 Let us introduce a new homology today and then relate it to the singular homology. This is called CW homology just like we had simple shell homology and this is applicable only for CW complexes of course, because simple shell complexes are CW complexes is applicable for simple shell complexes also, but not for arbitrary spaces. So, before that let me recall a result that we have put as an exercise earlier, exercise 3.5. Now I will state it as a theorem and indicate a proof also which is one line proof actually. On the category of topological triples XAB, there is a functorial long exact sequence of singular homology modules HN of AB, HN of XB, HN of XA and then the delta homomorphism back to HN minus 1 of AB and so on. So, I have not bothered to write down what are the maps, slowly you will know by the context itself what are the homomorphisms there. Remember what is the meaning of topological triple just like topological pair consists of two spaces A subsets of A subspace of X. Now here you have to take B is a subspace of A subspace of X. So, X is topological space and then you are taking two subsets one inside the other. So, smaller things are written later for XAB. So, X contains A contains B. So, that is a triplet. So, this forms a category with morphisms of continuous functions. So, X going to Y, A going to say A prime and B going to B prime correspondingly. So, that is the meaning of morphism inside the category of triples. So, once you have that, you take the singular homology of XA, singular homology of XB, singular homology of AB. So, these three things will fit together into a long homology exact sequence. So, this is not any mystery or anything, it is a one line proof from whenever you have a short exact sequence of chain complexes, we have assigned through snake lemma a long exact sequence of homologies. And that is what you have to do namely look at S of B, S of A and S of X. One sitting inside the other, SX is the largest, contains SA as a sub complex, contains SB as a sub complex. Therefore, SA modulo SB becomes a chain complex on its one, the quotient complex. Similarly, SX modulo SB is also chain complex. Similarly, SX mod SA is also chain complex, but by the ordinary usual isomorphism theorem of Abelian groups, SA by SB will be a sub complex of SX by SB. So, there is an inclusion map here, inclusion induced homomorphism from SA to SB goes to SA modulo SB to SX modulo SB. Similarly, SX is SX here, SB is a sub module of SA. So, quotient thing out this one, then further quotient, this will be another quotient. So, they are subject. And the whatever isomorphism theorem, Moietus isomorphism theorem, I do not know first or second or whatever, that tells you that the kernel of this one is precisely SA by SB, okay. So, you have such a short exact sequence. Now, if you take the long exact sequence, what you get? HI of this, which is nothing but HI of A comma B. Then this is HI of X comma B. And then this is HI of X comma A and then HN minus 1 of this. So, the theorem and the proof is over. So, this is what we are going to use now quite heavily. So, I do not want to leave it as unnoticed or unproved and so on, okay. So, this was the preparation there. Now, let us start with as usual, instead of directly attacking the CW complex, let us start with the one stage, namely X is obtained from Y by attaching only one kind of cells, namely N cells, a family of N cells, that family could be finite or infinite, we do not care. Okay, X be attached by, X be obtained by attaching N cells E and J to Y, while attaching maps alpha J, which are defined on the boundary of D N to Y. Okay, once we have this collection, let me put alpha, a homomorphism from direct sum of copies of Z. How many copies I have take? As many as the indexing set here, J and J. Okay, take direct sum of infinite cyclic groups as many as indexing set here to Y N minus 1, H N minus 1 of Y. This group, what are this homomorphism? The homomorphism can be defined provided you define it on each component here, namely on each generator of this component, whatever, which when restricted to the Jth summoned here is equal to the alpha J star, alpha J is from S N minus 1 to Y. So when you pass on to the homology, H N minus 1 of S N minus 1 is infinite cyclic. That copy, I am identifying with the Jth copy here, alpha J here. Okay, so Z, this is H N minus 1 of S N minus 1 is Z. Okay, that is identified with Jth copy, take that homomorphism on the Jth copy. So once you have defined these things, alpha is completely defined. Okay, that is the sum total of all these maps. So we have defined what is this alpha. Now this alpha is going to play the role in the homology of X. Then the first statement A is the relative homology of this X comma Y is 0 except at K equal to N. At K equal to N, it is just a direct sum of infinitely many copies of Z. How many copies? I mean finite or infinite, I do not care. As many as the indexing set J for K equal to N. That is the first part. The second part is the inclusion map from Y to X. When you pass to the homology it is an isomorphism for all I not equal to N minus 1 and N. At N minus 1 and N, there are some disturbance because of the attaching cell. So that is the meaning of this one. Inclusion in this homomorphism is an isomorphism for all K not equal to N and N minus 1. So this must be K here, okay. The third condition C, there is another statement, it is not at all C and D are there. I star from H N minus 1 to H N minus 1 of X. So this was not included here. When K equal to N minus 1, K equal to N, those two what happens, that is what C and D tell you. So I star H N minus 1 Y to H N minus 1. This is a onto map, it is a rejection and its kernel is precisely equal to the image of alpha in H N minus 1. This kernel will be subgroup of this one and that is equal to the image of alpha, alpha is this map. Similarly, the statement D says that on entomology H N of Y to H N of X, it is injective, okay. So H N of Y will be a subgroup of H N of X and with H N of X itself is isomorphic to this subgroup direct sum, the kernel of alpha. So this subgroup will come here, the subgroup of this direct sum, kernel of alpha is subgroup of that. That will be a component of H N of X that is a direct sum and what is the other component is image of I star, image of H N. So this gives you a complete picture of the homology of X, okay, in terms of the homology of Y and how you have attached the functions, how you have attached those things, attaching maps have a role to play here where in fact, exactly at one single place, everywhere it does not matter, namely wherever the homology of H N minus 1 was non-trivial, namely H N minus 1, okay. So look at the induced map, those alpha J stars will have, they have some say in the homology of X finally, okay. So let us go through the proof of this one. Proof of A, for each J choose a P J in the interior of this cell, think of this as a actually the 0 of the disc, okay. It can actually take the homeomorphism disc and take 0 and take this image for example, okay. For each J you take that point and put U equal to X minus all these points, X minus P J, J in J, throw away all the center points of all the attached cells, okay. Then whatever is left out, namely U, U will strongly deform the tract Y, Y is a deformation retract of U, actually strong deformation retract of U because all these cells can be collapsed to the F of image of F alpha or whatever attaching map, okay. F alpha or alpha, alpha J, what did I say? What are the attaching maps I have written down? Alpha J's. So alpha J of the image that will go, okay, they will go inside that one. So Y is a deformation retract of U. Therefore, from the long exact sequence of the homology of the triple X, U, Y, we studied X, A, B, right. Instead of A and B you put U and Y. What do you get? H i of U, Y to H i of X, Y. Then there is inclusion map I have, I want to pay, I want you to pay attention to this one. That's why I have put eta 1 star here. H i of X, U to H i minus 1 of U and so on, okay. This map becomes, this the front center becomes isomorphic. Why? Because both this and this are 0 because U is a, Y is a strongly retract of U. They are the homotopy square. And by the homology exact sequence of U comma Y itself, it will follow that this is 0 for all i. This is all, both of them are 0. So this is an isomorphism. So eta 1 is the inclusion map from X, Y to X, U, okay. Inducing isomorphism in all the homologies. This is the first thing you have to do, okay. So X, Y can be computed from computing X, U, homology of X, U. First we wanted to compute a homology of X, Y, right. X modulo Y. So it can be computed X modulo U, all right. Now, let us put V equal to the disjoint union of all the interiors of these cells. Then U is open in X, okay. U is what? U is X minus all the points, all the center points. That is also open. V is open. So U, V is an excessive couple. U, V union V is the whole space X, okay. Therefore, the inclusion map V, V intersection U to X, U induces isomorphism in homology groups. This is the excision theorem here, okay. Now look at the pair V comma V intersection U. V is the disjoint union of all these open cells. And V intersection U is throwing away all the center points, one of the interior points, okay. So this is the disjoint union of pairs like this, interior of Ej, interior of Ej minus Pj. What are these? These are, the homology of this one is nothing but by again by excision, homology of the closed disc D minus the closed disc D minus 0, I can say. Pj goes to 0 under this homeomorphism. You can write any other point also, no problem. It must be an interior point, okay. And that we know is the homology of the pair Dn Sn minus 1. And this we have computed. This is 0 except at n equal to, k equal to n. And then it is infinite cyclic, okay. So each of them is infinite cyclic. And this is a direct sum over J. So what I get is this growth, okay. Tired sum of these copies, okay. So this is where I have put. For k equal to n it is this one. Everywhere else it is 0. So statement A is complete, okay. So let us complete the other two statements. They are all easy now. B, C and D are easy consequence of long homologies sequence of the pair x, y, okay. If you look at this, I have not written down here, of the pair x, y, hi of x, y, we have determined, okay. This is 0 unless i equal to n minus 1 and i equal to n minus 1 and n. So hi of x, y, hi of x comes, hi of x, y comes, hi of, sorry, hi of y to hi of x to hi of x, y. The previous thing is hi of x, y. So this will be an isomorphism. So that was the first statement here, okay. Hi here, hi, hi is an isomorphism. Only when a k equal to n minus 1 and n, one of the either the top one or the left one or right one will not be 0. One of them is 0. One of them is 0 will mean that either it is injective in one case, in the last case and surjective in the first case. So this is B and C are, all these things are easy consequences of whatever we have proved inside here. And the description of what is alpha x. The long homology exist sequence of the pair you can apply. So what we have done is, we have done a good beginning in this lemma of understanding the homology of CW complex is one step at a time. We shall now launch a systematic study of homology of CW complex. Our theme is from parts to the whole, the main tools are just long homology sequence of you know homology exact sequences and the accession to around, okay. Just that is what we have used here. We will keep using that, okay. So let me just give you the definition. Maybe we will not proceed with this one today. Let X be a CW complex. Then each integer n greater than or equal to 0, we have the kth homology of the nth skeleton, modulo n minus 1 skeleton vanishes for k naught equal to n and is isomorphic to the free abelian group of rank equal to the number of n cells in X. Hk of Xn is 0 for k bigger than n. Inclusion map Xn to X induces isomorphism in homology for k less than n and surjection for k equal to n. All these things are easy consequences of what we have done in the earlier lemma once at a time. So this Xn minus 1 plays the role of Y here, okay. Xn is the place, plays the role of X because Xn is obtained by Xn minus 1 by attaching n cells. Therefore, kth homology will be 0 for k naught equal to n, okay. So for k equal to n is isomorphic to the free abelian group of rank equal to the number of n cells in X that is also part of this. Whatever we have, we have stated here. The extra thing will come there precisely, attaching maps alpha j, alpha from, so number of say DL cells are attached, right. So this Y becomes Xn minus 1 and alpha is the index, is the induced by the inclusion maps this alpha j star here, okay. Attaching maps are alpha j's, right. So the Xn plus 1 comma Xn minus 1 is precisely this one. So this is a statement, it is 0 or it is this many copies of that. So that is a part A, you have to read it here, okay. Part A here. Second one, HK of Xn is 0 for k bigger than n. Now let us look at this one. If k is bigger than n, okay, first you do it for here n scale, right. So this is a re-statement of that, that is all. HK of Xn is 0 for k bigger than n, okay. k is bigger than n, there is nothing there, there is no cell attached here. It is 0 all the time, okay. So 0, k not equal to n. So no matter how far you go, this Xn will be 0. The third thing, the inclusion map eta from Xn to X induces isomorphism k less than n and surjection for k equal to n. Let us forget about k equal to n, k less than n. What happens? Induce isomorphism homology, k is HK of Xn to Xn plus 1, it is an injection. Xn plus 1 to Xn plus 2 also is an injection, right. So it is a composite of injections. So therefore this whole map eta n from Xn to X will be an injection, okay. You have to argue this way because why are you wasting it is as a composite of infinitely many of them. Suppose eta n happens to be not injective. That means some element in HK here, some cycle here becomes a boundary when you go to X. That boundary itself will be in some case skeleton Xn plus r, let us say. But up to r you can compose. So that way you get that eta n itself is injection, okay. Similarly for k equal to n surjection comes from the part B, part D of this one, part A part C of this one. So let us stop here. Let us use this one next time to give a complete picture of the homology of CW complex and then out of that we are cooking up another chain complex for this X itself which is CW complex, okay. The homology of that will be isomorphic to the the homology of X. So that is the beauty of this one. Okay. Thank you.