Well, hello, everybody. I'm going to give what I hope is a very elementary talk. So you'll probably be very bored. But for those of you who don't know this or have forgotten a little bit, I'm just going to go over the problem of regularization and explain the traditional methods for dealing with it a little bit. So the question is this. We have a symplectic manifold. And it's got an almost complex structure, j, which is omega-tame. And we're looking at j-halomorphic curves. Now, to be simple, I'm just going to think of the domain being a sphere. So I'm going to write, say, m tilde top 0k of, well, decorated by all these things, j, a, say, which is going to be the set of maps. And now I'm just going to take the sphere as my domain into m, which are j-halomorphic, which represent the class a. So that's a class in the second homology of m. And oh, yeah, I'm going to have some mark point, z1 through zk, which is just underline z. And these points are different. And that's just a space of j-halomorphic maps. So these elements are maps. And the tilde means that the elements are maps. Now, does everybody know what all these words mean? Because I'm just assuming you do. And if so, if you don't, please ask. OK, it's up to you. And then we're interested in this space, but also we're interested in if you've got two of these elements, u, z, we can say one element is equivalent to another element, u prime, z prime, if there's a diagram of this kind. So we've got phi from S2 to S2, which is phi is a Namibius group. So that's a bi-halomorphic map. We've got u going there to m. We've got u prime going there. And the diagram's meant to commute. And say we've got z prime there, and that goes to phi z. So this would be something of the form. u prime would be phi composed with u, and this would be phi inverse of z. So we have this equivalence relation. And we're not interested really in the space of actual maps. We're more interested in the space of equivalence classes of maps. So that's going to be written as m top, 0k. So that's going to be m tilde top, 0k, divided out by this equivalence relation. So these are equivalence classes. So we're not really interested so much in the way they're parameterized, just that they have a parameterization. We're interested in the images, the geometric images in n. And I put top here because this is the top strata of some completed space. Now, so for example, well, so in the best case, this m tilde top, 0k of mja, is a smooth manifold, transversely cut out. Oh, I forgot. I said it was j holomorphic. Well, I forgot to say what that means is that the Cauchy-Riemann operator is 0. So del bar j of u is a half of du composed with plus j, du composed with little j. So that's the Cauchy-Riemann operator. And so this is 0, so an operator. And if you're saying it's transversely cut out, you mean that the derivative of the operator is subjective. I'm going to say more about that later on. But anyway, you want it to be transversely cut out. And the dimension of this thing then would be equal to the index of the operator is 2n plus 2c1 of a plus 2k, which is a number of marked points because they're varying in a two dimensional space. So that's a dimension. And why you want it to be transversely cut out is that there's a Fredholm theory. And that means that if you vary j, the space is not going to change very much. So that's a sort of basic space of j homomorphic maps. And so in the best case, the first situation is that this is a nice manifold. And then, as I say, what we're really interested in is the quotient space. So this is m top. That this is almost never compact, but it's compact enough. Why this space is almost never compact. You can see that it's almost never compact. Because one thing I'm insisting, I've got these points z1 through zk. And they're meant to be all different from each other. So two of them could come together. And of course, that would not be in this space. And then another problem about the compactness is that this group of equivalences, this group here, psl2c, is not a compact group. So that means you're quotienting out by something which is not compact, and that can cause problems. But what do I mean by compact enough? Well, sort of for the purposes, it's compact. So what are the purposes? Well, we're interested in the evaluation map from m tilde, well, really from m top now. And let me, well, ma, so just put an a in there. Mapping to m to the k. So we've got a pair uz, which is an equivalence class, so I can put brackets like that going to u evaluated at z1 up to u evaluated at zk. And that, of course, is defined on the space of maps, but it maps through the quotient. So there's a well-defined map here. And what you want for this is you want the image so that it represent a homology class. And for that, you see, I mean, if this were a compact manifold, then, of course, it would have a fundamental group, a fundamental class. So you could just look at the push forward of that fundamental class. But in fact, for it to represent a homology class, it doesn't have to have a compact image. What it needs is that, i.e., if you look at the closure of the image, so this is the image of the evaluation map minus the image itself, so this is the boundary of the image. This should have co-dimension at least two. So that means what we say is it represents a pseudo cycle. And to give you an example, well, let's take the very simplest example. I think it's a really simple example. Take the projective plane with a standard J and the class to be the class of a line. And then C1, the first Chang class of the line, is 3. And so the dimension of M, if we take two mark points, the dimension here is going to be 2n plus 2C1 of A plus 2K, which is 4, and then minus 6, which is the dimension of the reprametrization group that I've quotiented out by. So this is the dimension of P SL2C, the Mobius group. And this is 6. And this is 4. And so we get 8. And that, of course, is the same as the dimension of, if we take Cp2 times Cp2. So if we look at the evaluation map from M top 0, 2 in the class of a line, going to Cp2 times Cp2, we have UZ goes to UZ1 using 2. And what this is, remember, these things represent a line. And so the elements here represent, you've got an S2, you've got two points on it, which, in fact, we can normalize. We can take the first one to be at 0, and the second one to be at infinity, say, because the group is triply transitive. So we can fix these two points. And then we've still got an action of C star acting on these things. So this is M tilde top with a fixed point, set to 0 in infinity, and quotiented out by C star. So it's still something, a space of maps quotiented out by a non-compact group. And anyway, you've got this space of maps. You're looking at where they go. So you've got two points in Cp2. And you sort of take the lines through them. Well, those are unique lines through any two points. Therefore, this map is injective. If we take the standard complex structure, we know that that's injective. But of course, it's not surjective. And you can see it's not surjective because the two points it's going through are different. They have to be different because I'm assumed that these two points are different and allowed to come together. So the image is equal to Cp2 times Cp2 minus the diagonal. Very clearly. So it's not a compact image. You can see the space is not compact. On the other hand, the set of maps is somehow compact. That's by Gromov's theorem. And you can compactify this. I mean, this diagonal has co-dimension two. Real co-dimension four, I suppose, because it's where these two points are the diagonal here. And so it does represent a pseudo-cycle. It does represent a fundamental class because the image of the boundary is so small. I just understand that the words co-dimension of these two means inside the closure of the image or inside the ambient? Inside the nm to the k. So we're not talking about the structure of the domain here. I mean, in good cases, you can assume it's a manifold, but it's not going to be compact, you see. But you're saying, well, it's a non-compact manifold. But if you look at its image in here, it has a fundamental class because the boundary is so small. There really is a key co-dimension in addition to what? I'm going to give them a case of some huge thing. Co-dimension in the closure, right? Well, it's a co-dimension. It's a real co-dimension of at least two in n to the k. So whatever dimension that has. So in this case, I mean, this is trying to, in this case, what's wrong with it? It makes sense. What if n to the k was an enormous thing? And this was like one curve. We really care about the closure. Oh, I see. I see what you're saying. Yeah, OK. So what do I mean by co- yeah. So I'm saying this incorrectly. What I mean is that the boundary of the image minus the image should be a union of maps from some manifolds v. So this is some finite union, say. And the dimension of v should be less than or equal to the dimension of our original space, m top minus 2. That's what I mean. I'm sorry. You're right. Right. That makes sense. So it's co-dimension. It means it should have at least two dimensions less than the dimension of the image. And this is assuming m top has the expected dimension. That's assuming that m top has the expected dimension, which it's going to have if it's a, that was my first assumption that it's a transversely cut out manifold of the right dimension. So it means that it has a correct dimension. And then, right, that you're assuming that this thing is going to represent, then this is meant to represent a homology class of the right dimension, the dimension of the index. That's OK now. OK. So now, let's say, so there are several things here. I mean, the regularity problem is a problem of arranging the first thing that it's transversely cut out to find if you could find a j to make it a nice manifold. That's a regularity problem. But then there's also the compactness problem that you want to make sure that this image is compact enough, or how can you make sure about that? Now, the compactness is actually much easier to deal with than the regularity. So let me just say a few words about compactness. So there's Grumhoff's compactness theorem, which tells you that if you have a, you know, you look at these are sequences of spheres, right? And you look at these are spheres in M. Perhaps I should go to a different book. So we have these. So we've got some curves UK, say, from, say, S2 into M. We've got a sequence of curves. And they're in a fixed homology class. And so that means that the pullback of a mega on S2 is fixed. It's a mega of A, the integral. And this is actually, analytically, it's the energy. It's equal to, I'll probably get the formula wrong, pullback of du squared over S2. That's the energy of u. So it's basically the L2 norm of u, measured with respect to the correct. I mean, you have to choose a metric on there and stuff. Measure it with the correct things, and you get this is the energy. So this is a sequence of curves. So these are C infinity maps, the holomorphic curves. And they have bounded L2 norm. Well, they're actually W12 norm. This is the L2 norm of the first derivative. So this is what's called the, it's a sort of boundary case. If we had, because these are the holomorphic maps on a two dimensional space, if it were a bounded W1 P norm for P bigger than two, then there's standard sort of compactness arguments that tell you there would be a compact subsequence, or a convergent subsequence. That's just standard elliptic theory.  So that means it's an equicontinuous family and the C0 norm. So you use sort of standard results about equicontinuity to extract a convergent subsequence. But what we have is this borderline case where it's a two, not bigger than two. So in fact, it's possible for this thing not to have a convergent subsequence, but if it doesn't have a convergent subsequence, what's happening is that the energy is somehow concentrating near a point and you can rescale it and get a convergent subsequence. So, and that comes down to the fact that, for example, in this case here, we're quotienting out even here by a non-compact group. That's a non-compact group there. And so that sort of affects the convergence. But anyway, this is really well understood and what Gromov proved was that there exists a subsequence you also call UK say that converges to what Gromov called a cusp curve or what we now call stable map. And what a stable map is, you see bubbles can develop. You, I'll explain a little bit in this case later on, but what it's going to be is a domain now, sigma is a nodal sphere. So it's going to be a union of spheres joined at points. So it could look like this. It could be a bubbles, a set of spheres like that. That would be the domain. There'd be some mark points on here where, because your mark points would converge somewhere, but we've got three marked points, they converge somewhere like that. And the map, you would then map each component into n. So you'd have a sequence of, you'd have a set of maps and that's what, so for example, just suppose we were looking at a map from S2 into CP2, which had degree two. So that would be a quadric. But a quadric can degenerate into two lines. So that's in class two up. That's one degeneration that could happen. You can have, so that's a geometric degeneration where you have a sequence of things representing a quadric which degenerates to two lines. But you could also have a degeneration that happens because for example, suppose you're in this case and the mark points come together. Well, the mark points are not allowed to come together here. If they do come together, then you could have a situation like this where you actually have a bubble. And that bubble contains both marked points. And here, this is what's called a ghost component in the sense that the map you restricted to this component is constant. So what you have here is a situation where you have this goes, this is a line. So you've got a map of a line and your two mark points are sitting on the domain in this sphere which is called a ghost component. So I should give you a proper definition of a stable map so you can see you can put these all together into the space of stable maps. Now, is there any way of retrieving that board? Oh, wait a minute. What about this? No. It's got a very tall person. There's a rod. This thing. Doesn't have a, what's, oh, this is it, this is it. This one's got a hook. Okay, that's good enough. So look, it's going up by itself. Ah. Very obedient. Okay. When the most points you integrate, like the thing you integrate with the energy, did you develop poles there or is it just the same? Well, no, the energy just comes from the map. So it doesn't see where the mark points are. But you, so I wanted to give you a definition of a stable map. And of course, these can occur in any genus. They don't have to be genus zero. I was just restricting to genus zero for simplicity. But, so let's just do genus zero because the point about genus zero is that then you can think of your domain as a union of these two spheres and each two sphere you can parameterize in a nice way and give it a standard complex structure. So you don't have to worry about variations of the complex structure on the domain. Typically, you'd have a nodal surface which you'd have to allow the complex structure on the surface to vary integral space. You can put that in if you want, but that's just make it simple. So it's a, it's a, it's a nodal, nodal Riemann surface. So you have something and it's all, it's meant to, you know, you can have something like that, but it's meant to be connected in a tree diagram. So you're not allowed to have something like this. That would have genus. That would be a genus, genus one thing. So it's a tree, it's a tree here, a label tree. And you have a certain set of mark points. You sprinkle your points around. First we've got five mark points. You could sprinkle them around like that. The mark points are all distinct and so they're on particular spheres. So you have to say which sphere they're on. Then you have some maps, U, which is a one map for each component. And so we have, so that's, so we're looking at two poles, which is a form of sigma, complex structures understood here, Z, U. And then you put an equivalence relation. Well, the obvious equivalence relation that just like I had before, you just have some holomorphic, this is a holomorphic map here. So you're interested in these modulo reparameterization. And then there's an all important stability condition. Now you don't need to, when you're talking about a stable map, you doesn't have to be holomorphic. I mean, eventually we're interested in the holomorphic ones, but in general you can just look at arbitrary maps with nodal domains like this. But it's very important to have a stability condition and what that is telling you, this is equivalent to saying that for each element, if you have sigma, J, U, and you look at the group of automorphisms, so maps of that to itself, this is finite. So you never, you don't allow yourself to have an infinite number of automorphisms. And what that means is, now if say on a component like this component, the map is non-constant, then the map is either injective or it's a multiple covering and there could only be a finite number of automorphisms of that component. So if you have a component where the map is non-constant, there's no condition. But if you have a component, if the map U are constant here, then you see what would an automorphism as this would be? Well, it would have to fix Z0, Z1, it would have to fix this nodal point. But you see, if we've only got two, what are called special points, we've only got two special points on this component. There's the mark point and there's the nodal point. And so they're in principle as a one complex parameter family of automorphisms of this thing. So that would not be allowed. So the claim is that if U alpha restricted to the alpha's component is constant, S alpha to alpha has at least three special points. And special points are nodal points or mark points. That's the important condition. So that means for example, in this particular case, I could allow that to be, that could be a ghost component. So that one could be a ghost component. This one could be a ghost component. This one's not allowed to because there's only got two mark points. That one couldn't be. So this would be a possible degeneration. If you were looking at maps say from U from S2 to CP2 of degree two, so that it's in the class twice a line, then this one could represent a line, that one could represent a line and we've got five mark points. That would be a perfectly good stable map of this kind. And the claim is that if you look, so we define M bar, zero K, M, J, A to be the equivalence classes of stable maps in genus zero. So this is a genus thing and K mark points and representing the class A. And if you've got a J there, they're J holomorphic. Okay, so that's the space of J holomorphic stable maps and it could be, it's a stratified space. I mean, why I wrote top there is that top was its top component, if you like, where its domain just had a smooth sphere. But then it has all these other components where the domains are nodal curves. And some of them, if you like, is sort of artificial. Like this is an artificial one because the two mark points have come together. And so, nearby things to this would be something where you had a, and then you had this would still be constant say, this one could have a Z2 on it, but now we'd have three mark points on that component. So, and then when these two mark points come together, we'd have this bubble sort of artificially created in the domain. This is, this was Concevich's idea of the way of describing what's happening, way of describing a compactification. So the claim is that we have this space here, this is compact. It's a compact space. And if everything is regular, so in other words, if all the underlying maps, you see a regular thing would be for this, would be that this map was a J-holomorphic map transversely cut out, which I'm going to say more about later. So is that. And then all these intersections here, you've got a space of spheres here, a space of spheres here, these is kind of a constant map. All these intersections are meant to be transverse so that it all sort of comes out and it's all in the right dimension. And the strata, the formal dimension, if you like, of the strata is equal to the dimension of the original, of the top space minus twice the number of nodes. So each node gives you, contributes sort of a co-dimension two to the thing. And that's a formal dimension, that's the index. So that means that if you can arrange everything to be trans, all the things are transverse, that will be the dimension that it has. But of course, it doesn't have to be like that, but if it were, then you see if you're in a situation like this where you can arrange that all the lower strata have basically have the right dimension, then the top thing of this will represent a pseudo-cycle, because this is compact. So you can compact by the image of the map by just looking at the image of these things and then the boundary of it is going to be covered by the images of the higher dimension, the strata, the higher co-dimension strata here. And they're all going to have at least co-dimension two. So that's a sort of standard regularization picture that you compactify the image by putting in the images of the evaluation maps from the strata in this space of stable maps. If you can arrange, if you can control our dimensions reasonably, then they all have co-dimension at least two and then you get a pseudo-cycle. Now, of course, the kicker is that you can't always control a dimension. That's the whole problem. But if you could, then you'd solve this problem. You'd have got, because our aim, of course, is to try and use these spaces of maps to get some invariance. We want to get these homology classes out that we can then intersect and get to grumble for an invariance or something. And this is just for closed curves. But if you're doing open curves, you have boundaries and you put boundary conditions on. The formal structure is exactly the same. The equation looks a little bit more complicated because you have to deal with what happens near the non-compact parts of the domain. But formally, it's exactly this idea that you can always compactify the space you're looking at by making some space of stable maps or stable buildings or something. And then you just need to control the dimensions. In principle, all the boundary components you've added should have co-dimension at least two. You need to control that and make sure it does. So that's telling you all I was going to say about compactness. But you see, I must have erased. Here, I erased the first condition, which was that everything was transversely cut out. That's what I need to talk about because that's the hard thing. The compactness is sort of understood. Yeah? Why don't you have to worry about the unstable maps? Like, why aren't they somewhere in the computer? Well, the unstable maps would be, well, the thing is that they don't appear in the compactification because if you have an unstable component, suppose that were an unstable component. So that map was constant. Then you could just forget about that bubble and put that point there. You see, the unstable components are things which don't actually contribute to the image because the map on them is zero. It's just a constant map. And so you can just collapse any unstable components. And the thing is that it's, I mean, that's a good question. I do not understand. Can you say again? Yeah, I'm saying that, suppose we had this situation, but this sphere, it was a constant map there. Then that would be an unstable component. But the thing is that you, so I'm saying that that's not going to be in the compactification because we don't allow that in the compactification. What we do is, in that particular case, we just get rid of that completely and just put the mark point there. And there wouldn't be a bubble there at all. Because the image would be the same. If you have a sequence, I mean, in your top stratum, you can just define it so that it converges to a stable map with the stability condition. When you look at the compactness argument, what's happening when a bubble forms? Typically, in this particular case, if this is what we had, all that would have happened here is that some of the mark points would have come together. And the underlying map here, you just have a sequence of maps which was basically converging to a map, and all that would have happened to create this set of bubbles is that the mark points would have come together in a particular way, sort of these two points coming together first, and then these two points coming together second, and then they would form this bubble. And you can actually see that completely clearly, just by sort of rescaling and blowing up and looking at nearby this. You'd have, here would be the point where you'd be rescaling. This point is going to go into these bubbles. You'd have a Z4 and Z5 very, very close, and then sort of further away, you'd have a Z2 and Z3. So there'd be two scales of bubbling here. This would be the, these ones would be the furthest ones, and these ones would be the later ones. But the map itself would be constant. So the map to M would be constant, but you're talking about how the mark points are coming together.  If you just look at the way you prove the compactness, you can get, I mean, you could describe the, I mean, of course, you could put in something like that and allow that as a limit. But then you wouldn't have a sort of unique limit. And it doesn't give you a good structure. I mean, this was Komsovic's realization that the good structure is, because there would be an infinite automorphism group of this, which you wouldn't want. You wouldn't want. You'd have to divide out, you see, by a C star here. You want the set of automorphisms of all these things to be finite. That's very important. We are already doing that now. The other component that still has a C star automorphism. No, because you see, here the map is non-constant. So you've got some map U, and you're saying U is equal to U composed with phi. If you had an automorphism. So phi would have to be something that fixed these two points. And so you'd have to have phi, Z1 is Z1. Phi of the nodal point, let's say infinity is infinity. And it would have to satisfy this.  I mean, it could happen. Well, you would only have a finite number. You see, what you could have, you could be a map, so you could have U from, you could think of C union infinity to M. And it takes Z to say Z cubed. And then you could have an automorphism which took Z to E to the two pi I over three Z. That would be an order of almost seven, so that U of phi would equal phi. But it would be a finite group. You're not saying there aren't any automorphisms, you're saying it's a finite group of automorphisms. So if it was a multiply covered curve, it could have an automorphism. But if it's an injective curve, somewhere injective curve. So that's another thing, you have this idea of a somewhere injective curve, U from S2 to M. What that means is that there is some point Z naught such that if you look at its image and then its pre-image, that's just Z naught. So that it's a map, here is the domain as two, here's your Z naught on this little neighborhood, it's injective.  And that doesn't happen if you have a Mobius transformation that fixes three points, it's the identity. So really the stability condition, the sort of geometric way of understanding it is telling you that the group of automorphisms of this stable map is finite. That's what the stability condition is saying. And that's a very important condition that allows you to, I mean it's a fact that you could have automorphisms, means that you're actually, you know, the space M, the space here, the strata are typically overfalls rather than manifolds because you could have automorphisms. So you're modeled on an overfold, but it's very important that the finite does, you know, it's overfold with finite isotropic groups, yeah. It just occurred to me that, I think you didn't say that stable maps have to have finite energy which is what guarantees that the stability condition actually implies finite automorphism group. Okay, well I was saying that we're in some class A, right? So then omega of A is a number, right? So they do have finite energy, but you're right. I mean, if you're doing more general things, the finiteness of the energy is actually what makes this whole compactness serum work. You've got to have a bound on the energy, bound on the W12 norm to get started with the analysis. Excuse me. Yeah. How do you put the topology on M bar, is it okay? Oh. Something called a, how awkward of you. Something called the Gromov topology, which I don't particularly want to go into. I mean, it's each stratum, it's clear enough how you to apologize it. And then nearby, well, there's a structure you see when you have a stable map with, I mean, I have to mention this at some point if I'm doing sort of the elementary beginning anyway. If you want to know what's a neighborhood of something like this, you have to do gluing. So you have to say gluing is a way of if you've got a nodal domain, then sort of nearby there are things which nearly have the bubbles, nearly have the nodes, but don't quite. So nearby, so here we've got two nodes. So we've got one gluing parameter here and another gluing parameter there.  So suppose we've got a single node. So here's a, we've got some map. So this, we've got, let's have no mark points just for simplicity. This is going to some line and that's going to some line, so the non-trivial map. So this is going to a pair of lines. So the image would be a pair of lines. Well, nearby this pair of lines in CP2, you actually have, of course, a quadric. Impossible to draw. But how analytically do you see that? Well, you have this sphere, so you take, you take this sphere and you take with minus a disc, you take that sphere minus a disc. So you've got S2 minus a disc of size epsilon. You've got another one, S2 minus a disc of size epsilon. And then you sort of glue the boundaries together. You identify the boundaries so that you have something like this, right? I mean, that's what we'd smooth it out a little bit. So it's actually a sphere. Now, when you glue the boundaries, you have a twist here, you have a theta twist. So the parameters you have here, you have an epsilon which measures how big the disc is here and you have a theta. So that you can think of as A S epsilon times E to the two pi I theta. That is in C. That's a little complex gluing parameter. So when epsilon is zero is when you have the bubble and when epsilon's a little bit bigger than zero, you've joined the domains together. Now, of course, in practice, what you do is you use cylindrical coordinates here. So it sort of looks like an infinite end. So you think of something looking like this and something looking like this.  And then because a map, you see, how do you get a holomorphic map here? Well, the map you've got, U1 here and U2 there, say, well, it's near this point, it's essentially constant. So you can just on this glued together thing here, you just make your new map, you've got U tilde, say, which is U1, U2 glued together with some gluing A. It's basically equal to U1 here, it's equal to U2 here and then here you just patch together with some kind of bump function. You just interpolate. So this is almost holomorphic, not quite holomorphic. And then you use a Newton process to make it holomorphic. So that's an analytic thing called gluing. And again, Dietmar and I do a very, very simplest case of that in our J-holmophic curvebook. Now gluing, you can, the analysis of this is pretty well understood. It's horrible. It's real analysis, right?  But it does exist, it is well understood. But it can be done in different contexts. And of course, in the Fredholm, in the polyfoil context, they have a very beautiful way of doing the gluing.  I mean, there is something called the Gromov topology that you can measure, which I really don't want to go into. But basically, you can take, if you have a map that's sort of, the domain sort of looks like that, you can, you're basically, well, I'm just not, you have to rescale, you have to restrict the bits of the domain and rescale it and sort of measure it. I don't particularly want to go into that. All I want to say is it is a topology. Which, so you have these strata, and for each node, so if you have a strata with one node, like consisting of elements like that, then there's a, over that there's a line bundle consisting of the gluing parameters, and then there's a map which takes you that line bundle into sort of resolves it. So you can get a topology. Of course, Gromov, I mean, the Gromov topology, you don't need the gluing. You can describe it without the gluing. Anyway, I did want to say a little bit about the regularity question. What is the dimension of the strata? Right. So you have this two times the number of nodes. So this two is just before each node, you have this complete parameter. Exactly, yeah. So that's exactly right where those parameters come from. But I think I should say something about regularity to justify the title of my talk. Okay, so I had the notion of a some more injective curve. It's actually right up here. Now, what does transversely cut out mean? Well, so we have the Cauchy-Riemann operator, del bar j of u is a half of du plus j composed with du, composed with little j. And this thing, well, du is a differential. So that we're thinking of as a one form. This is a one form on the domain with coefficients of the pullback of the tangent model. So it's an anti-holomorphic one form. And du itself is a one form. The fact that we add that to make it, it makes it anti-j-holomorphic. So it anti-commutes with j. And so we can think in general, we have x is some space of maps. Perhaps I should write w1p maps from s2 into m. We've got j as a space of tame almost complex structures. And we have a pair of elements and we have uj in there. So we have, and u doesn't, so that this is just a map and that's a j. And then we have this operator. Now, this operator lives in a one form that you see its image depends on u. So that means that we really have a bundle over here, e. And we're thinking of del bar j as a section of this bundle here. And the fiber of this bundle, e uj, so for each pair uj here, we get a map in there. That's precisely the space of, and actually if I'm completing this to w1p maps, I mean you can do this in the c infinity case, but if you want to do Fredholm theory or something you better put some balance spaces here. So then this had better be the, instead of the, these are c infinity things, let's have the lp, lp sections, because we've differentiated, so you get lp one form. So we have this bundle here. You can prove this is a nice bundle over this space. It's a balance bundle. You've got this operator. What we'd like to say transversely cut out is we'd like that we're interested in the zero set. And what we want is that for this, we want del bar j to be transverse to the zero section. So we want that the, at the solutions. So to say transverse, that means we have this operator. We can look at the linearization of uj, which is the linearization of this operator. I mean this is a map from a Banff manifold to a Banff manifold, so you can differentiate it. It should be c one. So you can differentiate it, and then you can project onto the fiber, and using a connection in general, but along the zero set it's well defined. And then the linearization is sort of the linearization of the, so if we evaluate that on a xi and a y, so xi is in the tangent space to the mapping space at u and y is in the tangent space at j to the space of almost complex structures. This linearized operator takes the form while it's the ordinary linearization of the, of the Cauchy-Riemann operator where you fix j and just linearize it. And then you add to it the sort of linear term coming from varying j. Well j actually appears here only linearly. So this is a half of y composed with gu composed with j. So you have the linearization because it's these two terms.  And then when it is, you see this operator here, if you just look at du, you fix j and look at that operator, this is a Fredholm operator. So that means it has finite dimensional kernel, has a closed range and a finite dimensional co-carnal. So to say it's subjective that it just has a kernel, the kernel has a fixed dimension equal to the index of the operator. And then it's a very, it's an open condition that to have a, to have a subjective linearization. So there's an implicit function theorem in this dynamic space context for Fredholm operators. So if it's subjective for all u here and you vary j a little bit, it'll remain subjective. So that's some kind of nice stable situation. So that's what you're looking for. And if so if it's subjective, that means the solution space is a manifold of the right dimension just using the implicit function theorem. But now how do you, what's the basic theorem in this setting? The basic theorem in this setting is that you can't unfortunately prove.  But that's not quite true. So the theorem is that if you look at m star, a j, so this is all somewhere injective, j holomorphic curves, u a, I'm out of time, so I'm just going to have five minutes. Okay, well that would be helpful. Thank you. So you look at all somewhere injective curves and you allow yourself to vary j, then this is a Banach's manifold. And I suppose I should actually, I shouldn't be a little bit more careful. J L would consist of C L almost complex structures because we better put ourselves in a situation of a Banach manifold, right? So we better, instead of lying C infinity J, we better complete them to be, they could be W one P, but you can W K P, but you can make them just C L almost complex structures. So that's a Banach space. So that's a Banach manifold. And then if you look at the map, just taking a pair uj to u, so j, sorry, that this is Fredholm for L large enough. Actually perhaps it's always Fredholm, but really what I want to say is if L is large enough, does a set sitting inside there of regular values. So that would be where the thing is transversely cut out. It's called regular if the thing is transversely cut out. And the point here is that this has, is a residual, residual in the sense of bare. Or I suppose you can say co meager. Or if you're in the first edition of the book with Deepmar, we said of second category, but that's actually a wrong thing. That doesn't mean, doesn't mean what we meant it to mean. But anyway, this is a large set. In other words, it's a countable intersection of dense open sets. So it's a large set, a co meager thing. It's not open and dense itself, but it's a countable intersection of open dense things. That means it's always non empty, lots of stuff in it.  So what that's telling you, and then so that means then that there's a large set in here of regular J where there are all the solutions of regular. And so that, but the kicker here is that the curves have to be somewhere injective. You can't do this. You see, if it's a multiply covered curve, you may not be able to make it regular. The difficulty is that if you have, suppose you have U from S2 to M and this could be regular, that one could be fine. So DU could be onto and it could be somewhere injective. But then we could take a map which is a composite of, we could take phi from S2 to itself of degree K and then look at phi composed with U. So perhaps I should call this psi because it's not a holomorphism. So this is now, this is a multiple covering because you're doing sort of this, for example, could be U at Z to the K or something. Now that, if you have a regular curve that exists and you perturb J, it's gonna still exist, they persist. So you may be able to have, you may not be able to, this may be a perfectly good regular curve. You perturb J, you always have U. Then you always have these multiple covers. But what's wrong with these multiple covers? The fact is that the index of this thing, well, it's 2N plus 2C1 now of K times A plus whatever, well, perhaps we don't have any mark points, so that's what it is. Now the difficulty here is that if C1A is negative, which it certainly could be. I mean, there's no reason, in order to have the index, so you see if the index of U is 2N plus 2C1 of A and because if we're just looking at a map from a sphere into M, it always, if it exists, there's always a sixth parameter family because we have all the re-parameterizations. So the index of this has to be bigger than or equal to six in order for this curve to be there. But suppose you're in a dimension of 2N is 10. You could suddenly have this being 10 and this being minus four until you have a curve in that class with a negative shown class. But then you see if you look at something like this, this thing could be so negative that this actual index could be negative. And yet this curve would always be there. You can't perturb it away because the underlying curve is always there and therefore it's multiple covers are there. So if you have a strata in the stable map so it actually contains something like this, I mean, you could have a curve in class, suppose you have a curve in class B and then it could decompose into something of class B minus KE plus KE where C1 of E is negative. It could decompose like that. Well then you see, you're not going to necessarily be able to get rid of the curves in class KE. They may always exist. There could be multiple covers of curves in class E which are allowed to exist. But then the index of this thing would be big because you see the first chunk class here of C minus KE, well, first chunk class of E is negative and so the first chunk class of B minus KE is bigger than the first chunk class of B. So that means you'd have a strata and the curves in class B minus KE would, in the best situation, even if you could make them all generic, they would live in a space which has got too high a dimension. So that means you lose control of the boundary and therefore you can't necessarily, you can't have simple methods of getting rid of this. So that's the problem. Why you can't use regular standard thing and why you need this to be injective? Well, I don't have time. If people don't know the argument I can explain to you, you can ask questions about, I don't have time to explain why it has to be somewhere injective here. But basically the reason is that you've got, here's your manifold and here's where you're varying J. So when you're varying, you see you're allowed to vary J here to get regularity but you're varying J on M. But the image of your curve, here's S2, where the domain of your map is, your map is going to some one form on S2. So you have to take your variation in J, which lives in M, and pull it back to S2. Now if it's somewhere injective, then there's going to be a little neighborhood here. You can control it, you can control J here. If it's somewhere injective, you can pull it back and control it there. But if it's not somewhere injective, when you pull back something from here, it would have several pre-images here. And so you can't, it doesn't give you so much control over what the variations in J do for you. So really as I explain, if it's not somewhere injective the curve, you can really have these bad components of the compactification and it is too big that can't be dealt with. Now I was going to say more about the analytic difficulties in dealing with this in the traditional way. You see what you have to do, I'll be explaining sort of geometric regularization. You have to have more complicated variations. I mean, you have to allow yourself more complicated perturbations, which don't just depend, I mean the perturbations I'm allowing myself here are just to vary J, which is say, it's sort of a big variation, that's just on the manifold, and you vary J on the manifold. Well, you can have much finer variations of J to correct, but then there are all kinds of problems which I luckily don't have time to get into. So there we are, okay. So since we have office hours this evening, I propose that long questions get asked then, but do people have any quick, shorter questions, producer? Does the students say that the regular locus is connected? Yes, it does. One parameter family is, because everything sort of happens in co-dimension two. Well, you don't actually have to know it's connected, the regular locus. What you need to know is that there's a regular family, a regular homotopies. And regular homotopies, you see if you have a one parameter family in here, that sort of gives you an extra dimension because you've got a variation of J in one-dimensional families. So you can have a one parameter family of things here, which doesn't actually consist of everywhere regular things, but consists of things where the co-dimension of the image is just one dimension, right? So yeah, the part of this theorem tells you if you have two regular values, you can find a regular homotopy in between and where the inverse image would be a manifold of the right dimension. But you can't necessarily pass between them through regular Js. No, you can't necessarily pass between two regular Js because if you could, it would mean that the space of J homomorphic curves for J0 and for J1 were isomorphic, basically, right? And basically, all you can say is that they're co-bordant. For spheres in certain situations, you can, but in many, you know, in general, you can't. Thanks for Dusso. Thank you, Dusso, for a nice introduction.


Interviews. Job interview. A job interview is a meeting between an employer and someone who wants a job. The purpose of the meeting is to find out if the person is good for the job. Location. The location is where something is. Before you leave for a job interview, you should check the location of the interview so that you do not get lost on your way. Clarification. Clarification is when a word or sentence makes something easier to understand. If you do not understand a question in an interview, you should ask for clarification. Strengths. Strengths are the soft or hard skills that someone is good at. Before you apply for a job, you should identify your strengths and include them on your resume. Weaknesses. Weaknesses are the skills that you need to improve. In an interview, the employer will probably ask about your weaknesses. Think of ways to improve your weaknesses. Describe a situation. To describe a situation, talk about something that can happen or happened in the past. In an interview, describe a situation where you used your strengths. Impression. An impression is how people feel about you. When you first meet someone, you can make a good impression or a bad impression. Make a good impression by being polite. Demonstrate. Demonstrate means to show that you can do something. Employers want to see you demonstrate that you can do that job before they hire you. Under pressure. Someone is under pressure when they have a lot to do in a little amount of time. Being able to work well and be organized when under pressure is a good skill to have. Attention to detail. Attention to detail is the ability to notice everything that needs to be done. Making checklists is one way to pay attention to details and will impress potential employers. Professional attire. Professional attire refers to the clothes you wear to an interview or other work-related event. Wearing professional attire can help you make a good impression. To be on time. When you are on time, it means that you have arrived at your destination a few minutes before you need to be there. You should always be on time for an interview. The end.


It's a video on St. John talking about the fact that he really get paid that much for his song with Usher. But I think there's some important points that I don't see people talking about enough. Check it out. The song came out at the time the song streamed 70 million. That's a lot. That's not 70 million dollars. And here's the. That's a song that he do with Usher. That's what he's referring to. And it did 70 million streams. Mathematics. 70 million streams. I own 25% of the songs and it's $3,000 per million stream. It's 3,000 times 70. So $210,000. All right. It's math. The numbers got to make sense. I'm waiting on the song for two years based on the royalty rate that Spotify pays the publishers. There's two types of money in records. It's publishing money. That's writing. And then it's performance money. That wasn't my voice. I'm just on the publishing side. So that $210,000 that that song made had to be bust down. My take on that $210,000 was $1,500. And if I to put that song out myself and only had 3 million streams out of me $10,000, I said I'm done. I washed my hands. I was like, this is done. You know how hard it is to get these songs on these people who put it out on their own time. The artists who have their own vision for it. So your livelihood to be dependent on somebody else that you don't control or influence impact or have a relationship with. And it's not even your worth. You're not even getting paid. You're waiting salt. I wrote a record for us. Whew. That thing was moving because he's talking and he hit so many points that people aren't addressing. Like people keep going back to the $1,500 for 70 million streams. We know that. We know that. Look, the stream rate, it ain't that much. Yeah. Right. Especially when you look at the percentage of the percentage. So one, I do love that he smartly was like, look, man, I could get less streams and make more money. That's the basic indie concept. Any artist been telling y'all that forever, man. Like I could not be signed, but get a higher percentage and do quote unquote worse, but make more money. Yeah. Right. That's something that I feel like people understand for the most part. Right. Now, still, I think the generalized thinking of, I wrote a hit with usher, even though 70 million isn't like a usher usher hit, you know what I mean? But saying I wrote a song that did well with usher and I only made $1,500 off of it. That's mind blowing for, I know, like a lot of consumers and just people in general. Yeah. Right. Cause we're not even talking about, oh, it was a work for hire. He actually was participating in the back end, right? True credits, et cetera. And he still didn't get paid that much. But when you look at two years, I got $1,500 after two years, golly. Yeah. You know what I mean? And now this is how you know, you could always tell like the artists that are like older, right?  Yeah, bro. $1,750 a year? What amount, $1,750 a year, right? Exactly. You just bust that down into the hourly raise.  You know what I'm saying? Yeah. So I mean, he even made that point, right? Like he's like, I would most rather just like take the bet on myself and make less there. So I mean, I don't know, man, this is always such like a thin line for me because like the split conversation and just how micro those breakdowns can get. $20 a month. That's what they came down to. It's crazy. Oh, you just did the math on it? Yeah.   But it's just because I feel like it's safe to say he's probably locked in some like crazy club situation. You know what I'm saying? For it to come down like that, but that's why I go back to it. I don't know. You know what I'm saying? I don't know. But 10 K isn't that much either. Right? Yeah. Do 10 mil or three million get 10 K. We know that's not all that much either, which is something that artists are struggling with. But to me, the biggest thing out of all this is what the writers are dealing with. Remember, I talk about me not necessarily wanting to manage artists early in my career because I'm like, I can't depend my whole life's success off this individual, right? Yeah. Him from a writer standpoint in this time, he's not even talking about individual that he's personally dealing with. It's one thing where I'm like, bro, yo, Jacory, you're the artist and bro, I got to like wake you up. You're not inspired enough and I have to deal with all this. And it's like, bro, you being lazy, but at least I feel like I got direct line of communication and I feel like I might be able to influence you a bit, right? Yeah. So I'm just talking about as a songwriter, just throwing your music out into the ether and sometimes and just hoping that it gets picked up, you pitch it to somebody and then like he said, the artist has their own vision. So they might like the song, but then it might not be in this project because it doesn't fit this project. So I have to wait just like any artist does, right? With their own music. Oh, if it's here, better than there. So he's talking about after two years getting 1500. I don't know if that's after it came out or whatever, but let's just pretend, which has happened in many songs. It took like five years for the song to come out, right?   So it might be like five years before you start to see that money. You add in like the amount of time to get them to accept it. You don't know if they're going to accept it at all because you don't have much influence over the situation in many of these situations. It might take a certain amount of time to get accepted. If it does get accepted, which is a probability to work with there, then you have the amount of time for it to come out. Then hopefully it performs well, right? Because that's not guaranteed.  Even the biggest artists have songs that don't do well. Yeah. And then we know how slow music payouts come and how sometimes music payouts don't come in the way they should. And somebody might end up having to audit and do all these other things for $1,500 over an extended period of time. He said, too, it could have been longer. So a lot of this in terms of artistry speaks to why, especially in today where the opportunity is there, control as much as you can. Just point blank, control as much as you can because it's already hard enough to win this. And the further away from the control that you get, you're dealing with not only increased probability of failure, but you're also dealing with lower percentages if you succeed. So the risk factor is high as hell. Let me take a quick second to say if you're an artist trying to blow your music up or if you're a manager, a music professional in general, trying to help an artist blow their music up, I have something that's a game changer for you and it's completely free. As you may know, we've helped multiple artists go from zero to hundreds of thousands of streams. We've helped multiple artists go from hundreds of thousands to millions of streams, chart on Billboard, GoViral, all of that stuff. And we've now made the way we've branded multiple artists and helped them go viral completely free, step-by-step in Brandman Network. All you have to do is check out brandmannetwork.com. You apply, it's completely free. But the thing is, we're not going to let everybody in forever. So the faster you apply, the better your chance of getting accepted. Brandmannetwork.com. Check it out. Back to the video. Yeah, yeah, yeah. I do say it from his side too, where it's much easier to get three million streams and 70 million, you know what I'm saying? Like three million is not a crazy goal to hit, especially for like an artist of his caliber. But I would, devil's advocate a little bit and say, Of course, of course. And this goes back to, I would love to know what his situation was like then, but I would imagine that having the credit for that song probably opened hella doors for him. Right? Yeah. Something that maybe wouldn't have happened if he put it out under his own umbrella. And I think just like, because, because what I don't want every artist to hear this and say is like, oh, this is why I never would song write.  Because I know a lot of arts that feel exactly like this. Like one, I don't want to give someone a hit that I feel like could have been for me. You know what I'm saying? Now I'm a little salty about it.  This is getting too far away from my hands and my throat. You know what I'm saying? Those are the two main reasons I hear RSA don't want to do it. But then I think about like all of the prolific artists who started out as like song writers, you know what I'm saying? And we're willing to give away that shit and probably go through similar situations like this because of the doors that open just from being able to say like, hey man, I got an usher song. It got 70 million, right? Well, I got a whoever, a Beyoncé, some walk or whatever song. And then, you know what I'm saying?  So I do think like, the other thing I do kind of wish he would have bought a little clarity on like what part of his career, what point in his career was this? You know what I'm saying? Because I don't know like, do you know if you wrote that song for us before? Like he started popping as a solo artist? Or if it was like in the midst of it or? I think it was before he was starting popping and that was kind of why he made a point. I could try to look up what song it might have been that he's referring to specifically. Yeah. But who else did he write for? I remember, we were working with some people when they were like, St. John used to write for... I fight hella people. Oh yeah. So the song, he worked, Kaiser, he worked with her, which is like he had all these random credits that people didn't know him for. Yeah. Right? And you would never expect the artist. He might have had something on hideaway. I know a lot of people might not even know that song or whatever. But like he, St. John has some credits like for real, for real. Yeah, like he out here. Yeah. With different types of artists. Know what? It's not on hideaway, but it is with that artist. So I think the happiest medium that I can think of, and of course everybody do their own thing, like whatever makes sense to you. But like me, I would probably just strategically, if I was an artist, compartmentalize it. Like, am I writing a song for Usher? If I was in a position, right? And I had somebody connected where I could truly say, hey, I'm trying to write one and pitch it for Usher, Rihanna, whoever, whoever. So I'm not writing this for me anyway. You get what I'm saying? Yeah. Because I know some writers who actually do that, right? They're writing in somebody else's voice, but then their music is their own voice and is their own experience. Right? Yeah. So like that probably might help some of the dissension between like, oh man, I could have just kept this for myself. Yada, yada, yada. Beyond that, yeah, I think it's a fact that yes, there have been people who have been songwriters who have transitioned to artists. However, there is a grave of songwriters who, you know, never made it to that artist side. However, there's also a grave yard of rich songwriters who never became artists. Exactly. Exactly.  I mean, I'm chasing that up, man. 20, there was $20 a day, man. You get about 30 of them, you know what I'm saying? All they does is remember what it was all about. We talk about $20 a month, by the way. Yeah, I know what I'm saying.  Maybe about 300 of them, you know what I'm saying? Like 300 of them at 30. Yeah, like 300 of them, man. You know? Okay.  Like just be out here collecting songwriting credits. Production credits. Quincy is a different animal right there. That's a different animal.  We're like, man, he didn't even do nothing. All he did was, you know, push the button, one button, and were to say turn it louder, and now he on the credits. You know, I don't have no stake in that horse. I don't understand what officially qualifies as production. And I, and I wasn't there. But that, but even if it was that, I can now understand, right? The incentive to do that. Let me get my hands on as many of these things as possible.  One ain't enough. I'm saying, all the residual checks, but 2% of everything, bro. You know, that 2% started stocking up. Hey, that's a fact, bro. All I need is 52% and I'll buy that honey. All right, next.


So, in the previous lecture we saw that given a topological space given a locally compact topological space space X if it is not compact then we can compactify it. So, what did it mean by saying that we can compactify it meant that. So, we can construct a compact topological space X X hat such that X is sitting inside X hat as dense. So, recall that when we say compact topological space X in our definition of compactness we need when we say compact topological space X hat in our definition of compactness we need X hat to be host of and we had proved that X hat is host of as a dense subset as a dense open subset and in fact X hat was obtained is just X disjoint union a point P naught. So, there is a picture we can keep in mind while we. So, let us take the real line the real line is locally compact but it is not compact and we can take this point P naught over here. So, when we compactify the real line let us see what we get. So, instead of making the picture like this we will make the real line like this. So, recall that there were two kinds of open subsets one kind was type A open subsets these were open subsets of X right and the type B open subsets were as follows we take any compact subset of R. So, for instance we can take this union of two compact subsets of the real line. So, that is going to be compact and we just take the complement subspaces. Compliments in X hat compact subspaces or in other words what we have specified is what are the open subsets the open subsets which contain P naught are exactly complements in X hat of compact subspace of X. So, in the same way we can make see another example of R 2. So, instead of making R 2 like a plane we can make it as a sphere. So, so from this in the picture it is clear that one point compactification it seems that it should be S 1. And similarly if we take R 2 make a dotted line. So, this is a R 2 and we can imagine the point P 0 over here. So, R 2 is also a locally compact and therefore, we add this point P 0. So, the one point. So, what are the neighbourhoods open subsets which contain P 0 these are the open subsets of type B. So, these will be we can take any compact subsets subset of R 2. Let us say this is D. So, then D will be somewhere over here is D and we just take the complement of D. So, this X hat minus D. So, it is useful to keep this picture in mind. So, let us begin today's lecture. So, let us write down what we proved last time. Let X be locally compact topological space which is not compact. There is a compact topological space X hat along with the inclusion from X to X hat such that X hat is just I f X union a point which we call P 0 I is continuous I f X is an open subset. So, let us make the following easy remark I from X to I f X is a homomorphism. So, we have already seen that I is continuous and the image of I is lands inside I f X and it is a bijection. So, we have already seen we know that I is a bijective.   So, thus I is a homomorphism. So, this theorem we had proved last time and in the previous lecture I had mentioned that this particular one point there could be many one there could be many compactifications of a locally compact topological space, but this particular compactification has a very nice and special property. So, which is what we will see in this lecture. So, in the topological space X hat has the following property proposition that T be a compact topological space which contains X as an open subset. So, what is the meaning of this sentence right. So, this means that there is an inclusion J from X to T it is an inclusion of sets such that J of X containing T is an open subset J from X to J of X is a homomorphism. So, there is a inclusion says that let me also write J is continuous ok. So, then there is a unique map. So, what we have is we have X over here and we have this inclusion J, J is a continuous map from X to T and the image. So, let us say the image is J of X and this is a subset of T there is an open subset and we have this map like this and we also have the map I also satisfies these properties right there is X hat there is also open an X hat right. So, then there is a unique map F from T to X hat is a unique map F from T to X hat such that the following two things happen F compose J is equal to I and F of T minus J of X is equal to this P naught. So, X hat has this pointed infinity which is P naught. So, let us try to see what this means by means of an example. So, let us say X is this open disk let us say I embed the open disk to this closed disk right. On the other hand the 1 point compactification is J this is I the 1 point compactification looks like. So, I have to add a point at infinity. So, it will look like something like this. So, this point is P 0 right. So, what is F doing? F is collapsing all of the boundary F. So, F collapses the set T minus J of X which in this case is this black boundary to the point P 0 right and F is completely determined at other points F compose J is equal to I. So, we want to say. So, this in this example F does this, but ok. So, let us say that let us try to prove this proposition which is that there is a unique such map F ok. So, of course F is continuous there is a unique continuous map that is the assertion. So, the map F has been defined set theoretically on all of T right. So, this this map has been defined set theoretically on all of T. So, how do we do it? So, on J of X define F to be. So, J of X J is a homeomorphism from X to J of X. So, we apply J inverse is X and then we apply I. So, this goes to I of X this makes sense as J from X to J of X is a homeomorphism and ok. So, we do not need homeomorphism right now. So, we just need bijective is a bijection right. We just need bijection right now because as of now we are really trying to say that F is defined set theoretically right and outside on T minus J of X we have defined F to be the constant map P naught. So, thus the map of sets F is completely determined right and so, it is unique. So, we only have to show that F is continuous. So, for this so, let V contained in X hat be an open subset. So, let us first assume that assume that P 0 P 0 does not belong to be right. So, in that case the open subset is contained somewhere over here right. So, so then it is easily checked this is an easy trivial check that F inverse of V is completely contained inside J of X. So, as a result so, this shows that F inverse of V is actually equal to F restricted to J of X inverse of V. Now, notice that so, note that F restricted to J of X is from J of X to X J inverse and this compose with I to I of X right and as J from X to J of X is a homomorphism this implies that J inverse is continuous this implies that I compose J inverse is continuous. So, this implies that F restricted to J of X is continuous. So, this implies that combining these we get that F inverse V is equal to F restricted to J of X inverse V which is open. So, thus if P 0 does not belong to V then F inverse of V is open right. So, now let us consider the situation if P 0 belongs to V then X hat minus V is compact and contained once again contained in I of X. So, once again F inverse of this since it is contained in I of X is equal to F restricted to J of X inverse of X hat minus V, but this is a compact subset I of X right and this set is in fact equal to F restricted to J of X is equal to I compose J inverse and both these are homomorphisms. So, this is from J of X to X J inverse and this is I of X right and both these are homomorphisms. So, this implies that F restricted to J of X is a homomorphism. So, which implies that F restricted to J of X inverse is a homomorphism is continuous which is continuity. So, this implies that F restricted to J of X inverse of X hat minus V is a compact subspace J of X. So, also so this shows that F inverse of X hat minus V is a compact subspace and so also a close subspace. So, this implies that, but F of X hat minus V F inverse of X hat as F inverse of X hat minus V is equal to T minus F inverse of V this implies that F inverse of V is an open subspace. So, this proves that. So, this completes proof. So, we will end this lecture here.