 So, carrying on with the proof of cellular approximation theorem, we have proved it for the case when both domain and co-domain have just one cell. So, I will recall that lemma and then immediately extend it to one step ahead. So, let us see what was the lemma? The lemma was that you have a map from Dn Sn minus, which is like a you know relative Cw complex with one cell, right. To Yb which is also relative Cw complex with one cell, dimension of this m is bigger than n, that is the crucial thing. Then this any map like this can be homotopped to a map which is completely another last patch of X1 is inside B, okay. It can be pushed inside B. So, this is the lemma. Now we want to do the same thing with Yb it will be generalized. The domain we keep the same. So, this is the first corollary here. That Yb be any Cw complex. That means it is a relative Cw complex. Then for every map alpha, Dn Sn minus 1 to Y, Yn minus 1 skeleton I am taking here, okay. You see Dn Sn minus 1, Y there is a homotopy H from Dn cross I to Y such that H of X0 is alpha X to begin with. H of Xt will be alpha X always on the boundary and H of X1 will be in the nth skeleton, Yn. Y is any Cw complex. If it is already lower dimension than n, there is nothing to prove. So, the point is I am implicitly assuming that there are cells of higher dimension, okay. And moreover, if already alpha is going inside Yn, there is nothing to prove, okay. So, it is implicitly I have to take the case where in the original map alpha is mapped on to some higher dimension cells here. May be lower also part of it is higher dimension so on. But this time there may be several cells here, okay. So, how to handle this case? The very first thing to notice is that alpha of Dn being compact set, it will meet only finitely many open cells of Y. So, this is what we have seen in every compact subset with the union of only finitely many open cells, okay. Amongst all these cells because finitely many, there will be some largest dimension one. And that largest dimension is less than or equal to n, there is nothing to prove. So, that largest dimension m, let us say assume that is bigger than n. And let us take one of those cells, there may be finitely many of them, maybe 5000 or whatever, but take one of them, fix one of them, okay. So, let em denote one of the cells of highest dimension. That means what I have assumed that alpha of Dn intersects this em in the interior of em, otherwise I am not interested in that, okay. Now what I do, I will look at this just local em and the rest of em is attached to something, okay. And the rest of the lower dimensional Y, I am not interested in the higher dimension skeleton of Y. So, I am looking at em skeleton of Y, Ym, Y top em. Amongst that, I am just looking at just one cell. All the other cells are disjoint from it, okay. And right now I am concentrating on this m cell, okay. So, look at this situation, okay. Replace the old B in the lemma by Ym minus 1 and Y by just B union this em. Just look at that. So, that is the picture that we have, then the lemma gives a homotopy of alpha to a map alpha 1 relative to Sn minus 1, so that alpha of Dm can be pushed into, not exactly Ym minus 1. What I am taking is, there are other cells, okay. One of them, one at a time, okay. So, that is I am calling it Ym minus. In finitely many steps, you will actually get rid of all the top dimension things. Now, you have been inside Ym minus 1. Now, again do the same thing for, if M minus 1 is bigger than M, do the same thing for these cells, there will be finitely many. So, in finitely many steps repeated, what you will get is finally the map will be inside Ym, okay. So, that is the corollary. Map will be inside Ym minus, it is homotopic 2. Each time the homotopy is constant on the boundary of Sn minus 1, okay. So, one after another you keep homotoping. So, I am pushing it lower and lower till you hit only Yn, the n-scalator, okay. So, Dn was dimension n. Finally, the homotopy is inside the Yn n-scalator. So, that is a cellular map, okay. So, the cellular approximation theorem, when the domain has only one cell is completed, okay. Now, you can do it for finitely many, etcetera, that is not a big trick. But we have to do it for the entire thing finally, in finite case also. The infinite case in the domain is not so easy to get rid of. Whereas, in the co-domain we got it very easily. It will never come. Each time you have a compact thing. So, the image will be contained in a finitely many things. So, we are working in a finite case, okay. So, domain you have to be more careful. But for this also we have a readymade thing. When we proved local contractability of CW complexes, we showed how to compose, how to compose infinitely many homotopies, okay. So, that we are going to use now. In between, we have to use one more technical result also, okay. Now, let us complete the Puru Sava theorem. Inductively, we shall construct a family h n from x cross i to y b such that starting point h naught, h 1, h 2 and so on. h naught of x 0 is f x, okay. h n of x t is always f x for x in x prime. Remember, x prime was what? A sub complex on which f is already cellular. So, we do not want disturb this one. So, this I may not say every again and again, but this is to be assumed all the time. We are not, we are not going to disturb this part at all. We are going to disturb, we are going to change the map by homotopy only on cells which are not in x prime, okay. So, it is a h naught, how do you are going to do etc., you see. Next thing is h n of x 0 will be h n minus 1 of x 1, okay. This is where it is going to help us by looking at like path compositions. So, this composition is not, it is concatenation. It is not a composition of maps. It is a composition of paths. So, starting point of h n at 0 is the end point of h n minus 1. So, this is needed for composition, okay. Let us have this also, namely this is needed for passing to infinity, namely h n of x t is always h n minus 1 of x t for all x in x n minus 1. It is actually h n of x 1, it is constant map. It is just some, the map like f 1, f 2, f 3 and so on. I do not want to have another notation here. The deformed thing of original f, so that will not be disturbed on h n minus 1 when constructing h n, that is the meaning of this one. So, these h n's are extensions of, natural extensions of h n minus 1, okay. Finally, the end point of h n is a cellular on x n, okay. It means the is skeleton will go inside is skeleton only for points of x n. Beyond that, we have not yet done, okay. So, this is the inductive step we want to construct. Having constructed h naught, we must be able to construct h 1, then h 2 and so on. This is just a proposition. We have to do this one. We have to construct this, okay, all right. Once we have done that, let us see why we have done all this. Once we have done that, it is immediate consequence of the old thing, namely composing various h n's in one single shot, namely h of x t we are going to take like this. Remember, this is what we did, but in there I did it towards 0, instead now I am doing it towards 1. That is the only difference. Deliberately I have done it, so that you will know both of them. So, between n minus 1 by n and n divided by n plus 1, okay, you use h n minus 1, appropriately parameterized. So, you are composing h n minus 1, then h n minus h n and so on, okay. Finally, h n if it is not in this for every t between 0 and open 1, it will lie here, okay. But t equal to 1 itself is not here. For t equal to 1, what you do? Take h n of x t equal to h n of x, h of x t equal to h n of x t, which is already defined. What is this h n? That will depend upon where x is, x is inside x, okay, x is some inside n, so you can take it. So, this is what is going to help us because suppose x is already in x n minus 1, then this will coincide with h n of h n minus 1 of x 0 or x 4 t or whatever earlier map. So, this map, this is well defined, okay. Suppose we are already h n minus 2 and so on, your choice of n is not quite unique here, but the definition does not depend upon that. So, that is the consequence of this, the third hypothesis here, okay. The second hypothesis allows us to define like this, you know, take a t, there is a unique n, use that n to define this one. When you have the n minus 1 by n will coincide with the previous one, so the two definitions will coincide, h n minus 1 of x 0 will be equal to h n minus 2 of x 1 and so on. So, they will coincide, there is no problem, that is because of 2 here. And all the time h n of x t, if x is inside x prime, it is always f x, okay. So, each of them have that property, so this property will be here also, okay. So, we have to do this job, the 1, 2, 3, 4 we have to do. So, this we will do one by one. So, first we have to do h naught, so how do we do the h naught, okay. Anything which I have inside x y, x prime, I am not going to touch. Therefore, take over at x, the 0 cell which is not in x minus 1, this x minus 1 is just x prime in the, remember x, x prime if you take, you think of x prime as the relative part here, so x minus 1 you can write for that, okay. Then you have attached 0 cell, h h 1 cell and so on. To choose a path omega x from f x to some point in y naught, the entire y Cw complex okay has to be built up on the 0 cells, okay or the relative part. In y naught we are including the relative part of them. Therefore, every point has to be joined to one of the points of y naught, okay. This is an elementary result for Cw complexes. The number of connected components of any Cw complex, okay in each of them there will be always one 0 cell or a part of the relative part, relative cell has to be there in each of the common connector components, path connected components. So, you can join this f x to y naught, that path itself is a homotopy for each point. Put them all together because this x naught minus x 1 is a discrete set, so just put them together it will be a homotopy. So, h naught, little h naught of x t I have defined f x for x in x prime no disturbance for all x which is not in x prime, so x naught is omega this path. So, this is a homotopy, this is a continuous solution, okay. And what is it the last value t equal to 1, t equal to 1, this will be a point of y naught. So, it is cellular map x naught goes to y naught rest of x prime always already inside y prime whatever is cellular there I do not care about that. So, since now comes the point this is only defined on x naught is x naught cross i, but I want this h i's capital h i's here all of them defined on the whole of x cross i. So, how do I do that? Here I use the property that every sub complex of a CW complex is the inclusion map is a co-fibration it has a homotopy extension property, okay. Just for your benefit I will recall it from part one a homotopy extension property was this kind of a diagram start with a continuous at x these are subspaces a is a subspace of some space the inclusion map is there eta this is called a co-fibration if whenever you have a map like this one g from x to y and restricted to a there is a homotopy of that a cross i to y this diagram is coming to this means g I have taken then I am restricted g to this part a that part the restricted part is homotopy to there is a homotopy itself take the homotopy itself by this data a cross i is eta cross identity x cross 0 is also inclusion x cross 0 goes to x cross 0 okay this part there is nothing for all suppose this much is given then there is a map capital H here which makes this entire diagram committed that means restricted to x cross 0 this capital H with a homotopy is the homotopy of this g okay and on the restriction on the subspace there is a homotopy that homotopy this homotopy capital S extends this f the restriction this restriction is f that means H is an extension of that so this is called homotopy extension this is homotopy on a subspace is an extension of this homotopy of the extended map here this map itself is defined on the whole of g okay so that is called homotopy extension property if a is a sub complex of x this will be always true this is what we have proved okay we have proved it just two days back here by using another tool that we are done in part one namely what was that picture this picture x cross i deforms to retracts to x cross 0 union a cross i if a to x is a cofibration okay using this trick we have proved that so I am going to use that property started with a I have constructed h little h0 on x0 cross i now I extend it to the whole of x cross i okay I mean I extend means there is an extension that is a theorem I am using that so I get a so yeah so I get a this little h0 just extended to capital H0 having all these properties this property one here there is nothing more than that now you assume that you have done with hn minus 1 so hn minus 1 you extend it to hn exactly same way how do you do that first you will get little hn by our corollary how do you do that one by one you get the extensions okay so the whole of nth skeleton you can define that map we can patch it up because on the n minus 1 skeleton wherever you are attaching they agree therefore you get a whole map from x x n cross i to y okay so I will repeat that part inductively suppose we have defined hn minus 1 with the property as specified specified means 1, 2, 3, 4 for each n cell enj okay in xn or ni or inj out xn minus x prime x prime you do not touch so it is n cell in xn but not in x prime okay with the characteristic map phi i from dn sn minus 1 to h1 just put alpha equal to whatever hn minus 1 we have composite with phi okay hn minus 1 itself is defined on the whole of x remember that each h i is defined in the whole of x so this makes sense okay and now use that corollary to get a homotopy hni on the cell en cross i to y such that the final result is inside yn okay and on hn minus 1 there is no disturbance on x prime of course there is no disturbance I mean the homotopy identity there so once you have for each i like this you put all of them together that will cover xn okay how do you take xn cross what is the definition take any point here which is not in xn minus 1 it will in one of the dn i's so there you define it as hn i that's all so they will patch up because on the boundary they all agree with hn minus 1 of the boundary of this we agree with hn here therefore we have little hn now you extend again by coefficient on the whole of x so that is capital H so all these properties we will be following for capital Hn also okay so the inductive step is over therefore the proof is over okay the following this lemma we will make one or two definitions and then very interesting results of this one now can be deduced by applying your mind your exercise so I will give you them as assignments or exercises to you to think about them and they are not difficult now I have prepared you so much so that you can do all those things okay so let me just introduce this notation in these definitions a topological pair is said to be n connected if for each 0 less than 2 k less than 10 every map from dk to sk minus 1 dk comma sk minus 1 to xa is homotopic relative to sk minus 1 to a map within a this was the part of lemma right for m bigger than n and so on with the hypothesis if this conclusion happens every time then you call this as n connected so remember so you have xa okay a map for every k less than equal to n this should happen you do not know what is we have proved it in some case namely where x is got by attaching one em cell that is what we have to write so that conclusion I am making it a hypothesis that is called n connected okay by convention or d0 you have to have some convention d0 is a singleton space the s minus 1 is empty set in particular if a is a single point where x0 is a base point okay x0 is a base point and you take that one then in that case we say x is n connected and you want to make it a what is special case a is never empty here I will take one single point then you do not mention that you just say x is n connected otherwise the pair is n connected the condition is the same okay that is n connected and next I want to define what is the meaning of this higher homotopy groups okay somewhat similar to the definition of the first homotopy group but some miraculous things happen here the definitions are more or less same way okay but miraculous things happen here from dimension 1 to dimension 2 3 and so on 2 3 etc quite different n equal to 1 is different okay so n greater than equal to 2 you define two maps f and g from i n comma boundary of i n you see I have chosen the square model here again rather than the round model okay so that is just a convention that is a convenient convention that is all so xx0 suppose you have two maps like this if n equal to 1 this is just a representation of a loops or something right you remember that a path is okay they are path is but boundary point both the boundary point go to x0 they are loops so how do you compose loops exactly same way I am going to do this define f star g from i n to x as follows both t1 and t2 are less than or equal to half you take f of 2 t1 2 t2 2 tn so on so I am just do it for n equal to 2 but you can do it for any number okay 2 t1 2 t2 2 tn okay so all the ti's are less than or half if all the ti's are between half and 1 then you take g of 2 t minus 1 2 t minus half if ti's 1 then this is precisely the loop composition of f and g okay but now there are all other points where in mixed things happen there you take the constant map x0 for all the points t1 tn will be just x0 okay take this definition verify that just like fundamental verify that it is homotopy associative it has two sided identity namely the constant map etc okay so that one more thing namely what what will be the inverse of f star g namely just try g equal to minus f of minus t1 in one of the coordinates just one of the coordinates that will be the homotopy inverse so this will become a group that group is going to be called as pi n nth homotopy group of x base point x0 so that is the definition okay so this is the inverse I have given f minus f t1 t2 just one of the coordinates you just invert 1 minus t1 right 1 minus t1 here okay so this is f of minus f operating on t1 t2 n is f of 1 minus t1 t2 tn okay so the group so obtained is denoted by pi n of x x0 it is nth homotopy group of x okay okay so one more definition n greater than equal to 1 x2i is an n equivalence like homotopy equivalence it is just n equivalence depends upon n if it induces by ejection of path components of x2i okay and on each component look like take a point x xx okay the ith homotopy group pi i of xx which we have defined just now to pi i of y comma fx this must be an isomorphism for all 1 2 3 up to n less than n and for i equal to n it must be surjective this is epimorphism and such thing we call it as n equivalence if this is n equivalence for every n then we call it a weak homotopy equivalence the idea is it is very easy to see that if it is a homotopy equivalence then it is a weak homotopy equivalence so it is very close to being homotopy equivalence okay there is a big theorem called whited's theorem which says that for cw complexes suppose x and y are cw complexes a weak homotopy equivalence is actually a homotopy equivalence so what I have done so far you know I can give this as an exercise to you it is not a one-step exercise a number of short short exercise all of them you can solve them so so that is what I feel that you if you have understood various parts of whatever I have done so far you will be able to do that such a profound preparation I have done for you thank you