 Today's topic is lifting properties of covering projections. Last time we introduced the concept of covering projection. Check for some basic properties like a covering projection is a local homeomorphism. Therefore, its fibers are all discreet and it shares all the local properties of the base space and the top space are the same. That is what we have seen. Before proceeding with lifting properties, let me take up a few more properties and a bit of example and so on, then come back and go ahead with the lifting properties. So, we had also established that local path connectivity plays a very important role in the study of covering spaces. So, often all authors blanketly assume that the base space is locally path connected to do anything. But just to keep the theory on, we can assume, we don't have to assume that. If you want to do some serious things, we may have to assume that. So, the key was this theorem namely, when you have locally path connected, then you can restrict the covering space to each path component of the top space and yet it will be a covering projection. So, you can study each of them separately and then put them put whatever observation you have made together. So, this is why this theorem is very important. So, this is what we have proved last time. Now, so this is what I have told you. We shall assume that both base and total space as a covering projection are path connected and connected unless specified otherwise or it is clear from the context. When you are discussing some examples and that example itself is not locally path connected, then it is clear that it is not locally path that is all. If we let something and then we are assuming this thing, this you have to make it. This is what I am saying. Now, for example, this covering projection word is used by different people in a different way, slightly different way. So, that is what one has to be careful especially in algebraic geometry. A covering projection is need not be the covering projection the way we understand it, but it is not too far. So, I will give an example. Look at the map C to C, z going to z power n. We have seen that z equal to 0 is a difficulty point. If you throw away that point, then this map C star to C star is a covering projection. On the point z, there is only one point. On every other point, there are finite many points. So, this kind of situation is acceptable to algebraic geometry. So, he allows a number of points, some subset, some curve, some variety on which the given map may not be covering projection. If you throw away that, then the rest of this will be covering projection on a large open set. Such things are called ramified coverings actually because the point z equal to 0 is a ramified point here, ramification point. So, they talk about ramification points and so on, but their covering itself may have ramification. So, they call it as covering projection. When they want to talk about the covering projection in our sense, then they say un ramified covering, there is no ramification. So, the more expressive thing un ramified covering is used for covering projection because covering projection for them is a larger thing which is more often used, that is the point. So, this un ramified covering projection is the word they use for the actual covering position that we use. So, here are a few exercises which are straightforward. You can try them, but I will not use them till you have mastered them. So, take a covering projection. If X is Hausdorff, then X bar will be Hausdorff. It is the other way around here. You know that P from X bar to X is a covering projection, then it is a quotient map. So, under quotient maps, if the top space X bar is Hausdorff, it does not imply that X is Hausdorff. There is a case here also, but suppose you have a finite to one covering projection, then X bar Hausdorff implies X is Hausdorff. If it is an infinite covering, then this is not quite true. So, we will have an example for this one, but if finite, then you should show that X bar is Hausdorff implies X is Hausdorff. This is straightforward point set topological conclusion. So, this example tells you it is R2 minus 00. We have seen this example and there is an action of the infinite cyclic group. The generator takes X comma Y to 1 by 2 comma twice Y. So, more generally the powers of generator will take X comma Y to 1 by 2 power n X comma 2 power n Y. So, these are all equivalence classes now. The quotient map is actually covering projection. So, this happens because we have already introduced the methodology that action could be even action. So, this action happens to be even action. So, this is a covering projection. Only thing is just like in the case of C star to C star that one is at 4 and here we have to throw away that 00. So, 0 is a single class. So, that creates problem. However, after you throw away 00, it is a covering projection all right and this is Hausdorff is also all right because it is a subspace of the Euclidean space. But look at the class of 1 comma 0 and 0 comma 1 in the quotient space X. These things, these two points cannot be separated by disjoint open subsets of X. I think we have seen this one if you have not seen yet you should check this one again. So, it is not Hausdorff all right. So, this previous example is in any case T1 space. So, if it is a T1 space, if it is regular it will be Hausdorff also. So, it is not regular neither normal because normal plus T1 will imply T2. So, it is neither regular nor normal all these things are true in R2 minus 0 being a metric space. But the quotient but this the bottom space fails to be satisfy all these separation properties. And here is an example this also you can do yourself. But when I want to use it I will approve this one. So, take it covering projection take any continuous map. If you have S from X to X bar which is a right inverse to P that means P composite S is identity then you call as a section of P. If we have just this one people also call it section but then it will not be continuous I want continuity because I am doing the collage that is all. So, a section is a continuous right inverse it may not be left inverse all right. So, that is otherwise then it would be then P would be a homomorphism. P is a surjective map. So, this is a left inverse. So, there are lots of synthetic left inverses continuous inverses are difficult to combine. Now, suppose you have a continuous left right inverse that is called a section. Let now X bar be connected X be locally connected or locally path connected. So, I am specifically saying these things here because this for this exercise is important that is all. Now, suppose that P is a local homomorphism and X bar is horsed off. So, I am not assuming it is a covering projection. The second part B I am assuming P is a covering projection no horsed offness. Either homomorphism and horsed offness of X bar or just P is a covering projection. Then any section of P like this S here is automatically a homomorphism on to X bar. It just means that it is now left inverse also it is a full inverse. So, P becomes a homomorphism. S is a homomorphism P is a homomorphism. The hint is you show that S x is both open and closed because X bar is connected it must be the whole of X, whole of X bar. All right. Now, let us carry on with lifting properties. Homotopy lifting property if you recall assess that certain maps exist. What are they? What are they? Namely if one map can be lifted out of a homotopy then the whole homotopy can be lifted. That existence of lift means lift means P composite whatever a new map you get is the old map. So, that is what we have we have seen that homotopy lifting property okay and just recalling this. In mathematics there is in a uniqueness whenever your uniqueness result which goes hand in hand with existence result quite often like first order differential equations solutions in a small neighborhood that exist and in a smaller neighborhood it is unique. Such things are always nicer and more much more applicable than just existence theorem okay. Indeed the uniqueness part actually solves actually helps to solve the existence part proving the existence part okay. So, this is the case here also. All right. So, we shall first have this uniqueness result then we will use this uniqueness result to prove the existence of these lifts homotopy liftings okay. So, recall that if f from y to x map and g from y to x bar is a lift of through P means what P composite g is f that is the meaning of lifts. Now, we are always concentrating upon map x bar to x and that is denoted by P okay. So, here is a need theorem take a covering projection now and a connected space y. Take a function f from y to x. Supposed by chance we have two lifts g1 and g2 okay of y to x bar are two lifts of this f and they agree at one point y belong to y g1 y equal to g2 y at one point they coincide then g1 is equal to g2. So, this is exactly similar to the first order differential equation if you specify the initial condition then it is unique. This is like initial criteria at one point they agree then the whole thing agree is just like integration theory okay integrals are are defined up to a constant additive constant right. There is no addition multiplication here at one point they agree continuity is there that is all they already agree. So, the key is that y is connected that is all y is connected of course P has to be a covering projection otherwise for arbitrary spaces this will not work all right arbitrary functions it will not work. So, let us let us do this is proof which is very straightforward look at all points in y at which g1 and g2 coincide that is subset of y it is given that this subset is non-empty there is at least one point right. So, if we show that z is open and closed in y then because of the connectivity y it follows that z is the entire y z is equal to y means g1 will be equal to g2 at all the points of y therefore what we want to prove now is that z is both open and closed. So, take a point y insert that we be an evenly covered open neighborhood of f y in x y is insert f y will be in x and x is covered all points of x x whole thing is covered by evenly covered by P that is why P is a covering projection. So, take a neighborhood which is evenly covered and this neighborhood is neighborhood of f y let u be an open set of x bar mapped homomorphically on to v because v is evenly covered P inverse of v is a disjoint union of open sets each of them comes to v by a homomorphism namely under a pre restriction of P. So, you choose one of them let u be an open subset of x bar mapped homomorphically on to v by P look at g1 of y which is equal to g2 of y it must be in one of these open sets and choose that one to which it belongs to. Now, choose w an open subset of y in y such that g1 and g2 of this w goes inside u because one point is going inside then by continuity this is open set some neighborhood of that point must go there of course I will make it different neighborhoods for g1 and g2 but then I can take the intersection of those two w1 w2 g1 w1 contains a u2 g2 w2 contains a u2 contains a u by continuity of g1 and g2 but now you take w equal to w1 intersection w2 you will get this one. Then look at P g1 of any z equal to fz equal to P g2 of z for every z inside w because P g1 is equal to f equal to P g2 therefore P restricts P sorry P restricts u is an injective mapping it is a homeomorphism P of this one equal to P of these two therefore g1 must be equal to zero to for every z inside w by definition this w is contained inside z by definition w by choice w is a open subset around the single term point whatever point you have taken y so what we have proved is z is open okay if x bar were house door then the set of points where in two functions continuous functions agrees automatically a closed set but we do not want to do that we can do without the assumption that x bar is whole door okay for proving that that the set z is closed take a point that invites a set g1 of z is not equal to z we have to show that this is also open see this is the complement of Z so I want to show that complement is also open now so to show that what do I do let vpn evenly covered open neighborhood of P composite g1 it is equal to P composite g2 g g1 z is not equal to g2 of z but P g1 is equal to P g2 right that is f this equal to f is always the case therefore what is happening is this g1 and g2 g1 z and g2 z they are the same fiber call this as x P inverse of x they belong to okay so we can find open neighborhood ui around this point and this point okay on which P is a homeomorphism and such that intersection is empty because they are in different fibers all that I have to do is I take this point g1 z P of g1 z x and then take a evenly covered neighborhood of that look at the inverse image g1 will be in one of them g1 z and g2 z will be in another one of them two of them are not they are coming to the same point and they cannot be the same open set above of course you get evenly on u2 which are not which are disjoint okay by continuity now g1 and g2 are continuous function you will find a neighborhood w of z such that gi of w is in ui same thing again you can take first different than the intersect okay oh sorry I should not use this cover all right it follows that w is an open neighborhood of z not intersecting that at all because ui are disjoint g1 of any point here will not be equal to g2 of the same point because they will be going to disjoint open half sets here so the entire w is in the complement of z therefore the complement of z is open hence that is closed okay so covering projection is the only thing which is used critically here okay the next step is to prove homotopy lifting property of covering projections for singleton spaces for a singleton space what is the meaning of homotopy we have seen it is a path so this is called path lifting property okay path lifting property for for our covering projections first we prove that one first point wise we will prove then we shall prove it for space wise take a covering projection okay given a path from i to x that is omega and a point x bar in x upper bar x capital x bar such that p of x bar comms to omega 0 this is the starting point it is like initial value condition okay take a point above always points can be lifted right because p is a surjective value okay the starting point you lift it up in x bar then there exists a path say omega twiddle from i to x bar such that p composite omega bar is omega which is lifted the starting point of this omega bar is just with the point which you have chosen x bar okay we have already proved that such a thing will be unique if there is another path omega 1 with the same properties then agreeing at one point then omega 1 equal to omega at all points but we don't know whether this exists but the uniqueness of this will help to prove this one now okay and it is not difficult let's go through this one similar proof like this one we have used several times defines that to be all points inside i okay such that omega bar is defined up to 0 to t omega bar of 0 it has to be this x bar that we have chosen then it is defined up to t means what there is a function 0 t to omega x bar such that p composite that is equal to omega observe that z is a sub interval because why if some point belongs to that 0 to t will belong to that okay so interval and contains 0 0 is already there because x bar is there okay 0 to 0 is just closed interval is is omega is defined already okay now take t0 with the least upper bound of z this exists because after all this z is a subset of the inter closed interval 0 1 and so it is a it is a bounded above let take this one's least upper bound or what is called as the supremum I want to show that t0 itself is inside first thing second thing t0 is actually equal to 1 essentially this is similar to showing that 0 and using that 0 and is is actually an interval namely it is connected but we will hide it and make it easier just use the super existence of supremum and then you have that supremum has to be inside z which is equivalent to showing that z is closed okay and t0 has to be whole of 1 equivalent to showing that z is open okay start with an evenly covered neighborhood of omega t0 up to t0 the map is defined okay up to t0 at t0 t0 is a supremum if you take anything a little smaller than t0 the function is defined now I want to do it at t0 so take a evenly covered neighborhood of omega t0 okay for 0 less than epsilon less than 1 you put this notation namely I am only interested in actually no open interval but I do not know one part may be closed and so on so I will take a close interval no problem epsilon is positive that is good enough 0 less than 1 I epsilon is t0 minus epsilon comma t0 plus epsilon this interval may go out of I right it may be bigger bigger than I and some part so I do not want that so intersect it okay now choose epsilon so that omega this is notation now choose epsilon so that this omega of this I epsilon it is a neighborhood of t0 okay that is contained inside of v v is an open subset containing omega omega t0 omega t0 so by continuity of omega some neighborhood will be contained inside v yes or this is this condition is just omega is continuous now let ui be an open neighborhood of omega bar of t0 minus epsilon by 2 take t0 minus epsilon by 2 just t0 to the left of that that point of omega bar is already defined because t0 is supremum so omega bar makes sense there okay so take this one to be an open neighborhood of this one that is mapped homeomorphically on to v okay because v is evenly covered this must belong to somewhere inverse image of v take that complete that component of this that is my ui here okay ui contains this point put lambda equal to now p inverse of omega okay that is lift off is a lift off this becomes a lift off omega on I epsilon see this is the homeomorphism p restricted to ui to v is homeomorphism so I can take p inverse there okay p inverse is taken on this subset then composite with omega omega composite p inverse so this becomes a continuous function if you take p of this it will be omega therefore this is a lift off omega inside the whole of I epsilon because this whole of I epsilon is contained inside v omega of that one is contained f for omega of I epsilon p inverse makes sense now look at lambda of t0 minus epsilon by 2 okay which is by definition omega of t0 minus epsilon by 2 okay this is already there up to there we have the path already therefore by the uniqueness theorem this lambda t must be put omega bar of t for every t inside this one this omega bar is already defined now I have got lambda they both of them agree at one point so in this neighbor in this whole interval which they connected subset will be it will be what it will be unique so lambda t must be put omega bar therefore omega bar can be extended from all the way from 0 to t0 plus epsilon this must be plus epsilon by 2 that is the type of here t0 plus epsilon by 2 okay so therefore two lifts can be patched up that means omega bar can be lifted up to 0 to t0 plus epsilon by intersection I if this epsilon is still less than one then what you have shown is some point t0 plus epsilon is already inside z but t0 was supremum therefore this t0 plus this epsilon intersection this one is just to 0 to t0 so this just means that t0 is equal to one and t0 belongs to that simultaneously t0 less than one will contradict will contradict the definition of t0 that it is supremum so t0 must be one and once it is there omega is defined up to there so it is over oh let us stop here and the true lifting property will be done next time thank you