 with the proofs of various statements that we have postponed. So, now let us take one by one, this is one of the important results namely, accession theorem for singular homology ok. So, this is central to our theme, singular homology is central to our theme. So, what is accession? If X is union of two subspace is X1 and X2 such that their interiors X1 and X2 interiors of that cover the whole space, then X1 and X2 is an accessive couple. In particular if X1 and X2 are open then this will be automatically satisfied, then it should be an accessive couple is obvious once from this one. Often what you may get is not exactly opens of sets, but their interiors themselves cover is good enough is what this says. So, this is a very useful statement ok. So, what is the meaning of accession? You take the subspace, the subchain complex namely s dot x1 plus s of x2 inclusion map into the whole of s of x ok. Take that map induced by the inclusion on the homology h star of this to h star of x. This h star of x is nothing but h star of s dot x, the full chain complex, singular chain complex. We want to show that this is an isomorphism ok, this is a bijection that is what we want to show. Given any singular and simple x, one of the generators for the chain complex here, a sigma is from mod delta n to x, remember x is written as interior of x1 and interior of x2. Take the inverse image of these open subsets ok. They will give you two open subsets in mod delta n which will cover mod delta n and mod delta n is a compact thing. Whenever you have an open cover by a compact over a of a compact metric space, you have your Lebesgue number and so on. Once you have a Lebesgue number, you can cut down this delta n by various centric subdivisions as often as you want so that the final pieces will be all as small as you want, small means diameter smaller. So putting all these together, I mean I am just recalling the proofs of various statements here in short. What you get is an iterated subdivision, sd remember is the Paris centric subdivision. Power k is it is repeated k times of delta n each star of each vertex, if you take star of each vertex that will be contained in either this open set or in that open set that is meaning of finer than this covering. So sd of k denotes now capital sd denotes remember we have already introduced this. This is the subdivision chain map. This little sd is the operation of cutting down. This is the Paris centric subdivision. This is the subdivision chain map. You take the composite of this chain map with itself k times that is sd capital k. So what we have proved is that capital sd induces the same isomorphism, same identity isomorphism on the homology. So it follows that the singular n chain sd k of sigma belongs to the subgroup s s dot of x1 plus s dot of x2. So that is the final part here. Each piece is contained inside either in x1 or in x2 and then you are taking the sum total of that. So it will be inside this one. The original thing you know part of it may lie in x1 or part of it may lie in n t ref x2. But now when you have cut down each piece will be either in x1 or x2. So it is some total will be the sum of these two chain complexes. Thus if tau is a singular n chain then what is a singular n chain? It is the sum of summation n i sigma i finite sum. For each sigma i i have a k i. Right? So you take the maximum of all these k i's on that k when you take that many subdivisions each of the parts will be either here. There is a sum of these two. Therefore sd of the chain itself now, chain is sum of these two. If tau is a chain sd k of tau itself will be inside this one. Of course this is not this case it is not the same as this one k. There is some other k. Okay? So you can take the maximum of all the k i's where each k i is for corresponding sigma i that does it. Okay? So this sd k of tau is an element of x1, x2. This implies also always this is an obvious thing because of all canonical thing. That is boundary. Boundary is what? Boundary is some summation of various things subjected to restriction of the boundary. The image will be always in x1 or x2. So therefore image is also, this boundary is also inside x1 and x2. So that is not a problem. Further note that sd and capital D was the chain map, chain homotopy. Okay? The chain group map, the chain group group s1, sx2, they are always, once you are inside x1 and x2, x1 or x2, sd will be also inside x1. So it will never change the code domain of those things. So they are always respecting subspaces. So if something is from s1 plus xx2, it will always in sx1 plus sx2. Okay? And sd induces identity homomorphism on homology. Okay? Because it is already there. In particular, for any cycle tau in sx1 plus sx2, sd of k of tau and tau represent the same element in hx. If you already inside sx1 plus sx2 and taking a sd of k, representing the same element because of tau is a cycle. This will be a cycle and homology elements will be the same. This much we have seen for each capital Sd of c. By repeating it k times, this will be true for resource. Therefore, given an n cycle in sx, there is a k in n such that sdk of tau is in sx1 plus sx2. Okay? This much we have seen. Now pass it to homology, phi of this one, phi is just inclusion map. It will be tau because these two represent the same element. So this element represents the same element as this one. So this proves surjectivity of phi. Okay? After cutting down, it is ready. This one is solved. Essentially, all this is built up in the idea when you take a path and write it as composite of two of its parts. You cut a path into two parts, two paths, then you take it as a composite. So the composition of those two, you know, path composition, represents the same chain element up to chain equivalence. This is what we have seen long, long back. Okay? So that theme is usually always here in a much more strong sense. Okay? So these are all technical details here. That is all. Now to prove the injectivity is also similar. Suppose phi, we prove the injectivity, suppose tau is any n cycle in the sx1 plus sx2 itself. And alpha is an n plus one chain in sx such that boundary of alpha is tau. Okay? What I mean by this one? You have started with an element here. Okay? When you go to the larger thing, it represents a zero element. Okay? It is boundary. In the homology, it represents zero element. I want to show that it is boundary inside this sx1 plus sx2, inside this sub module. That is all I have to do. Right? So this again, for large k, sd of this alpha now, applied to alpha, that will be inside sx1 plus sx2. And boundary of sd of alpha is nothing but sd of the boundary of alpha. And boundary of alpha is tau is sd of k. Okay? Therefore, tau is zero in hx1 plus sx2 because tau, the bracket tau is the same thing as bracket sd of tau, sdk of tau, but that is the boundary. So this proves our injectivity. Okay? So most of these things are very easy and become strategy, but we have to develop these terminologies and concepts. And each concept is very simple. So those simple things we have to verify all the time. So this completes the proof of the big theorem, namely, excision theorem. The people are afraid of that, but this is very simple. Okay? So next thing is not all that simple. Okay? Little more technical details are here. The next thing is, you take a simple shell complex k and a sub complex l, take the relative singular chain complex. Take the relative singular simplicial chain complex. That is a sub complex. Inclusion map is a chain homotopic covalence. That was the statement of theorem 4.1. We are supposed to prove this, but we shall not prove this statement itself, but slightly a weaker statement, which will be the more useful for us, I mean more central to us, namely induced map on the homology is an isomorphism. So we shall be satisfied with the proof of this statement for 10 years. Okay? The actual statement, the proof of actual statement, this one will come in two steps, namely first of all, instead of k comma l, mod k comma l prime, I will just prove it for the pure thing here, right? Absolute thing here. But once you have done this one, it will prove for l also, right? Once you prove for these two things, it is inclusion map, they are both of them. You can put a long homology exact sequence for the chain complex for s dot of k comma l and s of k comma l. Okay? So in the ladder of this long exact homology sequences, each, you know, two of the arrows will be isomorphisms. So third arrow you do not know. Again, two are isomorphisms and so on. So out of the five of them, you will have the middle one you do not know. The other two, two here, two on the other side, they are isomorphisms. By the five lemma, it will follow that the middle one is also an isomorphism. The middle ones are precisely the relative things. Okay? So once you prove it for absolute thing, relative thing follows. So you can just concentrate on this one. Do not worry about the relative part. Okay? After that, the original statement about homology equivalence comes by another general statement on, you know, homological algebra on arbitrary chain complexes, which has free chain complexes. If you have a chain map of free chain complexes, which induces isomorphism homology, okay, then it must be a chain equivalence. So that statement is purely homological algebra. It is not at all difficult, but time consuming. Therefore, I am skipping it. Okay? So in principle, I am only going to prove this part now. Statement 40. So another simplification. You want to prove homology isomorphism. You can do that by first proving it only for the finite case, case finite. Then usually you can take a direct limit argument and then the proof will be true for all k. Okay? Indeed, you do not need direct limit as strong as you want if you do not understand it. It is very simple, just like we did for just now Paris Enter and so on. What you have to do? Suppose you want to prove this one. So you want to prove surjectivity. Take a chain here. Every chain, every actual cycle here, represented by representing a homology element, every chain is supported on a finite simplification complex. So that will be sub-complex here, which will contain that. Okay? So instead of k, you use that sub-complex, say k prime. Then from k prime here, s dot of k prime to that one, you have already proved it is surjective. And go to the inclusion map into this one because everything is inclusion map here, inclusion induced map. So that will be surjective. Similarly, suppose some homology, some cycle here becomes trivial here. Okay? What is my trivial? It is a boundary of something. The boundary is a chain. The chain and boundary of that I have taken. Again that is supported on a finite simplification complex. So take that simplification complex. So there you have already proved that if this is zero here, it is zero here itself. Zero here in this world, there is a boundary here. So that will be the same thing as in the larger one. So even if you don't know, direct limits and so on, as much as, of course we have done it very thoroughly. But direct limit in the homology case can be done by this method, that is all. But it is easy to remember what is happening for finite once you take a direct limit for homology. Homology commutes a direct limit here. So it will go through. For each finite simple complex isomorphism, for every k it will be isomorphism. Now, so how do we prove it for finite simplification complex? Here we use induction. Induction on the number of simplices inside k. If the number is one, what is k? k has to be singleton vertex, that is all. For a singleton vertex, both s of, now ss of singleton and ss of singleton, they are what? They are identical. Every map from delta into single vertex is simply shell already. So they are identical, so this is an isomorphism. So k equal to 1, for r equal to 1, the statement is obvious. Now you assume that it is true for all simple shell complexes, which total number of simplices is less than r and r is some positive number. k is a finite simple shell complex, therefore there is always some maximal simplex. Maximum means what? It is not contained in another larger simplex. So let s be a maximal simplex. Now delete that maximal simplex only. So suppose s is v0, v1, vn, you do not put v0, v1, vn, that is an n simplex. To delete that means rest of them you have to keep, namely all the boundary of this s will be there. So only one simplex you will be deleted. So let l be the sub complex of k consisting of all simplex is singleton other than s itself. And the induction hypothesis, what happens now, the inclusion map from s star of l to s star of s l, this is an isomorphism because it has smaller number of simplices. Not only that, the same inclusion map if you restrict it to the s dot, s dot is what? The boundary of whatever you have deleted s, that is already inside l. So this s dot is inside l that also restrict to an isomorphism here. So this is also induction hypothesis that this is also an isomorphism. Here s dot denotes the boundary of simplex s, we also know that both h star of, this is the boundary of simplex s and h star of s of s n. This is simplex itself. This vanishes in positive dimensions. This we have seen as particular argument. The homology of delta n, this is a simplex. This is contractible thing. So this is trivial, we have no, except the 0 dimension it is infinite cyclic. In the positive dimension it is 0. This is also true. This is also contractible. This part we have directly proved in example 2.4. So they are 0 in the positive dimension and identity this h naught to h naught, this is just number of components in both sides. This is an isomorphism always. It is actually identity isomorphism. Since any pair k 1, k 2 of sub complex of a simple complex is excessive. This we have seen for singular and simplexial chains. For both simplexial and singlex chains we have seen these x isocouples. We get, now we can apply meritary sequences. I can write the homologies for these smaller things as you know on k 1 and k 2. So here we are using the, what are we using here tell me. We are using the excision, excision theorem for k 1 and k 2 which we have already proved earlier. We get two corresponding meritary sequences like this. So this is for L union S and singular homology. This is for simplexial singular homology. I have put a hat here. This is not twiddled, this is hat. It is just a notation, temporary notation because this is large and if I put that one it will go out of the slide itself. So this is the first row is here either for a simplexial and the second row is a singular. So this is the union. This is the intersection. This is one of S dot or something. It is h n of S and h n of model is a union here. So this k is a union. This is a direct sum. This is the intersection. This must be something more here which is missing. What should be this one? H n minus 1 of S direct sum S dot here. This may be a sequence after all. Union k, k prior to coming here is no problem. But remember what we have got here. This is this plus this. So this is an isomorphism. This is an isomorphism. This is an isomorphism because induction hypothesis. This is an isomorphism and this is an isomorphism. So only thing that we do not know is this one. So that will be an isomorphism by silent. That is all. So we have one more thing here. Equivalence of singular simplexial and simplexial homology. S dot S double S and C. C is the simplest singular homology, the simplexial homology. So that also you have to see. Here this time we do not have inclusion map but we have quotient map. We shall present a proof for Poincare. For this, just for the proof of that, we would better do it with a total order on the vertices of k. Then for each n, each n simplex sigma, we can display it by writing V i naught V i n in the increasing sequence. V i naught i 1 i n are increasing because I have taken one single order on the vertices. V i naught will occur before V i 1 and so on. That is the meaning of this one. So that will define a unique element of Sn of k. Sn of k includes all the simplexial maps. So this is one simplexial map, 0 going to V i naught, 1 going to E naught, even E n going to corresponding V i naught V i n. So once you have that, this assignment extends linearly, defines a splitting of Sn of k into Sn of k. See this is an element of Sn of k. These are generators here. And this is some element here. This is too large. Actually, we have defined Sn k as a quotient of Sn of k. To define the splitting here, we need to have a order of that map. I am calling it as alpha n. Splitting means what? It is a quotient map here. You come back by quotient map. It is an identity. The quotient map is free from, far free from Sn of k to Sn of k. So if you put all these alpha n's together, that is the totality of this one. That is a chain map. And that chain map follows because of the uniqueness of this, you know, unique simplex and so on. Some seed or sort of a chain map verification is not difficult. Such that it is phi, compute alpha, identity of Sn of k. This is just that it is splitting. That is the meaning of splitting. So we have a splitting which is chain map. So that is the first thing to observe. Now, it remains to define a chain homotopy on the opposite side. See, phi, compute alpha is identity already. Alpha, compute phi is not identity. But I want to say it is chain homotopy to identity of a bigger complex. If we define chain homotopy, then this alpha star will be the inverse of phi star. So we wanted to prove that phi star is an isomorphism. So that will be over. So how to prove, how to define a chain homotopy from alpha, compute phi to identity of Sn? So both phi and alpha preserve sub complexes of k. In other words, if we have sub-complex here, phi of that will go inside Sn of that sub-complex. Similarly, if k is sub-complex of l, say l is sub-complex of k, Sn of l will go into Sn of l. These are all canonical things here, okay. So those things are easy to verify. So sub-complex, namely l is sub-complex of k, p of Sn is c of l, alpha of Sn is s of l. The chain homotopy that we are going to construct will also have this property that it will preserve the sub-complexes. Hence, we can easily pass on to relative chain complexes. So relative version will also come automatically is what I wanted to emphasize here. Okay, so anyway, so we have defined this chain homotopy h. Chain homotopy means what? It is a map from Sn to Sn plus 1 homomorphism so that this boundary composite h plus h composite boundary is the difference alpha composite phi minus identity. This is the way h has to be defined. As usual in many other cases we have seen, we will do this by induction on n, okay. The first thing is alpha naught, alpha naught from c naught of k to Sn of k, okay. He is the identity map. What is Sn of k? 0 delta naught to anything, there is only unique maps, right, where it goes, whichever text it has to take, that is all. It is the c naught of k is also same thing. So c naught of k and Sn of k is identity, the quotient map is identity, so alpha naught is identity. Therefore, there is no problem at this level. You can start with h sigma equal to 0 for all 0 complexes, okay. At the 0th level, that is 0 simplex level, 0 skeleton level, h sigma also can be taken. Now, suppose we have defined h from Sn minus 1 of k to Sn minus Sn of k, so that this is true, okay. Then we have to define it on Sn to have the same property in the next stage, okay. This is induction hypothesis and such that for each n minus 1 singular simplex k, the support of h sigma is contained in the support of sigma. So why I am saying that I want to construct it this way. So that is a part of the induction hypothesis so that it will help me to prove the next step. At the induction h sigma is 0, so here there is induction hypothesis is obvious, okay. So I will assume I have constructed it in this way namely all the singular n simplex sigma, h sigma will be contained in the support of sigma, okay. Now let tau be any singular n simplex in k, okay. So remember we have defined h on only the generators here, okay. Then we take linear combinations, extended linearly. So we are talking only about what is happening on the generator. The both sides here are linear. Therefore, if you verify this identity over the generator, then it will be verified for all of them, okay. So let tau be any singular n simplex in k and s equal to support of tau, okay. We need to find an n plus 1 chain h tau supported on the same s and such that h of daba tau, you see what we are defining here is h in the next stage. So h of this daba of this part, this is already defined. So we take it on the other side. It is alpha composite phi minus tau minus h of daba tau. So daba tau is one dimension lower thing, h of that is already defined. So the left and right hand side here is already defined, okay. And we want to define h here. It is a daba of this one is this one. So this is possible only if this entire thing is daba of something, right. So you have to show that this left hand side, right hand side is actually daba of something. It is in the image of daba. Now that is all one has to show, okay. Obviously, it is necessary that the chain on the RHS of the above equation must be a cycle. If daba of that is 0, then there is a chance. Then it can be daba of something, right. If we should daba of something, daba of that will have to be 0. So it must be a cycle first of all. So, okay, that is if you take daba of this thing, I am taking daba of alpha composite phi minus tau minus h of daba tau. What is this? It is alpha of phi of daba tau, right. This daba is computed. It is no definition. It is alpha phi of daba tau minus daba tau whatever minus daba of h of daba tau, okay. Now why this is 0? This is by induction step. So that we have already this equation on 45 of what we took, okay. So you have to verify that. It is also clear that the support of this cycle is contained inside as we are not going out of a singular simplex x at all completely. Since the homology of h n of s of delta q is 0, the whole thing is seen as single singular simplex. And any cycle which you have shown cycle is what is trivial, it is a boundary. So finally this is where we have used topology. Till then all the time we are using just you know algebra in some sense. So this is an example that we have already done in the case of s of delta q is 0 in positive dimension. So we are already in the high one dimension higher in the positive dimension we are we are argued it in different way. So it follows that this cycle is a boundary and hence there is a chain h tau belonging to s n plus 1 of x, but s is contains an of k satisfying our requirement mainly daba of that is equal to this one, okay. So this because this is a this is first of all it is a cycle but this cycle is inside s therefore it is 0 in the homology which means it is the boundary, okay. So that completes the construction of h and complete the proof of that singular simplex chain complex is equivalent to the simplex chain complex itself, okay. The methods employed here there is a pattern and that pattern is generalized to what is called as methods of acyclic models. Something is free, something is acyclic, this acyclicity these two things will be put together, okay. My idea is to explain these things in the rudiment of this in the in atomic things in this in the simplest case how these things work. So that if you learn more algebraic topology then you have to learn these higher machineries wherein you know neither the books nor the authors nor a teacher will have any time or patience to explain. That is why I have explained all these things separately here instead of using the high machinery right in the beginning, okay. So this completes one aspect of whatever you wanted to study namely the homology, okay. Whatever we have aimed up till here that we have completed. So we have completed all the results long back now whatever statements that we have postponed most of them we have completed proofs except things like this homological algebra here that we have that I told you. So except for a few things like this where in clear reference has been also given we have completed all the proofs, okay. So next time we will stop take a different topic namely topology of manifolds. Thank you.