We start a new chapter, the new topic, Covering Spaces and Fundamental Group. Of course, Fundamental Group we have seen before and now we want to study Fundamental Group in relation with Covering Spaces. On the other hand, Covering Spaces is one of the oldest or the simplest or the very beginning concepts in algebraic topology. It is a link between point set topology and algebraic topology, it has lot of point set topology. It slowly passes on to give you the algebraic topology result as we have already seen in the calculation of the fundamental group of S1. What we have used is the exponential map and that is a prototype, a beautiful example of a Covering Space. Covering Space here is closely related to another concept namely this is a classical approach, what they used to call discontinuous groups, okay. So the discontinuous group action, group actions you know we will be also studying that to some extent. Why are some extent each subject here is very vast, okay in applied in almost all branches of mathematics. Classically again these concepts occurred in the reverse order namely the discontinuous groups were the ones which were studied, the notion of Covering Spaces etc came much later. So like you and one can say cite examples of Gauss and then theory of electric functions by Abel and Weistras and so on. During the time of Riemann the notion of Covering Space started taking place. In fact you may say Riemann is the one who introduced even the manifolds, okay and then this Covering Space also in his Riemann surfaces of some algebraic functions and so on. The fundamental group appeared much later in Poincaré's work, okay. So one can give a big credit to Poincaré to put various things together and invent something, some other way of looking at it at all, okay. Nowadays these three notions have taken deep root amongst all branches of mathematics. They have been found useful and in any case make a very, very enjoyable delightful study of mathematics. So Covering Spaces. So let us begin with simple definitions and then consequences of these definitions and so on. Later we will give some examples also, okay. So that is what we will do for today. So start with a continuous function from one space to another space which is also surjective, okay. So the first space I am returning by x bar is nothing to do with the closure and so on. This is a different notation and x is a space, y, x bar is another space, p is a map from x bar. This is standard notation for a Covering Projection, okay. Take a surjective function, we say a subset V of x, usually this x is called the bottom space and this is called total space by the way. I will introduce those terminology also. So in the bottom space you take a subset V, okay. An open subset, this should be called evenly covered by p. If p inverse of V is covering is not inside x now, everything is very strange wordings here, okay. So p inverse of V, you have come to come back to x bar here, okay, p inverse of V is a disjoint union of open subsets of x bar. So let us write p inverse of V as disjoint union of UIs indexed by subset. Now you can restrict p to each of the UIs and come back to V because this whole thing is inside p inverse of V, right. So when you apply p to any of UIs, you come back to V. This map, the restriction map of UI to V, this must be a homeomorphism for every I. In other words, each UI is a homeomorphic copy of this V and each of them is disjoint from each other. So total of all of them will be the full inverse image of V. So if this happens, V is called evenly covered by p, okay. And look at the cardinality of this indexing set, okay, that will be called number of sheets. Each UI is called a sheet. Sheet means what? It is just a copy of V, copy in the sense it is homeomorphic, okay. If x can be covered by open subsets, each of which is evenly covered by p, then we call p is a covering projection, okay. Now this covering is precisely that, what I mean, this word covering here, this is that the union of all such V is equal to x, okay. So this is usual covering of a space by open subsets. These open subsets, each of them should be evenly covered by p, okay. So that means that inverse image of each of this set p under p is disjoint union like this. This condition should be satisfied. So if you vary V with this condition, satisfying for all the V's and if you get the whole of x, then this p will be called a covering projection, alright. Strictly speaking, each time we mention the word covering projection or a covering space, we should not only mention the two spaces x bar and x, but also the covering projection. It's like similar to a quotient space. When you say quotient space, you have to specifically tell what is that quotient map from x bar to x. So it's like that p should be mentioned. So strictly speaking, it should be x bar, p, x, the triple is a covering, it's a covering projection, but that is too much to write. Like even if you don't write x, tau as a topology, right. Just topology is never written, like explain topology as we say. Similarly, we have to do with this shortening terminology that's all, okay. However often this will be clear from the context, which function we are taking and so on. So for simplicity of language, we will merely see x bar is a covering projection of x, okay. We also say that x bar is a total space and x is the base space. This I already told you. Whenever you have a covering projection p, it comes with a total space and a base space, okay. Every covering projection turns out to be a local homeomorphism. Recall what is a local homeomorphism? For every x in the domain, namely x bar inside x bar, you must have a neighborhood view of x bar. And that f restricted to you is a homeomorphism onto f u, which is an open subset of x. Then you call it, then if this happens for every x, then you call f is a local homeomorphism. And that is precisely happening here if x is covered by open sets like this, inverse image will be covered by open sets like this. And these are disjoint in your view, so x will be inside one of them, right. And from ui to v, it will be homeomorphism and this vi is open, this v is open. Therefore, every covering projection is a local homeomorphism, alright. This property, local homeomorphism has a tremendous influence. Whatever local property of x is there, it will be there on x bar also. For example, if this is locally path connected, then x bar will be locally path connected and conversely. If this is locally compared, then x bar will be locally compared and conversely. If this is a first countable, then this will be first countable and conversely and so on, okay. More than that, there are structures like this is just smooth manifold, this will be smooth manifold and conversely. Such things are also true, okay. So, I have given you local compactness, locally connectedness, locally path connectedness, T1-ness, locally contractible, locally Euclidean at its manifold, etc., okay, if and only x bar is there. So, however, you have to be very careful in extending this kind of list. I said T1, the next thing T2, it is not true. So, this will actually tell you that T2-ness is not a, strictly speaking, local property. It's local global. It's about two points. Whereas T1-ness is about one single point. At each point something happens. However, the horse dorseness, the next one is regularity, normality, etc., none of these is a local property and if x bar has it, x may not have it, x has it, x may not, x bar may not have it. Either way, it can. I don't know, right? We can't say. So, there are exercises about that one, alright. Every local homomorphism is automatically an open mapping. Is that clear? Because take an open set, for each point, you have a neighborhood, such an image is open. Right? So, this open set is union of such open sets. So, image will be union of those open sets. Therefore, image of every open set is open. So, it is automatically an open mapping. Therefore, a continuous rejection which is in open mapping is also a quotient map. Therefore, x will be a quotient space of x bar. Okay? In general, if you have a map f from x to y and a point y in y, we call the set f inverse of y the fiber over f. This is a general notation. This is general terminology. The fiber of a map means the inverse image of that one. If f is a local homomorphism, the fibers of f are always discrete. Look at the fiber. Take two points there. Each of them has a neighborhood, homomorphic to a neighborhood below of the point y. Right? Each of them. If they had intersection, then this won't have happened because under f, the two of them are coming to the same thing. That will be a problem. Okay? The same f restricted to this open set, restricted to this open set is a homomorphism. Therefore, the two open subsets that you have taken above must be disjoint automatically. So, this happens for every pair of points, which just means that, you know, you can do for a single point and everywhere else. So, this just means that f inverse is f inverse of a single point is f. It's a discrete set. Okay? This is a subspace to probably f inverse y. It's a discrete space. In particular, the fibers of a covering projection are discrete. This fact is going to play a very important role in what we are going to study. These are some few immediate consequences of the definition that we have done. Namely, the entire X is covered by evenly covered open sets. Okay? So, I am repeating a few of them here. Given a covering projection, the cardinality of p inverse of X is a constant as X varies inside one open set. One evenly covered open set. Because for that open set, you look at the inverse image, disjoint union of open sets. Each of them coming bijectively to the set U. Therefore, for each point, they are exactly as many points as the indexing set for this disjoint union. So, what does this mean? It's that the cardinality of p inverse of X is locally a constant function. In an open subset, it's a constant. Okay? Suppose now X is connected. This kind of topology you must be familiar with already that every locally constant function on a connected space is a constant. Okay? The connectivity has to be used strongly here. A locally constant function will be constant if X is connected. So, in particular, this just means that if you start with a connected space X and then take a covering namely p, then the inverse image of every point has the same cardinality. That cardinality is called number of sheets of p. Okay? If this cardinality happens to be finite, then we call it p is a finite covering. Okay? So, here is an example Z going to Z power n. We have studied it earlier as a question perhaps, self-question, right? Of S1 onto itself is a typical example of a finite covering where the total space and base space are the same. But this map is from n to 1. n points are in S1 which go to the same point. No matter what point you take. Okay? Take any point in S1, in the inverse image there are exactly n points. That doesn't prove that it is a covering but you can show that Z going to Z power n is a covering. So, now let us work out to more examples properly. The simplest thing is if you take an identity map, it is a covering. Next, you can take any homeomorphism that is also a covering. Okay? Those things are trivial coverings. They don't give you much. A typical example of covering projection is already familiar to you namely the exponential map. R to S1 which you can write it as theta going to be power 2 pi i theta. Just if 2 pi is e raised to i theta also. It is according to your fancy. 2 pi writing 2 pi is just a normalizing factor that's all. Okay? Fix a theta between 0 and 2 pi. 2 pi omitted. Then if u equal to S1 minus just 1 theta I have fixed. To remove that point exponential function this one is the disjoint union of intervals theta plus 2 n pi less than t to theta plus 2 n plus 2 pi. So, it is of period 2 pi. I have just taken exponential theta going to e power i theta. That's why I have to take 2 pi here. If I write theta going to e raised to 2 pi i theta then this would have been 0 less than theta less than 1. Okay? Restricted to any of these intervals X is a homeomorphism. If you throw away this point also then you get both open success on both sides. Keep shifting this theta at various levels that will give you the covering. The whole covering you will get covered of the entire S1. Therefore, this will tell you if you have verified it is already and we have used this property before. So this shows that exponential function is a covering projection. And this picture also perhaps is familiar to you. Here I have thrown away this point, let us say. In fact, I have taken only half circle here. What are the inverse image of this one? This will be half interval here then jump by another half, another half interval, jump by another half interval and so on. The inverse image of this point, this is 1 comma 0 will be all integers. So exponential function restricted to this interval to this arc is a homeomorphism. From here also it is a homeomorphism. From here also it is all. The full inverse image of this arc will be union of all these intervals, half intervals. Okay? This part we have already in a similar way and it is not hard to do this one. Namely, Z going to Z power n defines a covering projection of C star to C star itself. C star is what? Non-zero complex number. Restricted to circle, it will go into circle itself. Namely, unit circle, mod Z is 1, mod Z power n is also 1. If you do not put that one, any non-zero thing will go to a non-zero thing and it will be n to 1 mapping. Okay? You can work out neighbors, how they, how small neighbors should be taken here and then look at the inverse image. There will be n copies of that, each of them mapped on to the same open subset here. Okay? If Y is any subspace of X, I can take inverse image of Y under P that will be subspace of X bar. Now you take the P restricted to P inverse of Y to Y. That itself will be a covering projection. What you have to do? Take an evenly covered set inside, open set inside X. Take intersection of that with the Y. That's all. That will be even covered by this restricted function. Okay? On the other hand, you cannot do this by taking a subspace of the top space X bar. Then you have to be careful. Okay? In general, if I take X bar to X covering projection and Y bar is subspace of X bar, restriction map may not be a covering projection. Okay? So when it will happen, that will become interesting result. Okay? We will see it later on. To construct local homomorphisms, we said covering projection is always a local homomorphism. To construct a local homomorphism which is not a covering projection is very easy. Okay? So just local homomorphism, on-to-ness, etc., won't give you covering space, covering projection. Okay? So if we take the restriction of a covering projection X bar track to any open set U inside X bar, it will be automatically a local homomorphism. However, by choosing U very badly in some way, okay? We can destroy the covering space property. The even lake awareness, we can destroy. In many ways, to retain it is more difficult. Destroying is automatically happens. Namely, all that you have to do is, for example, just omit one point from X bar. That is an open set. Normally if X singleton X is a, you know, if it is a T one space, this will be an open set. Okay? So now if you restrict, this will never be a covering projection. Starting with a covering projection X bar to X, to omit a point, the restricted thing will never be a covering projection. Okay? Verify this one. Then you will understand what is why covering projections happen. So I want to tell you some more thing about the role of path connectivity here. Path connectivity, local path connectivity are part and parcel of algebraic topology. We assume that one quite a bit. Okay? So here, path connectivity is very important in the case of covering projections. So that is what I want to tell you. Okay? So this is the theorem. Take a covering projection, take a continuous map from P bar to P, P from X bar to X, where X is now locally path connected. Then something nice happens. Namely, the map P is a covering projection if and only if on each component C of X, component means path component. The restriction map P from P inverse of C to C is a covering projection. This is from C is a path component of X. If you take P inverse of C to C, it should be covering projection. If P is a covering projection, then for each component C bar of X bar, now see again I am taking path component. The map P restricts C bar, P C bar is a covering projection. So under restriction map, the covering property is not destroyed if C bar is a path component. And in that case, P C bar will be a path component of X. The point is that you have to take a component. If you restrict just to a path connected space, it is not true. Just like the other way, if something is path connected by removing a point, it may still be path connected, then it will not be a covering projection. But if you remove a point, it will not be a path component in the original thing. You have removed something, right? So it won't be path component. So that kind of contract example is not, but this is a theorem actually you have to prove this. Namely, the second part is very important. First part is easy because you are coming from bottom.  So this is not so difficult. This part is something which you have to doubt and so you have to pay attention to the proof of this one, okay? So here is the proof, okay? So first part, if P is a covering projection for any subspace Y of X, the restriction is a covering projection, okay? So one part is easy. Conversely, for each component P, C, P from P inverse of C, C is a covering projection for each component, I have to say. The crucial thing here is that since X is locally path connected, each C is open. Therefore, for any point in X, consider the path component of C, path component of, path component C which contains X. Every point is inside a path component and that is an open set. So if U is open in C, U will be open in X also. But this is a covering projection so I can choose U to be evenly covered. Then since this P inverse of C is the full inverse image, it will be same thing inside X also. So the same U will be evenly covered, neighborhood of X. Since every point is covered by an evenly covered neighborhood, P from X to X bar to X is a covering projection. So this proves part one here, okay? That is the easy part. The second part, first of all, look at P C bar, okay? T C is an open set. Why? Because X is locally path connected, X bar is also locally path connected, path connected components are open because they are locally path connected. And P is an open mapping. So P C bar equals C is an open set. Given X belonging to C, let V be a connected open neighborhood of X which is evenly covered by P, okay? Put P inverse of V could disjoint union of UIs. Then each UI, each UI, because I have started the connected open neighborhood of this locally path connected. So you can choose this one to be locally path connected neighborhood. It's connected neighborhood. And these things are each UI is homomorphic to V, right? Therefore, they are each of them is connected. And hence, either when you write, when you write a component, UI is inside C bar or UI intersects C bar is empty, okay? Therefore, it follows that when you take restrictions, some of these UIs will go away, the rest of them will be remaining, okay? Fully. And then they are contained in no part. No part of UI will be there. Either it is full or it is none. So periods of Cs will be also a homomorphic, corresponding to that, okay? Finally, the converse of this one. To show that C, sorry, one more thing, C is a component. P C bar is C here. So it is connected. It is path connected because it is path connected. Why it is a component I have to show. Okay? So let X be a point in the closure of, in the closure. And V be an open neighborhood of X as above. Then one of the UIs, okay, has to intersect C bar. Because union of all U bars is a neighborhood of this one. So it is in the closure. So one of the UIs has to intersect C bar. The whole thing intersects. So at least one of them has to intersect. That is the point. Which in turn means that that particular UI is contained inside C bar. Because C bar is a component. Okay? So these components that I am working inside a larger space X. And U is an open substrate in larger space. And hence V itself is inside C. Because V is P of C. P of C bar. Sorry. C is P of V bar. So this intersect is one of them. So it is inside C bar. So V is inside C. So if V is inside C, it shows that C itself is open as well. A component, you know, it is closed. Okay? Now I have shown that every point has a neighborhood which is contained inside C. So it must be, it is closed as well as open. Actually I started with a closure point. Then show that the whole thing is inside, in a whole neighborhood inside C. Right? So C must be open as well. If it is open and closed and connected, it must be a component. So that completes the theorem. I think we will stop here today. It will take you some time to understand this one. Let us go more things next time. Thank you.