 Hello everybody. So, today's topic is solution of lifting problem. We have already solved this problem in the simplest case, namely for the loop. A loop based data point X can be lifted through the covering projection. If we download the loop class represents an element of pi 1 of X X naught should be inside the image of P check where P is the covering projection. This is what we have seen. So, now we want to expand this one to arbitrary spaces. The point here is the space cannot be completely arbitrary. We need to assume some enough path is must be there inside the space. So, we assume that it is locally path connected. And of course, after that we can assume that it is connected that is not a restriction because we can always argue with component wise. So, let us assume that space is locally path connected. Take a map from this to X the base space where X bar to X is the covering projection. So, then we ask the question whether this map can be lifted. So, it will depend upon the character of this map F. So, what is that condition? The condition turns out to be completely in terms of algebra. So, this is the gist of the thing here. So, this is the question. F is a map from Y to X bar, Y to X, can you lift it to X bar such that P composite F, F hat is F. So, finding F hat is the problem. So, what we want to assume is that Y is locally path connected and connected. So, because of our earlier criteria on loops, lifting of loops, this problem can also be converted into purely algebra. So, this is the theorem, the lifting criteria. So, as usual P is a covering projection, Y is locally path connected and connected space. F from Y to X be any continuous function. We assume that Y belong to Y, X bar, you know X bar, all these are the set P of X bar equal to F of Y. So, Y is here, F of Y is here. So, you can think F of Y is X, then we want to X bar to be such that P of X bar equal to F Y. If this is the case, there exists a map F hat from Y to X bar as we want namely P composite F bar equal to F such that F bar of Y is the point X bar. If we do lift pi 1 of Y Y, pi 1 of Y at Y under F check goes inside P check of pi 1 of X bar X. Notice that left hand side and right hand side both are subgroups of pi 1 of X, X. F of Y goes to X, P of X bar also goes to X. So, both of them F check and P check are both maps into pi 1 of X, X. So, these two subgroups must have the property that this entire subgroup must be contained inside this subgroup. If this happens, there exists F bar and if F bar exists then this must be true. That is different only. The only part is very easy because F bar has the property P check F bar is F, right? P check F bar is F. Do you understand? So, if there is an F bar from pi 1 of Y, F bar of pi 1 of Y, F bar check of this one would have gone into pi 1 of X bar, X bar. Then you apply P check, it will become F here. So, that is very easy by functoriality property. The induced map from the composite is the composite of the induced Romeomorphisms. That is what we have to use. So, given F bar, I repeat that we have F check is P composite F bar check because P composite F bar is F. But this is nothing but P check composite F bar check. And hence F check of pi 1 of Y, Y is simply a P check of F bar check of pi 1 of Y, Y. But F bar check lands you inside pi 1 of X bar, X bar. So, you have to follow it by P check. P check of this one is the notation is K, which is K is the subgroup of pi 1 of X. This subgroup must be contained inside this subgroup. Convertly, suppose this is contained inside K. Now, we must construct F bar. So, here is the procedure which resembles more or less the primitive mapping theorem in two variable calculus that you must have learned. If you remember, you can see the analogy. Given any point Z belonging to Y, first we choose arbitrary path from Y to Z inside Y. This is possible because Y is path connected. Connected, locally path connected means it is path connected. This is what I meant by having enough path is. So, here all that I used is it is path connected. We will use locally path connectedness again. So, start with any Y is fixed. Start with any Z, join it to Y by a path. Take a lift of this path namely omega bar, lift of F composite omega. F composite omega is a path inside X starting at little x. So, take a lift of this one, lift will be a path inside X bar. Look at the end point. The end point I am going to define as F bar of Z. So, this kind of thing we have done right in the beginning in computing the fundamental loop of the circle. So, here for an arbitrary function we are defining F bar of Z as the end point omega bar of 1. Okay. Now, you have to verify many things. Y F bar is well defined. Remember, we have chosen some path is here Z to Y right or Y to Z right. There may be many paths. In fact, once a path connected there will be many paths. If I change the path the end point may be different. So, I have to ensure that irrespective of what path you take the end point is the same. Okay. So, this is the path independence of independence of this path is. Absolutely there is no homotopy or nothing in this time. Y is path connected, locally path connected. I have joined the two points. Take the image inside X and then lift it and I want the end point to be independent of what path I take. Only thing is it must be joining Y to Z. That is all. So, whatever we need it turns out to be it is just there. We have to first show that F bar is well defined. Okay. So, let gamma be another such path. Another such path means what? Y to Z. Gamma bar be the lift of F composite gamma at X bar. Then we have to show that gamma bar of 1 equal to omega bar of 1. That is what you have to show. Look at the loop gamma composite or omega composite gamma inverse. That is a loop in Y, loop at Y in Y. Now, if you take F check of that F check of omega composite omega star gamma inverse. This is an element of A is the hypothesis. The whole of F check of the fundamental group of Y goes inside K. Oh, that is the hypothesis. Therefore, by our lemma for lifting the loop, this loop F check of omega composite gamma inverse. Okay. The lift of this loop will be a loop inside. Inside where? Inside X bar at X little X bar. Okay. But that would mean that F of F composite omega and F composite gamma, these two are, this is the inverse of that one, right? F composite gamma. These two should have same endpoint. So, this is our lemma 7.2, which was purely geometry that we have seen. So, we are using it again here. Okay. So, these two have same endpoint. Okay. The lifts of this wire. Okay. Therefore, this gamma bar of 1 is same thing as omega bar of 1. So, theoretically, we have already formed a function. Now, we have to show that it is continuous. What is immediate is that P composite F bar is equal to F automatically. What is F bar? F bar of H z, whatever z I have chosen is omega bar endpoint of omega bar. When you take the P of that, it will be the endpoint of F composite omega. That will be nothing but F of z. That is what we want. So, F bar followed by P is P composite F bar is F. So, that is already there, theoretically. If F bar is continuous, it is continuous, but we have given F is continuous, P is continuous, we have to show F bar is continuous. Okay. So, we have to work hard on that. Okay. So, let us see why the continuity comes. This is where we have to use local path connectivity of Y. Take a point z belonging to Y at which I want to show F is F bar is continuous. So, take a neighborhood, U be a neighborhood of F bar of z. Okay. Then I must produce a neighborhood of z, which goes inside that neighborhood. That is a continuity. But I choose a neighborhood U, open neighborhood of F bar of z, which is perhaps homeomorphically, I want to evenly cover neighborhood. Inside every neighborhood, I can choose a smaller one such that its image in X is evenly covered. That means it is disjoint union of open sets, which are all mapped to the same set below under P. Okay. Takes an evenly covered neighborhood. Okay. By continuity of F, F is continuous. Right? And the local path connectivity of Y, you both use both of them, we get a path connected open neighborhood W of z in Y. So, set F of W is contained inside V. The continuity, you can choose F of W contained inside V. But inside W, whatever you have chosen, you can choose one such that this is path connected neighborhood because Y is locally path connected. So, this way you get a path connected neighborhood F of W contained inside V. Okay. So, this point, this W contains the point z. It is a neighborhood of z. Let omega be the path from Y to z chosen to define F bar of z. How F bar of z is defined? You have fixed some path is then take F of that and take the left and take the end point of that part. Right? So, look at that omega. Omega is coming from Y to z. Inside this W, W is path connected. I can take a path completely lying inside W from z to any other point. For each point A and W, we can choose a path gamma A suffix gamma, this is gamma A. Inside W joining z to A and then use the path. Now, gamma followed by, sorry, omega followed by gamma A, that will be a path from Y to A. Take F of that, lift it, take the end point. That will be the definition of F bar of A. Okay. Remember, in defining F bar of A, we are free to choose any path from Y to that A. So, we have shown this path. If gamma bar of A is a lift of F composite gamma A, see gamma A is just a portion completely lying inside W. But that can also be lifted at F bar of z bar, F bar of z. Then this gamma bar star gamma A, if we apply P of this, this will be precisely F of omega composite gamma A. Therefore, this is a lift of F of omega composite gamma A. Okay. Therefore, F bar of A is nothing but the end point of this one, which is the same thing as the end point of this one, namely gamma bar of A. Gamma bar of A operating upon 1. Okay. So, where is it? Where is this point is the question. If this point must be inside our V, then I am done. Then I have shown that you know F bar of this W goes inside V. I have chosen W going inside this one. Now, I want to say that F bar of z. Right. On the other hand, since V is evenly covered, that is what we have to use now. And P is maps U, homeomorphs if you want to V. Okay. It follows that the entire path gamma bar A is connected. It cannot be a portion here and portion here and portion here. Wherever the point is starting point, starting point of gamma bar of A is nothing but omega bar of 1, meaning which is nothing but F bar of z. So, whatever component contains that must contain the entire of gamma A. But that is our choice. Our choice is that this is, U is one of the components of homeomorphs to be mapped onto an evenly covered open cover. So, it must be inside this. Okay. So, that completes the proof of that this is, this is continuous. Rest of them you have already done. All right. So, you see effortlessly, almost effortlessly small, small steps. We have proved a big theorem now. Namely, when to lift, when an arbitrary continuous map can be lifted. Okay. From now on, there will be several important results derived from this one. Just like we derived several things by just looking at what happens to a lift of a single point, single path. So, this is what I am telling you. Apart from application within covering space theory, which we will study in the next section, this result has many applications elsewhere also. Here is just a sample classically in complex analysis, various versions of it go under the name monodrame theorem. And that is the application I am going to give. Okay. So, the monodrame theorem you may read may be, may be, may look like different. But if you understand that one correctly, what I am saying is and what I, what we said in the elsewhere, wherever you read, they will be the same principle. Okay. There are different versions of this one. That is what I wanted to tell you. Okay. I cannot cover all of them. So, I will state thing which is immediately follows from what we have seen. Take Y to be a locally path connected and simply connected space. Okay. When you take open subsets which are disks or some such thing or simply connected space in complex analysis, this will be automatically satisfied because open subsets of RNA are locally path connected. Take P from X bar to X, a covering projection. This is some, any covering projection. Then every map F from Y to X, okay, has a lift F hat to F hat from Y to X bar. Why? Because it is simply connected. Fundament group is trivial. The F check of that is trivial. So, trivial group is contained inside every subgroup. So, it is contained inside P check of this also namely K. So, whatever criteria we need is trivially satisfied. Therefore, every map can be lifted to the covering. In particular, look at a map from Y to S1. It can be lifted to exponential function or through exponential function to R. That is the meaning of this because Y is simply connected. Okay. It can be lifted to the exponential function R. In R, R is contractible. So, any map into R is what? It is null homotopy. Take null homotopy, compose it with exponential function. You get null homotopy of the map here. So, every map from a simply connected space to A to the circle is null homotopy. Of course, I am assuming locally path connected, connected and so on. Okay. So, this is what it is. The first part is obvious because Y simply connected. Zero, the trivial group is contained inside every subgroup. So, criterion is satisfied. Second part, you lift it to through exponential function, lift the map to R. R is contractible. So, that is null homotopy. Push it back to S1 by taking the exponential function. Okay. So, in the remaining time, I will do a little bit of new constructions, how to construct non-trivial coverings. If you take identity map, it is a covering. Exponential map is a covering. But you may not have seen many examples of covering projections. I have given you only some group action and so on. There are other ways of construct, start the space and constructing coverings. Not the other way around. In other way around group action and you quotient, that is covering projection is a different way. Okay. Let us construct some non-trivial covering maps for next part. Starting with X, non-trivial covering means every loop if you can be lifted, then it will not be a non-trivial covering. Some loop should not be, you should not be able to lift at least at different points. That means you have to break certain loops in the space X. So, that is, you know, heuristically saying, obviously these loops must be representing non-trivial elements in the fundamental group. For instance, let us do very simple thing, namely take the space S2, the two sphere along with one of the diameters. Namely, let us say P is 1 0 0, take minus P minus 1 0 0 and join them. So, that is a diameter. So, now your space is a two sphere and a diameter. In the center, you have a handle. It is like that. Okay. If you throw away 0 0 of R3, which is the center of this diameter, what do you get? You get S2 and two open arcs, half open arcs. Those arcs can be deformed back to S2, which just means that this X minus 0 deforms back to S2. S2 is a strong deformation tract of X minus S2. Okay. But we know the X, S2 is simply connected. S2 is simply connected. Therefore, X minus 0 must be also simply connected because these two have the same amount of time. One point if you remove, it is simply connected. All right. Now, your theorem says that if you take the inclusion map, shrink of this as Y, X minus 0 as Y, included in X. X minus 0 is simply connected, connected, locally path connected also. Therefore, the inclusion map, you must be able to lift it up. You choose a point somewhere, say 1 0 0. Above that, there will be many points because it is a covering projection. At each point, you will have a copy of X cross X minus 0 sitting inside X bar. Okay. So this is the picture you have. Then the missing points will be also there. A number of missing points will be there. If you fill them, they fill them up, you get X bar, full of X bar. Right? So how is this to be visualized? So I will give you a simple example of this one. This much is logic. Now, you need some visualization. So at least two copies of X minus 0, X minus 0. And then you fill up the 0s somehow. And then you must get a one single connected space. You see, so that X bar to X is a covering projection. So this is what you have achieved. All right? So it is not very difficult. Okay. So all this I have done. So missing point, you should also be careful. Namely, just filling point is not right. The neighborhood of this point is an arc that is also simply connected. So all these points, missing points will have neighborhoods equal to an arc, an open arc. Okay. So what you have to do is, you have to take care of X minus 0, several copies of that, equal number of copies of intervals. Then glue them together neatly. Okay. When you glue them together, only the extra point in that neighborhood must be at the center point 0 of the arc. So that is what you have to do. Okay. So arc I have taken minus p by 2 to plus p by 2, which is an open subset of of the whole thing. And it is neighborhood of 0. All right? So I will show you the picture here now. So this is the picture. This is the starting S2 with a diameter. So here what I have done? I have taken several copies of this one. This is one copy. The inside portion I have drawn outside, that is all. Okay. I have joined it to next one. Another one we will join it to next one. So this is an infinite cyclic covering, infinitely many copies. Okay. Just like in the case of infinite covering R to S1, exactly similar to that. Instead of these spheres here, if you put a single point here, what you say, the single point is the integer and these are intervals of length 1, what you get is R here. So I could have said, take the exponential function, each integer you replace it by two sphere. And in the bottom what you have to do? Take the circle, look at 1, 0, replace it by a small sphere. So that is the picture. Okay. So that is an infinite covering projection. I can do just a double covering projection. Take two copies of this with two viscars here, two viscars here. Then join them together like this, neatly. So this sphere, this sphere, both of them go to the same sphere here by a homeomorphism. In that homeomorphism, this part goes precisely to the central line here, the diameter. Similarly, this will be also going to central line. Now this is three of them. Okay. This is a triple covering. You can do triple covering. You can do a four fold covering. You can do n fold covering. Do you know why this is so simple? The fundamental group of this picture is nothing but infinite cycle. Covering space is pretend as if they are coverings of S1. Only one of the point is replaced by the big sphere there. That is exactly what is happening. Okay. Let me do one more construction which I had promised last time. Which we did not have time to do. Let me do that one also now. Obtaining a, yeah, this is the one. Obtaining a covering which is not a regular covering that will show that the fundamental group is not alien. Right? So this was the topic last time. So I start with S1 veg S1, this figure 8. The common point is A here. I oriented this one and labeled it as X, oriented the other sphere labeled it as Y. Okay. So how is the covering got? This A is copied here. This one circle is copied here, X. Okay. Another copy here, Y. But I have taken a arc here completely Y, open arc. So this open arc is mapped to the single circle here, namely this A3, A2, A1 are all mapped to the same point A. This arc goes to Y and the bottom arc here also goes to Y. But these two arcs go to X. There are three arcs and sorry, two arcs and a circle mapping onto this circle. Similarly, two arcs and a circle mapping onto this circle. So that is the covering protection. Okay. Now by the very nature, this is a loop at a X, denoted by X. It has lifted to a loop at A3. But at A1, it lifts to a open arc. It is not a loop. Okay. There is another loop here, A2, it goes back to A1. That is also an open arc. At A2, if you lift it, it comes to A1. At A1, it comes back to X, A2. But A3 comes back to A3. So this is a loop whereas these two are not loops. That means that this element is not in the normal subgroup of the image. The image is not a normal subgroup actually. Okay. So the image of this one is actually a subgroup of order 3. It is, I could not construct anything of order 2 because order index 2 subgroups are normal automatically. So I need at least three points in the covering projection, three-sheeted covering. Okay. Is that clear that this is a covering projection which is not a normal covering? Why? Because some loops, a particular loop lifts to a loop as well as to open arcs here. Okay. Thanks. We will continue this kind of discussion next time.