 Today's topic is homotopy lifting property. So the big theorem here is that every covering projection is a vibration. The proof is not at all new in the sense that the fundamental idea involved that we have already seen while studying the computation of the fundamental group of the circle. The exponential map had the property that path is can be lifted and homotopy of path is can also be lifted. That is homotopy path lifting property that we have proved. Whatever idea is there, it is the same idea will work here. The only thing is we are instead of the exponential map explicitly, we are taking any covering projection. And the second thing is instead of homotopy of path is now we have homotopy of functions from arbitrary spaces. So you will see that you have to work a little harder that is all. Start with the covering projection x bar to x, y is any topological space and h is a homotopy on y into x. g is a function into x bar such that when you take p composite g and come back to x, it is h of y0 just starting map of the homotopy. This is the homotopy lifting data if you recall. If you have homotopy lifting property, what we have to prove is that you must find a function g from y cross i to x bar continuous function such that p composite g is h and this g of y0 must be this little g. That is what we have to find. But path lifting property we have already proved not homotopy lifting property. Now you fix one point here at a time, y cross i is a path that can be lifted starting point what do you take? You take g of y, it is given to you already. g of y such that p composite g of y h of y0, y restricted to h restricted to little y cross i is a path. So that can be lifted all already. Do it for all the y. What you get is a function g with this property but this function may not be continuous as a function from y cross i to x, x bar. It is continuous whenever you restrict the first coordinate y to a single point. So that much you have already got and once you by this requirement as a function it is unique. There is no choice but we have to prove that this very g whatever we have got is actually continuous function from y cross i to x bar. If this is not true then the theorem will be false that is all. So we have to prove this one. There is no other choice we cannot there exist g but this g has to be this one. So this is the point of path lifting property. The homotopy of each point here is nothing but a path so that is what we have to do. So what we have to do is g is continuous as a function from y cross i to x. This is all we have to prove. Given any point in y we shall first construct an open neighborhood w y of y such that okay. Idea is suppose you prove a neighborhood open y such that h restricted to sorry g restricted to w y cross i is continuous. Then you are done because continuities after a local property for each neighborhood exist true then you are done alright. So I am finding a neighborhood w y of y such that there is a partition t0, t1, tn equal to 1 okay partition of the interval with the following property that w y cross ti to ti plus 1 h of this okay not g. h is a continuous function h of this one is contained in an evenly covered open subset of x okay. So this is what I want to find out. So how do we do that? First of all choose an evenly covered open subset v little y comma t for each point h of y y comma t by continuity of h for every t inside i okay. You can choose a neighborhood subset v i t which is evenly covered wherever it goes to and use the compactness of y cross i to find finitely many of these v i t i is to cover h of y cross i. Y cross i is compact h of y cross i is compact to choose evenly covered neighborhoods around these points inside x just that finitely many of them cover the entire path h of y cross i everything is inside x now okay. Once you have done that choose a partition there are only finitely many open subsets so this is a standard argument with a vague number for the interval. So the interval can be divided into finitely many sub intervals such that consecutive intervals y of y cross ti ti plus 1 h of that will go into one of the open sets okay. So this is similar to what we have done with the path lifting property also why even to lift path lifting we have done this one again using compactness of i okay i is compact now okay y a neighborhood of i inside x cross i you get inverse neighborhood you get a uniform neighborhood that is the alastera you can find an open neighborhood wi comma i okay of y such that this h of wi i cross ti ti plus 1 is contained v i for a way. Now take wi is equal to intersection of i into 1 to n wi i so you have got a uniform wi this is alastera okay such that partition okay and the partition of i as required such that one single wi cross ti to ti plus 1 h of that will be contained inside an even become a neighborhood that was our first m here the one single neighborhood of y such that these intervals keep changing consecutively all of them are contained inside okay from first we had various intervals here next we may make them finitely many then you took the intersection of this finitely many to get into one single neighborhood of the point y okay now the hard work is over so here i am again quoting the same picture that you had while pathlifting property for the exponential function this is your interval and this is your x here that is all x was also interval so here i am taking now you have to think of this as x okay so x is now covered by opens of sorry y y is covered by opens of sets wi instead of rectangles like this and so on but here on along the y axis you have you have subdivisions along the the i in the interval you have subdivisions here okay so this is what i have explained already so now put w equal to wi okay we shall see that if g restricted to w cross ti successive application of that okay we will produce g restricted to w to w ti plus 1 is a continuous for all i what i am trying to say start with on this one you have your function continuous okay therefore you get a continuous function along this one because this whole thing is contained inside a single even neighborhood so the inverse image makes sense because it is even the covered by the uniqueness whatever g you have what already there is no question this must be the same thing as that so continuity on on these parts like this one by one follows this on each of them it is given by a inverse image now you have continuity along say this is g cross on ti instead g cross is a y cross 0 to y cross ti use that again along these blocks to go to the next stage so ultimately in finite stage you will come for the whole thing is continuous okay continuity on each of them these are closed subsets it is of y cross i half cross actually open subsets so finitely we open subsets actually all along this way but this way it did not need not finitely many because there are wi's which cover the whole of so on each open subset it is continuous so it is continued okay so this part is same as in the case of exponential function and path of multiplicity so let us go to some little bit of exercise and so on what we are going to do in subsequent sections is this theorem we have applied again and again okay and then we will then we will produce lot of interesting results the fundamental result is that any from any space y a homotopy one of the mouth can be lifted whole entire homotopy can be lifted the homotopy extension prop homotopy lifting property of the covering projection okay this should be used again and again so let us look at this exercise take a covering projection over the space x cross i okay where x is locally path connected etc you have to assume and you can assume path connectivity also this is if you look restrict p inverse of x cross t at each level okay take p inverse of x cross t to x cross t okay this restriction map is a covering projection for each t is what you have to show remember if you take arbitrary subsets the inverse image okay full inverse image then that will be a covering projection okay so this is not a very difficult thing to prove the point of this one is all these covering projections will be in some sense later on will be same because x cross t is the same now x cross t or x cross s they are homomorphic features they are copies of x the above spaces will be also homomorphic features that in such a way that the projection map it commutes with that homomorphism so that will come later right now you have to just say this much okay this is the second part for each t comma s belong to i show that there are homomorphisms depending upon t and s theta t s p inverse of x cross t to p inverse x cross s this is a homomorphism this is identity map identity map on x comma t goes to x comma s this map is x comma t identity cross identity whatever t goes to s that is the difference here that is all so such a diagram is possible this is what you have to show looks crazy but you do not need anything more than what you have learned do not even need homotopy lifting property just the definition of our integrations okay okay so let us stop here thank you