then on the boundary, OK? S1 of B2. It's always a polynomial. So this means, so what is this? f, this is our mapping, f, extends to B2. That's G, OK? So f extends to B2, no? G. It's always a polynomial.  The use method is trivial. So let's see. f star goes from, here we see why it's true, OK, already. f star goes from pi1 of S1 with some base point 1 always to C pi1 of C minus 0 with some base point. I don't know which one. f of 1. So I didn't say what is this fundamental group, OK? This is z. Both are z. We don't need that necessary. But this is also the integers. And we have rotation number also here, OK? Around 0, here along S1, here around 0, OK? Also rotation numbers. But now you look at your polynomial. This is polynomial of degree n. This is not the proof. I mean, this is the idea, OK? You have a polynomial of degree n, OK? And now you take this S1, this S1. This is a unit, OK? But you take some S1, which is very large, OK, in C. Some other, which is very, very large, OK? Then what does it mean? Then the only thing which is important, if this S1 is very large, then this is completely dominated by this power. This get, OK, are no longer important. If the larger we get, OK, then these are not important. This x to the n is a big thing, OK? The other things you can neglect then, OK? Then we can suppose that our mapping is just this, OK? And forget this for a very large, OK? And then you say this would be trivial, OK? In any case, this would be trivial, f star, no? Also for a very large, no difference, OK? For a very large S1. This would be trivial. But x goes to x to the n. How does that look like? Now we have just x goes to x to the n, OK? You say this is not important for a very large x. Small, so forget. So x goes to x to the n, like x goes to x to the n. But what is the rotation number of x goes to x to the n for large? If you go one time around 0 in C and you have x goes to x to the n, you go n times around 0, right? n times around 0.  So 1 goes to n. That's not the trivial homomorphism. That cannot be the trivial homomorphism, no? You said it's trivial, but that's not possible. If you go to a very large S1, it shouldn't be trivial. It should be 1 goes to n, OK? That's all in some sense. That's your main idea, OK? If it extends, F star is trivial, OK? But this map should not be trivial. Because if you go one time around S1 here, then x goes to x to the n for getting the small stuff, OK? It goes n time around 0 here. And this is not the trivial homomorphism, as you will see. But that's the idea. That's the whole idea here, OK? That's the idea of the proof. That's not the proof. I will prove it now. That's a good question. I said we have to take this S1, very large, OK? This circle, then if x is not so large, then these are important, OK? If x is very large, then this is not so important. And the fact that you take S1, very, very large, corresponds. If you take S1, radius 1, then you take the coefficients very, very small. This is the correspondence, OK? In some sense. That's not the proof, OK? So very small coefficients, then this S1, the standard S1, seems large, OK? If the coefficients are arbitrary, then you have to go to a very large, OK? That's the connection uses. More or less, it's not the proof. Intuitively, it's so, no? If your polynomial is very small coefficients, then either you have a polynomial with very small coefficients, or you go to some S1, which is very, very large, no? This is sort of that's only the idea. But you can understand the idea, OK? Why the fundamental theme of algebra should be true, OK? All here, that's the idea, OK? If there's no 0, then you have this mapping, and it is trivial, because it extends, OK, to B2. And then it's a lemma, that's easy, OK? But this should not be trivial for this polynomial, because you have a rotation number here one time. You have here n times around 0, OK? And that's not a trivial. 1 goes to n, not to 0. That's the whole idea. No, I give the proof anyway, OK? I mean, we are in the proof. Where are we? So f extends to B2, OK?  I don't know if this is that. We can prove that, but it's not important here for the rest. Not too important. So I go on with the proof, no. There's a formal proof. So we want a contradiction. So the contradiction, now, we follow this informal proof, OK? We say these coefficients are not very important. The other ones, OK? So we must put them to 0. So what we do is we define k. Let k from s1 to c minus 0 be the mapping x goes to, well, I take just x to the n now, OK? That's the dominating first term of our polynomial. That's I call k, complex numbers. Then, and this is important, f is homotopic to k. And these are maps from s1 to c minus 0, OK? Not to c. To c doesn't help, because the fundamental group of c is trivial. So these are maps from to c minus 0. They have to go to c minus 0. Any two maps to r2 are homotopic. Any two continuous maps to c, to r2 are homotopic. Important is c of c minus 0. That's an important fact, OK? For c, it's a trivial fact. So I have to define a homotopy here, OK? And this uses this condition. I will use now this condition. So the homotopy is the following. The homotopy is the most trivial you can imagine. What is the most trivial you can imagine? So you define h of, I call it h, no, h is OK. H of xt, h of xt. So h should be a map from where to where? From s1 cross i to c minus 0, to c minus 0, OK? So what I say, I just take the most trivial, which I can think of, OK? So I put here some small, this t, and let the rest disappear. a n minus 1, xn minus 1, plus a 0. So for t equal 1, it's all polynomial, no? For t equal 0, it's just k, OK? So let this disappear. Then there's one point to check now, because this has to go to c minus 0, OK? In c, everything is trivial, but we need c minus 0. And this I will check now. This is our condition, OK? So h of xt, absolute value. So this is this computation, which is this is bigger or equal to x to the n. And now I subtract as much as I can. All the rest, t times the rest, a n minus 1, xn minus 1, plus a 0. Is that OK? Yes, I subtract as much as I can, OK? So this is smaller equal to, this is 1, anyway. We are on s1, right? We are on the unit circle. So it's 1 minus t. And now I make it also this bigger, as big as I can. So I take a n minus 1, xn minus 1, value, absolute plus. I make this even bigger, plus a 0, OK? What I subtract, I make this larger and larger, OK? So this is equal to 1 minus t. And now x has value 1, OK? So it remains this value, a n minus 1, plus a 0, right? And the hypothesis was that this is smaller than 1. That's the hypothesis. That's exactly what I wrote. This is smaller than 1, means that this is bigger than 0. This is 1 minus t is between 0 and 1, OK? So this is bigger than 0. And this implies that h of xt is not 0. It's not 0. So we avoid 0. So this is the condition. Here we have the condition of the first case. That we use here, only at this point. So we have almost finished with the first case. So f extends to b2, no? f extends to a map. Well, we have to always be careful. f from, no, what was it called? g, was it, right? From b2 to g minus 0. We have to take always g minus 0, otherwise everything becomes trivial. f extends to a map g, right? f is homotopic to k. Then that's why I proved the lemma before. This lemma, which we said, that k extends to b2, OK? That's the lemma. k extends to a map from b2 to c minus 0. This was x in the lemma. f extends to a map. f is homotopic to k. Then also k extends. This means k star is trivial, OK? This implies another lemma implies k star. So let me write from where to where. Pi 1 of b2, sorry, s1. No, no, no, b2. k goes from s1, no? k goes from s1 to c minus 0. Extends to a map from b2 to c minus 0, OK? So k star goes from pi 1 s1 with some base point 1 to pi 1 of c minus 0 with some base point, which I don't know. No, no, k, now I can take the same. k is just x goes to x to the n. So 1 to the n is 1, OK? The lemma implies k star is trivial. It's trivial. It's a trivial homomorphism, OK? But k is this very simple map here. And now I'll show you immediately that this is not a trivial homomorphism, OK? So that's the last step. So I'm interested in k star. So I take pi 1. I make a diagram. Pi 1 s1, 1, this is k star. So this goes to pi 1 c minus 0. I can take the same base point here because 1 goes to 1. And as I said, I don't know this. This is also z, the fundamental, but I don't do this here. So what I take is I have this retraction on 2s1. So z is r star. What was this map? C minus 0, 2s1. So I go to pi 1 of s1, also 1, OK? So r of x is equal to what? x over the norm of x, again, OK? That's the retraction. So I come back to this. What is this map? Here, I mean this k. k goes x goes to x to the n. This retraction, x to the n, goes to the same x to the n. Right? Doesn't change because it has norm 1 anyway. So this mapping here is induced by the mapping x goes to x to the n. Right? This is this mapping. Continuous, x goes to x to the n. That's k. x goes to x to the n. This doesn't change. Just we go from c minus 0 to s1. But we are anywhere in s1, OK? So this is also induced by the mapping x in s1 goes to x to the n. So this is z. This is z. Now, the generator here, you go one time around, right? Rotation number 1, the generator, no? One time around. So 1, now you go x, go to x to the n. You go n times around. So 1 goes to n. That's a rotation number. So 1 goes to n. 1 here goes to n. So this is multiplication with n. But n is not trivial, no? n is bigger than 0, OK? And bigger than 0, that's not a constant polynomial, OK? So n is bigger than 0 implies that this cannot be. This is not trivial, OK? Then this cannot be trivial. Otherwise, also this would be trivial, OK? Means k star cannot be trivial. k star is not trivial. Homomorphism. If this would be trivial, this is trivial, no? Because it's a composition, right? But this is not trivial. This is 1 goes to n. This is very easy, this k, OK? This is x goes to x to the n, and you see what happens. So it's not trivial, but it has to be trivial because it extended. So there's a contradiction. Where are we? Contradiction. So this is the end of the proof of the first case, OK? And the second case, well, I finished the second case. Two minutes, OK? The second case is not interesting, OK? But here, the idea is nice, OK? If you avoid 0, you extend to be 2. So it's in use map, it's trivial, OK, on fundamental groups. But this is rotation numbers, no? If you take this S1 very large, then it's like x goes to x to the n. If you go one time around, the image goes n time around, OK? And that's not the trivial homomorphism. That's all. So in two minutes, I give you the second case. It's really nothing, the second case. This is a general case, OK? So we have the equation xn plus an minus 1 xn minus 1 plus a0 is equal to 0. We want a solution of this equation, no? For some x. This is, and now what I take, I let, I call x equals c1. x equals cy. So c is bigger than 0, some real constant, big positive real constant, OK? I make a change of indeterminate. So then this becomes equivalent. This equation is equivalent. So I just put c to the y. cny to the n plus an minus 1 cn minus 1 yn minus 1 plus a0 is equal to 0, OK? Now in y. So I divide by c to the n, OK? y to the n plus an minus 1 c to the n. This is not 0. This is bigger than 0, OK? y to the n minus 1 plus a0 divided by c to the n equals 0. If you find a solution here, we have a solution here, we have a solution here. A root. But if you choose c large, clearly we are in the first case, OK? Because the sum of the coefficients c we can see as large as we want, OK? If you choose c large, we are in the first case. The sum of the absolute values of these coefficients becomes arbitrary small with large c, no? Clearly. If you choose c large, we are in the first case. We are in case i, in the first case. Whatever that was, i was in the first case, OK? So the last equation has a solution, has a root. So the last equation has a root but then also the first one. So this implies the last equation, so this one, this one. The last equation has a root, has a solution but then also the first one because it's just giving other names to some variables and so on. So also the first one has a solution. But this is a general case, OK? This is a general case. So this was a good question, no? You asked, why is this condition, no? This condition is, this polynomial, if you take the standard as 1, then the polynomial has to be small, the coefficients, OK? If you take an arbitrary polynomial, then what you could do is take a very large s1, OK? So this is exactly the condition. And if the coefficients are small, then if, or we take a very large one, then the coefficients of xn minus 1, xn minus 1 are not very important. It's completely dominated by x to the n, the nth power. That's important, OK? The rest. This is our homotopy, which makes this appear the rest, OK?  Only you have to be sure that you avoid 0, OK? One minute. I give one exercise, last exercise. You have exercises, right? I give the last exercise now. I have to write it because I didn't find it in the book. Just one exercise and then it's finished. I looked in the book today. I didn't find it. So p from e to b is the covering. p of e0 equal b0. You have to write, OK? p of e0 is equal to b0. And b connected. That's important. B connected. If let f, so we have a continuous map, f from x to b continues with f of x0, so we fix all of our points here, b0, OK? We have a continuous map from x to b. Then a lift, if it exists, then what you have to prove, a lift f tilde from x to e, sorry, to e, OK? Then a lift f tilde from x with, again, the base point that we have to preserve, f tilde of x0 is e0 now, OK? It's unique. The important thing is that b is connected. Then the lifting is unique. Then the lifting, OK? That's all. So we have this diagram, the usual diagram, which is this one. We don't say existence here, OK? It's only uniqueness. If it exists, it's unique, OK? So existence is not true always. So we have this diagram. We have e to b, the projection. We have x, the usual diagram, f, and here is the lifting, OK? And we preserve base points always, OK? So we have f tilde of x0 has to be e0, where e0 goes to b0 and f, OK? So f of x0 is b0, and obviously p of e0 is b0. So this is the situation schematically, OK? If it exists, it's unique, but it doesn't always exist. It exists if x is a path, if x is an interval, then it exists. That's a path lifting lemma. It exists if x is i cross i. That's a homotopy lifting lemma. But in general, it will not exist, OK? There's some condition, which we will not see. Because last week is the last week, OK? Next week is the last week. So this is the last exercise, OK? In the book, I didn't find so many. I mean, I made some exercises here, no? Some kind. Next week, I will do some exercises. But in the book, I don't know. There are many chapters, OK? If you look in the book, we are going diagonal a little bit, and there are not so many useful exercises. So next week, we will have some. OK, sorry. So that's it. Yeah, you can collect. There were three exercises, which I wrote also, right? Yes.


Hi everyone, Joe for Jazmy's casebreaks.com coming at you with 2023 Fanatics Underwraps Series 1 Autograph Baseball jerseys. Three box, random team number three.  Big thanks to this group for making it happen. Let's double you up like Sir Mix a lot. All 30 teams are in. Let's roll it. Let's randomize names and teams two and a six, eight times each. One, two, three, four, five, six, and a two, seven, and eight. Frank down in a cold, two and a six, eight times four of the teams. One, two, three, four, five, six, seven, and eight. Is that Freddie Freeman with the stolen base? Nice. Arizona down to Houston after eight. Now trades are allowed, but trade at your own risk. And there's no checklist for this, so I don't, I mean, who knows what's inside. Franklin, you got the diamondbacks. Michael with the Giants, Brian with the Brewers and Nationals. Frank with the Yankees. Eric with the Guardians. Rex, you got your Cubs. Eugene with the Royals. Brian with the Rockies. Rex with the A's. Brian with the Twins. Eugene with the Angels. Eric with the Cardinals. Nicole with the Blue Jays. Eric with the Padres. Mets. Brent with the Orioles. Eugene with the Braves and Pirates. Nicole with the White Sox and the Marlins. Michael with the Red Sox. Eric with the Rangers. David with the Reds and Tigers. Eugene Phillies and my Dodgers. Eric with the Mariners. Brent with the Rays and Nicole with the Houston Astros. Let's sort alphabetically by team, and I will pause the video just for a little bit. So feel free to negotiate some trades, and when we come back, we'll pop this fresh case open. We'll play a little bit of Who Am I, and we'll see who we pull. So stick around. We'll be right back. Hi, welcome back, everybody. Look at this. We got a deal done. Rex and Eugene were able to negotiate a trade here. Eugene is out of the pirate spot and into the A spot, put T there for trade, and then Rex is out of the Oakland A spot and into the pirate spot. Good luck to the both of you. Let's print and rib. Let's see what happens here. Who knows what's going to happen? I don't know. We'll see. Now, there's six boxes in a case, so we're doing a half case break. Let's run a randomizer here. Let's give it a little randomization here. So the original dice roll was eight times. Just for fun, we'll go whatever side on top is what we'll do. So if it ends up with the top side, I'll open it from the top and pull three boxes from the top, vice versa, if it's on the bottom. Good luck sides, one, two, three, four, five, six, seven, and eight. So we'll go right from the top. And again, big thanks to this group for making it happen. Also, I don't think we have anything else sold out, so we can play some Who Am I? America's favorite game. The clues to the player, and then you try to guess who that player is. So if you're not familiar with this, here's an idea of what we got here. Hopefully, there won't be any randomizing things that we got to do. So each box contains one randomly inserted MLB authenticated, autographs, all those print up there, autographed jersey, collect from the following collection stars, all star, inscribed, Hall of Fame. I guess there is a checklist right there. So there's still in this familiar blue fanatics jersey boxes, little tape on the bottom. I'm actually going to turn off that top camera right there, just so I won't reveal this jersey too early. Oh, okay, all right, pretty cool. Active starting pitcher in Major League Baseball. Pull up some stats here. Just writing down on the page here. He's on the IL at the moment, but he's not like he's a retired player or anything like that. Left-handed pitcher from Santa Monica, California, a California kid. He's got a career 56 and 26, win-loss record, a 305 earn-run average, and 642 strikeouts. He's an all-star, a World Series champion, an all MLB first-teamer, a two-time all MLB second-teamer, a three-time gold-glover, and a silver slugger when pitchers used to hit. You wish Justin Steele had 642 strikeouts and a 56 win record with the Cubs. He might get there someday, but no, not him. Also was drafted originally by the Padres in the first round and chose to, I didn't realize this. I chose to sign with the Padres for a $3 million bonus, that's a huge bonus for a pitcher, despite my commitment to UCLA, and then I made my professional debut for the Arizona League Padres in 2012 and spent the whole season there, blah, blah, blah, blah, played for the Fort Wayne 10 Caps where XU could have seen him in 2013, blah, blah, blah, ranked the 43rd prospect in the MLB pipeline, who are they trading for? Was number two prospect overall by MLB.com in 2014, injured for much of 2014, didn't make a season debut until July, then he went Tommy John and then missed the remainder of the 2014 season. In December of 2014, the Padres traded me, Jace Peterson, Dustin Peterson, and Malik Smith to the Braves for Justin Upton and Aaron Northcraft, and then Freed missed the 2015, wow, that was, I guess looking back, it's probably not a good deal. So Braves lefty with 56 wins and a World Series championship with a 305 earn run average and 642 career strikeouts. Wow, I didn't realize that, he was with the Braves. Now, not Zach Wheeler, not Justin Steele. I have a 94 to 96 mile per hour forest seamer. I got a 74 mile per hour plus curve ball, which he patterned after Sandy Kofax, an 84 mile per hour slider, a 93 mile per hour sinker, and an 84 mile per hour changeup with the curve ball resulting in the lowest batting average in 2021 with a 160 batting average against that curve ball. He was also part of that, there's a high school in Los Angeles called Harvard Westlake in Los Angeles and played with Lucas Geolito and Jack Flaherty. He's the third guy of that trio. Who am I? Come on, this has got to trigger some thoughts here. He wears number 54 with the Braves now. So which is going to go to Eugene, you have the Braves and he guesses Eugene before I show you what you got. And this is cool because I think according to the sticker that I'm looking at here, this is a Braves World Series patch jersey. No clue. Max Fried. Yeah, Nick P. got it, I think, got it in before I said it. All right, so there's his autograph right there. Yeah, he's working his way back from some injury. What's great about this, if you're wondering why are these brakes so pricey? Because these are officially licensed jerseys. You got the MLB logo right here. Should I show you, I think these are so nice, it's probably worth showing you the jersey, right? Got all the hologram stuff right there as well. And here's the World Series patch right here, which is really awesome. Look at that. That's pretty cool. You got the retail tag still on there as well. I mean, the jersey itself is probably like over $100, maybe $200. Eugene, I'm going to refold that nicely after the break. But congrats to you. You have the Braves card back there. Yeah, he is on the IL, unfortunately, but hopefully he'll be back soon. Jammin' up Gilo's fancy baseball team. Let's take a peek in here, see who we have. All right, we got a retired player here. Who's that going to go to? All right, who am I? My first name is Vincent, but that's not what people know me by. I've got a lifetime 250 batting average, 141 home runs and 415 runs batted in. Played for three teams my entire career. I'm retired now. And a one-time All-Star. Who am I? Now, you're thinking 250, 141 home runs and 415 runs batted in. Why would I want this scrub? Trust me, you do. Well, his first name is Vincent Rex, but that's not what people know me by. Mr. Belcher, there you go. Bo Jackson, that's Vincent Edward Bo Jackson. No one was getting out. I was going to read off his football stats as well. That is nice. Bo Jackson and his Royals here. So that's going to go to Eugene with another one. Look at that. Let's take a look at this jersey too. First off, we got his auto. You got all the authentication stickers as well. That's pretty sweet. Nice. This is part of the Cooperstown collection. It's a Mitchell and Ness jersey. You got the old KC right on the front. Gilo's going to be jealous. You've got the number on the side. I like the old trim right there. The old perforations that they had in the jerseys back then. There's the Mitchell and Ness tag and all that stuff. It's a good looking. Oh, you have his Raiders jersey too? Nice. The big KC logo right there, which looks awesome. Gilo's definitely jealous. He's a big Royals fan. All right, I'll fold this one up nicely too, Eugene. After this break. All right, here is the third and final one. The next three boxes in the store. So check it out. Is that right? Rex is saying all three breaks have had one person hit two jerseys. Well, keep joining these breaks, ladies and gentlemen. And Rex, you may be the next person to get two of the three jerseys. This could be three for three for Eugene. Hard to bust open here. Eugene does have the angels.  Let's see what we got here. All right, I am an active shortstop. Am I an active shortstop? Am I an inactive shortstop? Yeah, I'm still active in Major League Baseball. Lefty. I'm one of those bat left but throw right kind of players. A shortstop, lifetime 257 batting average. 999 hits, 134 home runs, 530 runs batted in. Played for four teams my entire career, currently on a minor league deal with the Mariners. But my jersey is the team previous to that. And a Dutch baseball player, that might give it away. He actually had some great moments. Yeah, it's D.D. Gregorius. And Eugene with a sweep. Eugene has the Phillies as well. Clean sweep for Eugene, congrats. Yeah, D.D. Gregorius. I feel like he had some nice years with the Yankees. But just the rest of the career, the rest of the time didn't really materialize. I felt like with the, let's see if we can get his autograph here. There's a short porch at Yankee Stadium. Probably would have helped him out. He should have stayed there. Yeah, not Yankee's edition, Eugene, but you got Phillies at the edition of D.D. Gregorius. There's his auto. There's his, oh, it goes this way, I think. Got the MLB logo. This is a cool Phillies jersey, though. You got the stars above the eyes. This is MLB player's choice, genuine merchandise, standard fit, Nike jersey. You got the Nike retail tag right here as well. The jersey itself, $135. Oh yeah, there's the Nike logo right up on top. So pretty cool stuff. There you go, ladies and gentlemen, hey, Eugene. Great break for you. Eugene sweeps the break. I'll fold this nicely as well. But thanks, everybody. We got another three boxes in the store right now, jaspyscasebrakes.com. Let's see what's in the rest of those boxes. Check it out, jaspyscasebrakes.com. I'm Joe. I'll see you next time for the next one. Bye-bye.


and the democracy community, there is a recognition that political participation is a human right. There's an element of democracy that has intrinsic value that contributes to human development, independent of any economic development, independent of any direct impact on economic development. Some donors such as the Swedish Sida define poverty in a way that's really about citizens and people having expanding choices, whether that be through greater economic access to greater economic resources, having more options to participate in economic life, but equally recognizing that kind of expanding a person's choice includes having greater political voice and greater political participation. So I think there's, as a starting point, I would say that I think there is a strong feeling by many in the donor community that you shouldn't only look at democracy support for its instrumental value in contributing to economic development, that it is in fact part of human development and that when you look at poverty it's really about constraining choice and alleviating poverty is about expanding choice for people in their daily lives and that includes both economic participation and political participation. But even after that, if you look at it from a more instrumental perspective, I think there are a couple of things. Over the long term, democratic institutions provide a stability and resilience that authoritarian systems lack. I think there are, you know, there is the possibility of having stronger short-term economic growth if the leader can unilaterally make economic reforms without any need for consultation and participation. But I think oftentimes the result is kind of an illusory stability that looks extremely stable until the regime collapses. So I think there's a, in looking at kind of the resilience and the sustainability of economic development over the long term, I think democracy contributes to that stability. And so that's an important point. But I think also in terms of having a vibrant marketplace and an engaged citizenry, I mean, you know, it involves engagement by people in the decisions that affect their lives, that affect the country. I think kind of the type of engaged citizenry that thrives under democracy also is much more productive. There is sometimes an argument that, you know, people will refer to the alternative Chinese model or authoritarian capitalism. You know, I think there are lots of examples of very strong sustained economic development in even relatively new democracies. If you look at what Indonesia has accomplished for its citizens since the Sogarto era, in terms of both expanding political participation and economic participation and economic growth, it's amazing. You know, it may be possible to sustain similar or higher levels of growth in non-democracies, but again, I would question the long-term sustainability. There's also a corruption angle, I think, in systems that lack accountability mechanisms, both vertical and horizontal. I think, you know, the corruption, which impacts the sustainability of foreign investment, which is often a much bigger driver of economic development than actual aid. I mean, I think all of those things also contribute to the business climate and kind of the ability to attract international investment in long-term development. So this is a question that's been debated in the academic literature for a long time, but I think there's increasing evidence to suggest that over the long term the contribution of democracy to economic development exists, but even more importantly, I think if you take the broad view of human development and not just look at a single macroeconomic indicator of human development, there's no question that democracy has to be a big part of that. From the practitioner experience that we've had with respect to parliamentary strengthening, I guess in my presentation I tried to make four kind of main points. The first is just thinking through the dynamics of parliamentary reform. It needs to be treated essentially as a political process rather than as a technocratic exercise. I think sometimes within the donor community there's a tendency to, you know, want a nice development, a log frame, particularly for people that have come perhaps from an economic development kind of background. And the reality is that political reform rarely happens according to a five-year log frame for a project. It has to be domestically driven. There's a political cycle in a country, oftentimes there are political scandals that may create opportunities for reform. So just being very aware that kind of parliamentary reform, whether you're talking about that in the context of the developed West or a transitional democracy, often involves kind of many different actors and involves the media, it involves civil society, competing political parties, and it can sometimes be, you know, a very sometimes messy political process rather than something that proceeds according to a nice donor developed log frame. So that's one point that I make. Secondly, I think it's important to kind of look at the question of parliamentary strengthening within a larger political system. I think sometimes there's a tendency by donors to kind of look at an institution, how can they strengthen it with respect to training and the like. Parliaments around the world play different roles, play different roles in the political system, face different challenges internally and externally. So I think it's important that when a parliamentary development program is designed that you do so in the context of a larger political analysis and that, you know, sometimes a parliamentary strengthening program may involve working with the media on covering parliament, may involve working with civil society group on monitoring parliament, may involve work with parliamentary candidates on a debates program prior to the elections. And so kind of thinking outside a very siloed institutional box, I think, is also very important. I think it's also important to look at how we engage with members of parliament, politicians in any country of the world. It's a very specific audience, and you have to engage with parliamentarians in a respectful way that recognizes their electoral mandates that they don't like to be trained particularly by the donor community. So I think it's important that in designing a parliamentary strengthening project it's really a process of accompanying the partners, a process of, you know, sharing comparative experience from around the world from places that are relevant, but doing so in a way that kind of responds to their political needs, their political timetable, I think sometimes programs that have done, you know, standalone trainings on oversight techniques or, you know, constituency relations in the abstract sometimes have limited impact and even more so, for example, parliamentary strengthening programs that will just do the procurement of a voting system and expect that behavioral automatically changes the results. So I think it's a process of engagement back and forth of accompanying partners. And it really takes kind of an on-the-ground presence to build the relationships of trust and confidence with politicians to actually get at some of the core problems rather than just treating symptoms perhaps. Other lesson learned is kind of the importance of focusing not on a specific rule reform or on a specific institutional reform within the legislature for a couple of reasons. One, I think, you know, specific changes in the parliament's rules, the content of particular laws, all that really has to be driven by kind of domestic political actors and has to emerge from the process. I think it's important to kind of focus on kind of the issues of kind of power relationships and kind of what the political culture is. So instead of simply buying a voting system or doing a standalone training and seeing what can be done to, again, accompany partners as they kind of develop a more democratic culture. So that may be working on issues of cross-party dialogue. And sometimes, you know, again, trusted international actors can be helpful in providing a neutral space for parties that are often engaged in conflict to kind of get access to international best practice and sort out solutions. So kind of sometimes it's as simple as helping convene or inform cross-party dialogue. But I think ultimately the things that contribute to stronger parliaments often involves as much kind of the norms of political behavior, the culture, the way in which kind of parties treat other parties as maybe political adversaries but as colleagues also that respect the institution of the legislature. And so kind of focusing on trying to build over a long-term time horizon a more democratic political culture rather than focusing too narrowly on a specific kind of process change that the international community might, in the abstract, like to see. NBI's model has always been to kind of, you know, recognize that no democracy is perfect, that it's, every country in the world can do things to strengthen their own democratic process. And that it isn't, it's never a question of exporting particular models. But that's that I think, you know, people that are in this journey of democratic development can often learn from experiences. And, you know, sometimes that experience is South-South, sometimes it's North-South, sometimes it's South-North. You know, at NBI we did a study mission that brought together some of the political party leaders in Northern Ireland and brought them to South Africa to learn from South Africa's transition experience. And I think a lot of parties to the Good Friday accorded still cite kind of their discussions with South Africans about their transition, their discussions with President Mandela. I think we're very helpful as they kind of look to resolve some of their conflicts. So it's not a question of, you know, assistance from kind of developed democracies to less developed democracies, but I think it's more kind of what we try and support is a conversation among kind of democratic actors that are trying to deal with common problems, whether that's political finance, certainly an issue in the U.S., whether it's improving parliamentary transparency. We just convened a conference of organizations that are working on more open government to expand that conversation to include more open parliaments. And, you know, it was a situation where the Sunlight Foundation, the U.S. NGO, is pushing very hard on issues of parliamentary transparency or U.S. congressional transparency could share their experiences and also learn from the experiences of parliamentary monitoring organizations that are doing similar work around the world. So it's definitely more facilitating exchange and dialogue than providing assistance. I think that there are more flat platforms that engage kind of what people might call more developed democracies or transitional democracies on a level playing field. For example, there's an intergovernmental organization called the Community of Democracies that includes both transitional democracies, longstanding democracies. And, again, it's designed to kind of show democratic solidarity among democratic countries, regardless of the level of development, and to share that with countries in transition or countries that, you know, where there may be democratic activists in very closed system that need support from the international community. And just that moral support, that kind of democratic solidarity is very important. There's also something called the Open Government Partnership that was an initiative of eight different countries. It's currently chaired by Brazil. But, again, that's a situation where, you know, countries like U.K., which is the next chair, or the U.S. or Mexico or Brazil, kind of collectively work together to advance open government share strategies. So, again, it's not a question of, you know, assistance provider and assistance recipient. It's just all governments have to deal with these challenges. And I think there's an increasing number of fora that allow governments and political actors to share that experience kind of without any distinction between provider and recipient. So, under the Community of Democracies Initiative, there are a number of countries that are a part of the Community of Democracies Initiative that are supporting kind of countries that are in transition through the Community of Democracies mechanism. So there is a certain amount of financial support that's being provided to facilitate that sharing and that learning. But, again, it's kind of on a more flat playing field because I think, again, all countries, you know, even the most long-standing democracies can do a lot to strengthen their own internal democracies. And so just making sure that that, you know, you can have a dialogue that recognizes that and you can all learn from each other. Democracies can learn from each other. It's important that that happen kind of in all sorts of directions. South-South, South-North, North-South. I think a lot of Western democracies developed in a very different historical era. And I think some of the newer innovations in parliamentary development are happening in institutions that have less historical baggage and are less tradition bound. So, you know, there are experiments with, for example, the Parliament in Uganda using SMS as a way of kind of engaging citizens and doing constituency relations. So, you know, I think there are lots of innovations out there that are transferable, the Brazilian experience with citizen participation. There's a lot of excellent examples of citizen engagement by the legislature and engaging citizens, allowing kind of citizen input into the process that can be shared. So, I think there is, there's no doubt that there's a lot of information to be transferred and shared. Of course, it always needs to be adapted to the local context. People have to have the freedom to, you know, adopt the models that work for them. But there's no doubt that we all have a lot to learn from each other. On the scalability question, maybe there I would provide a bit of a contrarian response by questioning whether that should be the right question. I think sometimes donors are looking for a silver bullet that they can magically scale up and solve the world's problems. And sometimes, you know, democracy happens at the small scale. It's not about a huge program. It's about lots of different groups of citizens participating in the process. And so, the goals shouldn't necessarily be, you know, finding an idea that can be scaled up for donors to roll out in lots of countries. It shouldn't be kind of, you know, what works best in a particular context, what responds to the needs of local citizens, you know, where is their buy-in for a project. And sometimes the most effective investments a donor can make are really small scale. Sometimes the small scale interventions are things that are sustainable that can be sustained with local resources in a developing economy much more easily. So, I think the question shouldn't necessarily be what can be scaled up. The question should be what works and how do we learn from what works in more places.


The Mark Munson Memorial School of Rock took place this week and ended with a concert at Tornstorm Auditorium in Brainerd Thursday night. Rachel Johnson has more. Rock and Roll was alive this week at the Mark Munson Memorial School of Rock Camp where kids got to learn from airy musicians the ins and outs of what it takes to be a classic rock and roll band. They just learn how to put a band together. So at first they don't even have any idea and then you have to pick what songs we're going to do and then they just try and learn it and put it all together. The camp was founded after Mark Munson, a retired chiropractor in the area and former musician was diagnosed with pancreatic cancer. He hoped to share his love for music by establishing the camp before he died. He recognized that music was something that made him feel good but nothing else did. And one thing led to another and he partnered with the Crossing Arts Alliance which he was a board member at the time and we came up with this idea to leave a legacy called the Mark Munson Memorial School of Rock. The week led up to a final big community concert where the campers now rock band members got to show off their skills. Let's check it out. Cap it off with a real live for keeps performance on a stage with a complete sound system lighting staging the works just like any professional band. It's the real deal. And you've only got a couple days is what it feels like and then when it happens it's just amazing and it's mind-blowing that they can even do it but they do. The camp which is only in its second year has been huge as successful in bringing young kids together in their love for rock music and it hopes to continue well into the future. If you really love music then you should definitely do this because it's all just playing music and just sharing your love. Next year take a crack at it we got your back we'll get you through it and you will have more fun than you have ever had in your life and you know rock and roll forever reporting from Brainerd Rachel Johnson Lakeland news now for more information on the school of rock or to sign up for next year's camp you can visit the story on our website at lptv.org slash news if you enjoyed this segment of Lakeland news please consider making a tax deductible contribution to Lakeland PBS


Hey everybody, this is Brian. I just wanted to drop a quick note. Let me move some of this clutter out of the way real quick here. VoidDromes.com has been successfully funded for yet another year. Yay, thank you very much. This site is 100% funded from your donations. You've noticed I've been doing, most of my tutorials have been on free open source software. So it really, if you've been learning along, it really hasn't cost you a dime. And a lot of people have worked very hard and they've helped and they've donated. Thank you very much. I get a lot of emails saying, hey, I wish I could donate some money, but I'm flat broke. Well that's great, but everybody can donate something. So if you can't donate money, please do. If you can't, well don't.  Instead, donate your time. Teach someone else how to program or make a few videos yourself. I mean, you never know. Some guy out in Michigan that makes videos, you know. Speaking of collaboration and people getting along and donating, if you have no money to donate, you want to donate to a community. In Facebook, I made a group called VoidDromes. You can just find it by searching VoidDromes and go ahead and request membership and I'll add you. Wow, we got 16 new people already. It's growing fast. Anyways, this group, I'm hoping people just start collaborating and there's a lot of that going on where people are saying, hey, I got a question and other people jump in and start answering. Really cool. I love seeing that. Also, if my voice sounds a little different, I did buy a new headset. I should note this was not with donations. This was with my paycheck. This was kind of an early Christmas present to myself because I had two headsets, one that had amazing sound quality and one that had a microphone. And I couldn't find one that had both. And I got sick of the wire. I had this wire that the volume control was on the wire so every time I turned my head, it would turn the sound up or down. It just drove me crazy. So I did a lot of research and I purchased the Logitech G930 wireless gaming headset. 7.1 digital surround sound. As you can see, I'm using it in Linux right now. Highly recommend it. I literally plugged it in, started the computer and it just auto-detected everything. Should note that if you're in-plane with the sound settings and you hit test sound and you go to test it, the sound will stop working. It's not till something else in the system tries to generate a sound that you'll actually get it to work like Pandora. But yeah, I haven't tested it in Windows yet but I was more concerned whether or not I could make videos with it. Really cool headset. It's got this little thing that flips up and when it flips up it automatically mutes and I'm really geek because I haven't had a decent headset in, well, ever. Expect more tutorials. I've been kind of focused on Yi lately. I've been really bad about getting them up here. I've had a lot of family things going on that I've had to tend to so my time has been very limited. Appreciate the feedback. Let me know what you guys think of the sound quality in this video. I'm kind of curious.


Hi, and welcome to our session. Let us discuss the following question. The question says, write the negation of the calling statements. Second statement is, all cats scratch. Before solving this question, we should know that what is meant by a negation of a statement. The denial of a statement is called the negation of a statement. For example, if the statement is, u deli is a city, then its negation will be, u deli is not a city, it is called that u deli is a city, or it is not the case that u deli is a city. So, if we form the negation of a statement by using the phrases, it is not the case, it is called that of just writing not in the statement. So, in this question, we will form the negation of this statement by using these phrases. So, keeping this in mind, let us now begin with the solution. Now, the given statement is, all cats scratch. The negation of the statement, it is false but can be rewritten as not scratch. So, this concludes the session. Bye and take care.


For 25 pvp in SOD, rogues are most likely going to be running 3f3 improved gage, 2f2 improved sinister strike as it also affects ship, 3f3 improved backstab, 4f5 precision, 2f2 endurance, and 2f2 improved sprint.


Hi, this is Osopal Bhartya and welcome to a special edition of TO4 Let's Talk here at KubeCon in Amsterdam in the lion's center. Today we have with us once again Haseeb Boudhani, CEO and co-founder of Rapids Systems. Haseeb is great to have you on the show here in person. Good to see you in person actually for the first time I think it's nice to be here on the show again. Yeah and meeting in person is a totally different you know chemistry totally different experience which also brings to the point of you know you are here at KubeCon. What kind of you know experience you had here what kind of conversations you had here? Over the last I would say six months as a trend I'm seeing in our industry which is we're going from folks kind of doing Kubernetes in a in a sort of a vacuum small teams doing Kubernetes or some dev team doing Kubernetes to the enterprise standardizing on Kubernetes. So the rise of platform engineering teams right this is what we talked about before also offline. It's happening consistently. In fact the reason why I walked in seven minutes late to this conversation is because I was with a customer who is trying to bring the central team concept together in a very very large enterprise and we spent time talking about how they would do it. This is a consistent trend. Platform engineering is a is a is a new concept in our in our industry and it's a good idea for enterprises because they get to take the smartest people put them in a room and they come up with the right ideas and deliver real value to developers so the developers can focus on writing applications. This this was always going to happen. The question was when and the fact that it's happening now speaks to the maturity of Kubernetes right and the adoption of Kubernetes in the enterprise. This is great. And when you do talk about platform engineering how different is it in your opinion from DevOps because you know DevOps is here it will stay there but I think as the organizations evolve over time they do have to embrace platform engineering approach but how different is it from your perspective. So here's what I'm seeing you know what I what I think perhaps is not as important as what I'm seeing and I'll talk about that. So I'm finding that the platform engineering teams seem to be becoming the owners of the infrastructure right so the landing zone in the cloud the Kubernetes namespace that's that's the function that they are adopting so they're building sort of these environments for developers then separately there are DevOps organizations that are focused on the deployment of the applications. So they own the pipelines right the CD platform and potentially the CI platform they own the the backstage right so they enter the IDP for the developers is owned by the DevOps organizations in some cases and this is how lines are being drawn infrastructure versus application delivery and then the developers are then you know sort of kind of customers of both DevOps and platform engineering right and the more I think about it I mean seems like a really good idea but you create kind of this triumvirate of developers and DevOps and platform engineering I think this is a good way to allocate resources for enterprises where they can actually make fastest progress with the fewer number of people because look the biggest problem in the industry is not platform engineering versus DevOps it continues to be a skill set problem right we just don't have enough people so let's specialize making the few people we have do everything that is unfair on our on our team members right let them specialize some people want to focus yes you let them do that some people really enjoy writing automation for for cloud deployments let them do that and then some people write applications that's what actually makes money we are here at KubeCon from once again Rafa's perspective when we look at the emergence of KubeCon like new technology but now when we look at hey it's quite old because in today's world you know technologies immerse so quickly what kind of adoption you are seeing of you know platform engineering approach cloud native approach Kubernetes in the enterprise space of course you know when we walk around it's look like you know all the big vendors small players they're all using Kubernetes all these technologies in production so that is not even the question now but what kind of adoption you're seeing and when this adoption grows you know companies start running into problems you know cost is becoming a big challenge you know cloud cost because of potential recession layoffs teams are getting a smaller so let's talk about the adoption in the enterprise space and what kind of challenges enterprise are seeing when they're adopting these technologies we've been coming to KubeCon for a long time a really long time at this point and because we are in it we feel like it's been around forever I would say that true enterprise adoption of Kubernetes happened at some point in the last 18 months or so for the majority of the world yes there's some shops who have been doing this for five six seven years yes of course they are but for the majority of the world it's new it's still very new for the developers and the same companies it's actually not new they've been working with Kubernetes a long time but for you know generally speaking the IT function right platform engineering is a subset of IT this is a new concept this is maybe a year to a year and a half or so I think they're also learning what to do what not to do right so so that's to me is a adoption yes it seems it's been here forever it's new we are at best in our second innings of Kubernetes but then in terms of challenges the challenges are arising because we think it's mature but actually it's new right so we all think that well we've been doing this long enough so all the things that should have been learned all the lessons that should have been learned had been learned already no we're learning them now because now people are taking real applications to production and then they find all kinds of challenges and well then they are learning on the job right and that actually is the biggest challenge I see in in enterprise they don't know what they don't know till they hit these crazy problems and then the you know leadership is saying hey maybe we should look at options that are off the shelves for some of the problems that we're trying to solve and that is by the way very very good right that again speaks to maturity when IT functions start buying technologies versus you know letting people thinker this is how IT works right that's when you know IT feels this is really mission critical I need to put some governance around this I need God wills around this that it feels like it's constricting actually it's great it means that this is serious right enterprises if it's a serious thing they're not gonna just let you do whatever you want right they're gonna want to have some control over it and control doesn't mean a bad thing right and the I find that the the real beauty of our industry is that this these people learn so fast right so now enterprises have one control so they said no problem we'll give you control but then we'll be the best possible developer interface backstage behind backstage is all this control that is possible in our world so so this this is how in my opinion the market is moving now when we go back to reference number you folks what role are you folks playing in this space mm-hmm so our job is actually quite simple you adopt Kubernetes and as an enterprise you have an expectation of management governance controls card rails ready these kinds of things that's what we do for our customers the analogy that I use with my customers is look back in the day when you're using the m's right so from BMR you would buy a stack which would have ESXi which is the you know sort of the hypervisor which is arguably like the orchestrator right in thematically and then you'd have vSphere vCenter to manage so Kubernetes is thematically not technically but thematically like you know ESXi right because it's on which intention so the question is what is your vSphere strategy what's your vCenter strategy for Kubernetes well today the strategy is every enterprise is building one right so what is it that they're building they're building they're solving for multi-tenancy access management audits provisioning of clusters add-on management policies for Kubernetes network policies that is a different problem how do I have visibility everywhere to have the right dashboards for the right people oh by the way I need to do this across multiple clouds so I want to use EKS and GKE so that means I may have to build three different sets of tools to build these things and I have to upgrade them and this is a lot of work right this is on top of Kubernetes Rafa is not a Kubernetes company Rafa is a Kubernetes management company that's the problem we solve for customers and the interesting part is something that if Rafa doesn't exist and this is a conversation we have a lot right whether you buy from me or not you're going to solve this problem anyway that is not up for discussion that you have to do these things you're an enterprise you have to do these things the question is how will you get there there are two paths in front of you you can go down this journey and it's a it's a panelist journey it takes a lot of time it takes a lot of effort it takes a lot of people or you can buy the best practices framework that we sell as a product and you can be up and running with the right governance and the right control and the right automation most importantly now and we have a lot of very heavy customers and who are these customers who do you like know what is the ecosystem that you are serving yeah so we have a very large number of financial services customers we have a number of health care sort of you know former customers and then we have a bunch of high-tech customers on the platform what's really interesting about these customers is there's one thing that is common across all of them which is the customer for us is consistently a platform engineering team and then they have internal customers my customer has multiple customers my customers not the developer my customer's customer is the developer so we understand what problem we're solving for and for whom my customers platform engineering they have to solve for multi-tenancy they have to solve for repeat repetition and in in configuration management they need for the standardization some of their customers are in AWS some of their customers are in Azure they can't say no you have to solve for all this when you have clarity on your your ICP your your your buyer a lot of things become very obvious to you what do you have to work our roadmap is easy these people tell us what they want and we just built for them it's as if they they have a new person on their team I said this a lot it's like you hired a guy named Rafi you're hiring people you hired a guy named Rafi but this guy named Rafi comes with 200 people right and we're gonna bring all this automation governance that you would have solved for anyway and when you're talking about these industries some of these industries are very well regulated they have to so does you know Rafi play any role in there as well or that is also the scope I mean we're not a compliance company per se but in terms of configuration management and standardization that these things help right so one of the things that enterprises do is you know they have a center of excellence organization right so you know it's got people from DevOps and development and security and platform and these folks come together and come up with a blueprint of what things must run on a cluster who has access to what I want to make short people can see this versus that so the the implementation of those things is done with Rafi's platform so so we're just the doer right I mean in the grand scheme of things right they create the policies because we don't create policies our enterprises customers know they know exactly what they need but then the manifestation of their of their asks right Rafi is the tooling that they use to apply those in the Kubernetes context and that's our job right we understand what a job is we're the doer right and we do it really really really well and we do it as a SaaS product if you look at Rafi systems if you see you know that as you said the adoption is in the phase two what does it mean for the company what are the things in your pipeline what are the things when you come to these events when you as you said these are not your direct customers but you know they give you the pulse you know where the industry is heading so what does it mean for Rafi what kind of things we should be looking at then look at your company yeah so there's two demos we're doing this week there are new ideas for us and we you know we're doing them because we want feedback from the from the from the developer customer right one is we built a number of backstage plugins and we're showing them because yeah we want to make sure because end of the day they'll be the one using it they will not be the ones buying from us but their platform engineering will be buying from us but they'll be the ones using it are we doing all the right things I hope you are and if you're not no problem we're going to go make it better right the second thing we're showing is what we call an environment manager so the idea is that okay you built a Kubernetes cluster but your application is not just in Kubernetes you have RDS you have kinesis you have an advocate where you have to set up all these things that live around Kubernetes to make your application maybe use lambda I don't know right so how do we help a platform team create a templatizable pipeline to deliver infrastructure like a full landing zone in one how do you do that right Kubernetes is a part of that it's an important part but it's it's not the whole thing it's one thing so we want to start kind of thinking about a broader sort of cloud landing zone concept that we can help our customer solve this platform is designed to plug in nicely with Terraform Cloud or MZ or any of these sort of Terraform framework companies which are there are multiple of them and they're all doing very well this should plug into that and help my customer the platform team deliver an end to an environment based on the identity of the user if you're a springboard developer your environment is different if you're a data scientist your environment is different but you should be able to go to a backstage page and say I need an environment and the platform will know exactly what that means and just make it happen for you so we're we're we're doing a preview right now usually we write new code we keep it in preview for two three months and then we make a GA so we can get feedback and we're eager to get feedback right so folks watching this if you if you won't mind please watch our demos of our environment manager platform if you want to see a demo in a call happy to do it because I'm convinced that this is a problem I see every enterprise struggling with us maybe we can help Haseeb thank you so much for taking time out there and talk about the company the whole you know evolution of the you know platform engineering and where that option is going and what RAP is planning so thanks for sharing all those insights and as usual I would love to have you back on the show thank you I look forward to doing this again thank you for having me today