 So, we would like to construct the CW chain complex of a CW complex. Let me recall what we did last time, the last lemma is very, very crucial for this construction. So, it will be worthwhile to recall that whole thing and also vote to the proof very carefully so that you do not have any doubt left there. So, let me begin with this lemma that we did, X is a CW complex and for each n positive we have the homology, the kth homology of Xn relative to Xn minus 1 vanishes for k naught equal to n. So, it is like now atomizer at one single k and at k equal to n, it is isomorphic to the free abelian group of rank equal to the number of n cells in X, ok. This is a consequence of the previous lemma that we have done, I am summing it up here. The second thing and third thing are slightly need some proof and they follow from this one of course. So, the second thing is Hk of Xn is 0 if k is bigger than n. That means, if you have n dimensional simple shell complex, CW complex, then all the homology beyond n n plus 1 n plus 2 etcetera, they are all 0. Inclusion map, let us, let us denote the eta n because it is starting from the n skeleton to the whole of SpaceX, induces isomorphism homology, Hk to Hk where k is less than n and surjection for k equal to n, ok. Let us look at the proofs. Part a is actually direct consequence of previous lemma that we have proved. Namely, for attaching cells, if you attach n cells to a space y, then the relative homology, the nth homology is precisely equal to the free abelian group over the number of cells you have attached. And everywhere else it is 0, you apply that one. It is a special case Xn is obtained by attaching n cells to Xn minus 1, ok. So, b let us do it by inductively. For X0 which is discrete space, Hk of X0 is 0 for all k positive. Therefore, we can start the induction namely on n this Xn. Suppose this is true for Xn minus 1, namely for all lower indices, the homologies are 0. Let us prove that one. Then we want to prove it for Xn, ok. So, inductively assume that the statement be holds for Xn minus 1. At kb greater than n, then I want to show that Hk of Xn is 0, right. In the long homology exact sequence, Hk of Xn minus 1, this is 0. Hk of Xn, Xn minus 1, this is also 0 by a. Therefore, the middle term must be 0. So, that is all, ok. So, b is also proved. Now, to prove c, what does c say? This composition from Xn to the whole of space X induces isomorphism for k less than n and surjection for k equal to n, ok. So, this is what we want to prove. First, you assume X is finite dimension, say dimension of X is m. If this n is bigger than m, then the nth skeleton is the whole of X and the inclusion map is the identity map and there is nothing to prove, ok. So, you may assume that n is smaller than m and Xn is a proper closed subset of the whole of X. X is now, X is equal to Xm because dimension of X is equal to m, ok. Then for each i greater than or equal to 0 and k is less than n, we have the inclusion needs to map n plus i skeleton to n plus i plus 1 skeleton, ok. The inclusion induced map, these are all isomorphisms from b. Take composition n plus i, n to n plus 1 to n plus 1 to n plus 2 and so on, ok. You take up to Xm, we did hit Xm. So, all the inclusion maps are isomorphisms. If k is less than n, if k is less than n, k is less than n plus 1 also n plus 2 also, ok. These finitely many compositions of isomorphism, they are isomorphisms, ok. So, the composite of finite it is fine. For k equal to n, what happens? Very first one is not an isomorphism, ok. Xn, Hn of Xn to Xn plus 1 here is not an isomorphism, but it is only surjective. After that, all other things are isomorphisms. Therefore, the composite will be also surjective, ok. So, only the first stage is not an isomorphism, but that is actually surjective. So, that is what we have to see, because Hn of Xn plus 1 Xn, the entomology of this one is 0 in the long homology exact sequence, ok. So, that will take care of the finite dimensional case, ok. Now, comes the infinite dimensional case. Suppose X is infinite dimensional. Now, suppose the inclusion map eta n from Xn to X, where I am taking k to be smaller than n, ok. Suppose this is not injective, we want to show that it is injective. First of all, then we want to show that it is an isomorphism. Suppose it is not injective, ok. Let C be a k cycle in Xn which is a boundary in X. That means, something here goes to triviality here under the inclusion map. So, it is boundary is inside X. So, C is boundary of Y. Y is a k plus 1 chain in X, but Y is a finite combination of k plus 1 singular simplicis. Each of them is compact. So, the entire support of the chain, any finite chain, finite after all they are, which is support is a compact subset, ok. A compact subset of any Cw complex is contained in a finite skeleton, ok. So, it is contained in some Xm, alright. So, what does, what does this mean? This means that this C is boundary of Y, where Y itself is in Xm, which means that the inclusion map from Xn to Xm itself is, is not injective and that contradicts our whatever we have already proved, ok. Therefore, hn of this eta n at kth level, hk of Xn to hk of X is injective. Exactly similarly, we can prove the surjectivity also, ok. Using the compactness of any support of any chain and passing on to the finite, finite skeleton here, ok. Now, having recalled this one, we can now define the, the chain modules, modules of chain and then the chain map also, the boundary operator also, ok. So, let us have some elaborate notations here for, for this particular part only we will have these notations fixed, ok. So, In is Xn to Xn plus 1 inclusion map, but I will denote In the same thing instead of going on writing star and so on, In for inclusion induced map here, induced homomorphism here. Similar Jn I will write for the inclusion induced homomorphism of Xn into the pair Xn, Xn minus 1, ok. Think of Xn as Xn comma empty set that will be included in the Xn, Xn minus 1. That inclusion gives you a homomorphism at the homology level. So, once you have these notations, let us put the nth module or nth group whatever just a billion group it is of C star Cw top X that is the notation, the Cw chain complex of X. It is the homology module of, the nth homology module of the relative pair namely Xn, Xn minus 1. Recall that this is actually a free abelian group over the number of n cells attached to Xn minus 1 to get Xn. It is a free chain complex it will be finally. So, these things are free abelian groups and the boundary operator dn there will be Jn minus 1 composed delta n where delta n itself is the connecting homomorphism of this pair homology pair. Remember for any X comma a we have a long exact sequence homology groups and there are connecting homomorphisms. Hn of Xn comma Xn minus 1 to Hn minus 1 of Xn. So, take that nth one and write it as delta n. So, that is the connecting homomorphism and this Jn I have defined already Jn, Jn minus 1 or JK all of them I have defined here because they are inclusion induced homomorphisms. So, composite Jn minus 1 composed delta n. This I am taking it as delta n. That is a homomorphism. What we have to verify? We have to verify that dn composite tn minus 1 is 0. Once we do that this will be a chain complex. So, the following lemma tends to dn and dn plus 1 is 0 or you can write it as first dn plus 1 goes to that one then dn or dn minus 1 composite dn whichever one for all n and hence this becomes a chain complex. This chain complex will be called cellular chain complex associated to X. So, we will just write as Cw of X. Cw of X. The homology of this chain complex will be denoted will be called cellular homology of X. You can have another notation H Cw bracket X etc. But soon we will prove that this homology is the same thing as the singular homology of X and therefore you will not need an extra notation for this one at all. You can only use this as a description of back to the singular homology itself. The homology of this chain complex is called cellular homology of X temporarily. So, first let us prove that the composites here are 0. d square is 0 is what we have to show. So, I will show you a commutative diagram here from which all these things will be clear. What is the diagram? Here the horizontal arrow, these horizontal sequence is the homology exact sequence of the pair Xn Xn minus 1. Only a part of the time taken, only 4 terms I have taken Hn of Xn, Hn of Xn, Xn minus 1. This is remember, this is my definition of Cn now. And this is Hn minus 1 of Xn minus 1 and this is the connecting homomorphism. What is the term here? It is Hn of Xn minus 1. Hn of Xn minus 1, just in the previous lemma we have shown that that is 0. So, these 4 terms are being the horizontal one. There is a vertical one which is also the homology exact sequence of a pair namely Xn plus 1 to Xn. So, this is delta n plus 1, Hn of Xn, then Hn of Xn plus 1, then Hn of the relative thing. Hn of the relative thing, n being smaller than the skeleton dimension is 0. So, these are the things you have taken. Even this one is the homology exact sequence of some pair. What is that? You get Hn of, this is what is what is called as the of the triple Xn plus 1, Xn, Xn minus 1. Homology exact sequence of the triples which I have done last time because we needed this one. So, Hn plus 1 of Xn plus 1 Xn and then the dn plus 1, Hn of Xn minus 1, I can take it as actually, sorry, I can just dn plus 1, I just declare it to be this composition here. Sorry, this is just taken here. There is here one more vertical one. This is again exact sequence of Xn minus 1, Xn minus 2. Previous term is Hn minus 1 of Xn minus 1 and this is the inclusion map. What is the term here? It is Hn minus 1 of Xn minus 1 of Xn minus 2. So, Hn minus 1 being one dimension higher than Xn minus 2, that is 0. So, this is just that part. Having done that one, dn plus 1, this diagonal arrow, I am just defining it as this composition. That is my definition. This is my Cn plus 1, this is Cn, this is Cn minus 1 and this is dn dn plus 1. I want to show that this composite is 0. This composite, this map is this composite, this one. This map is this composite, this one. Therefore, if you take dn and dn plus 1 composite, the same thing as starting from here, come down, go to the right, go to the right and come down. But when you take this composite, this is an exact sequence. So, this composite itself is 0. Therefore, this whole thing is 0. So, that proves this lemma. So, dn composite dn plus 1 is 4 of them and then in the middle you get something which is 0. So, the whole thing is 0. Now, the next thing is to prove that the kernel of this by the image of that is actually homomorphic to Hn of X. I want to show that it is isomorphic to Hn of X itself. That is the statement here. Hn of Cw of X is Hn of X. What is the definition of on the left hand side? It is kernel of dn, modulo the image of dn plus 1. Now, look at this one. What is dn? dn is this composite this. I am taking the kernel of this, but this is an injective mapping. Therefore, the kernel of dn is the same thing as kernel of delta n. The kernel of delta n by the exact sequence is equal to image of Jn. So, therefore, my group on the left hand side is quotient of image of Jn by the image of delta n, dn plus 1. See, this Jn is a subgroup of this is injective memory to take the subgroup here Hn of Xn and apply Jn of that. And then take the image of that. Do you understand that? So, what is the image of this? Image of this is image of this composite this. So, instead of going under image of Jn there, I can identify that those groups under this one. It is Hn of Xn, the whole of Hn of Xn, modulo image of delta n plus 1. Under Jn, it will be isomorphic to the kernel of delta n, modulo image of dn plus 1. So, what we have, what is the entomology group here is nothing but Hn of Xn, the whole Hn of Xn, modulo image of this delta n plus 1. Therefore, it is a co-kernel. Therefore, it is Hn of Xn plus 1. But we have seen that this is isomorphic to Hn of X in the previous lemma under the inclusion induced map. So, that completes the proof. So, under the inclusion induced map only, we get finally this group is isomorphic to Hn of Xn. So, that is the proof here. So, I repeat kernel of dn here is kernel of delta n because this is injective. So, that is equal to image of Jn because this is exact, this is exact. Therefore, image of, so this is what image of Jn and that is isomorphic to Hn of Xn because Jn is injective. Therefore, Hn of Cn is equal to kernel of dn by image of dn plus 1 which is Hn of Xn divided by image of delta n plus 1. Because dn plus 1 is Jn compared to n plus 1. So, you can work out this one here itself. This by the image is same thing as a co-kernel here which is which is this one, this is this is injective mapping here. This is co-kernel of delta n plus 1. So, what we have done, we have done a big thing. See small, small things we have collected and finally what we have done is you check out the n minus 1 skeleton to get the n skeleton whatever is remaining n skeleton. They are only the cells that gives you a free abelian group over the cells. So, those things form a chain complex. The homology of that chain complex is the homology of X itself. So, it is very easy to understand the chain modules there. You do not have to work out all the singular homologies, all continuous functions and all chains and so on. So, this is quite similar to what we have done for simple shell complexes, the simple shell homology. In principle it is similar, but we have to work harder here because we are not doing linear things and so on. It is not combinatorial. It is not completely combinatorial. It retains a little bit of combinatorial nature of the simple shell complexes, but it combines a lot of topological information. The most difficult thing is to compute these delta n plus 1s or dns sorry. The new notation is how to compute these dns. These dns were very, very easy and you know combinatorial data itself gave you in the case of simple shell complexes. Here it has to be given by the attaching maps. So, that is still purely topological. So, you may say that the catch is there, but even there it helps, it brings a lot of simplification from the arbitrary singular chain complex of an arbitrary topological space. So, let us take this example now. The homology of the complex project spaces. We have already given the complex project space a CW structure consisting of 1 0 cell, 1 2 cell, 1 4 cell and so on. The odd dimension cells are missing there. There are no odd dimension cells there. And in each dimension, in each even dimension there is exactly one cell. So, how does it help? One of the simplest thing is suppose a CW complex does not have any n complexes, then its CN CW will be identically 0. Therefore, HK will be also identically 0 because after HK is the kernel of that homomorphism from that 0s, 0 homodule. So, sub of 0 module is 0, quotient will be also 0 and so on. So, corresponding homologies are all 0 there. So, immediately you can tell that all the odd dimensional homology of CPN are all 0. There is more to come. What happens to even dimension? In fact, look at the chain complex. This is CW of CPN. So, how does it look like? The 0, there is one cell. So, it is z. There is no one cell. So, this is 0. There is a one cell which is the two dimensional cell. That is why it is z and so on, alternatively 0 and 0. You do not have to know the boundary maps. Boundary maps are automatically 0 here because this module is 0. This module is 0, so this map is 0. So, they are all 0. The kernel will be the whole space for whole module and the image will be the 0 module. So, each homology is either z in the even dimension or when you go to odd dimension it is 0. So, to sum up, HI of CPN is z if I equal to 2k in 0 otherwise. Very neat result. In order to exploit cellular homology further, we should try to understand the boundary operator homomorphisms that I told you. So, here is a description, a complete description of what you have to do, but I do not have any examples right now for that. My examples were too simple like complex friction space and so on. But we will have that when you study the length spaces. So, let us make a preparation for that ok. Let phi alpha from dn Sn minus 1 to xn xn minus 1 denote the collection of all characteristic maps of n-cells. And restricted to the boundary you will get the corresponding attaching maps of alpha. We know that I am just writing Cn it is Cn Cw of x ok. It is just Cn for me just now. It is hn of xn module xn minus 1 right relative to xn minus 1. It is freely generated over the basis this phi alpha. Phi alphas are maps. So, they are one chain, one simplices. So, you can think of this one simplex is generating the entire group. A chain will be a linear combination of these things freely ok, integral combinations of these things. Therefore, in order to determine dn, we have only to find the expression what happens to dn of phi alpha, dn of phi alpha by very definition is an element of hn minus 1 of xn minus 1 xn minus 2. So, this is again a free module over what? Over the cells of dimension n minus 1 inside xn minus 1. So, let us denote the collection of all characteristic maps of n minus 1 cells by psi betas just like phi alphas here. Let us denote psi beta these are for n minus 1 dimension. Observe that the quotient space when you collapse the n minus 2 skeleton to single point then the all the attaching maps should be a constant functions when you when you take the quotient space. Therefore, each cell n minus 1 cell when you attach it will become a sphere ok. The entire boundary of the cell boundary of the dn minus 1 is collapsed to a single point. So, what is the all these points are common point one single point what is what that one single point it is the image of xn minus 2 right. So, this will look like the bouquet of sn minus 1 beta I am writing this one to denote the n minus 1 cells that their images sn minus 1 beta is a sphere. So, bouquet of this one over beta. So, take all these collection distort union take a one point identification that is the bouquet of sphere right. So, this says you know there was an earlier exercise what is the homology of this one? This homology of this one of course H naught is this is a connected space. So, H naught is infinite cyclic all other homologies are 0 except n minus 1 homology which is again the free module over this beta the indexing set beta. And what are the generator take hn of sn minus 1 beta hn minus 1 of hn minus 1 of sn minus 1 beta that is infinite cyclic take generator there. So, those will be the generator you can call them as t beta or minus t beta whatever ok. So, the indexing of these spheres is such that the n minus 1 cell psi beta is mapped onto the sphere sn minus 1 under the quotient map qn minus 1 from xn minus 1 to xn minus 1 modulo xn minus 2. Also this quotient map induce an isomorphism of cn both of them are free abelian groups over or the same indexing set beta. So, psi beta is the in the collection of of n minus 1 cells n minus 1 simplexies a singular simplexies that will go to the generator here of hn minus 1 of sn minus 1 beta beta corresponding to beta ok. So, that is the understanding. Now you look at the quotient map here the projection map here from the bouquet to one of the spheres that means you are folding all the other spheres to a single point and keeping one of them as identity ok that is the projection map ok. This is the projection map onto the beta component I am calling them component because this can be actually thought of as a subspace of the product infinite product of you know all these sn minus 1 beta over beta. So, that I will leave to you ok. So, it is beta component is a projection map then dn of phi alpha is anyway a linear combination of this psi betas which I have written psi betas right. They are generators here bracket psi beta can be thought of as a generator here. So, n alpha betas are from integers take the sum which is a finite sum though I have written all over beta most of the n alpha betas will be 0 except finite limit. So, it looks like this I do not know what are n alpha betas, but what are they I can tell you now if you take delta n of phi alpha ok this will be put f alpha the boundary of phi alpha ok is given by the attaching map f alpha. Therefore, what is f alpha doing is precisely what you have to understand follows that n alpha beta is nothing but the degree of this f alpha starting from sn minus 1 into xn minus 1 then you take the quotient xn minus 1 module xn minus 2 then you take phi beta the projection to the one single factor. What you will capture is all other components will disappear only psi beta will remain along with this integer and that tells you that that is nothing but the degree of this map this n alpha beta is the degree of this map. So, this will help you in understanding more complicated spaces it will help you a little bit about the same thing in the case of this singular homology. So, we shall do that singular homo that I what is it the CW homology of the length spaces that is our next topic. So, we shall do it next time. Thank you.