 Finally, we come to a section in which we would like to complete the proofs of various statements that we have made in this chapter and postpone their proofs. So let us take them one by one, the proofs of several statements. The very first thing about homotopy, it should remind you of Poincaré's lemma in differential topology, if you have studied it. Indeed people got the idea from there and now they have made it completely generalized. So let us first concentrate on singular homology. And then the homotopy axiom will be proved by proving the lemma 3.7 in which we promise that we will have, we will construct the prism operator. The prism operator is going to be the homotopy between eta 1 and eta 2. This is going to be a map from sq2, sq plus 1, sq of x to sq plus 1 of x cross i and the relative, relative version also, okay. Eta 0 and eta 1 are inclusion maps of x in x cross 0 and x cross 1. And these are the maps which are induced at the level of chain level. And this h we promised that will be factorial. In a way, by demanding that it should be factorial, actually helps us in construction of h itself. So what is the factorial here? Suppose you have a continuous function of the pairs from x a to y b, any continuous function, then h from x a to s of x cross i, similarly, the h from y b to s of y cross b, they should be compatible under this alpha. Alpha is here, obviously what I have to take here is alpha cross identity, identity factor on the interval, interval factor its identity, alpha cross i. Take alpha cross identity star here, this diagram must be commutative, okay. So this is what we demand, though it is not, it would not have been necessary for us to prove the homotopy invariance. But then if sigma is a singular n-simplex, a singular n-simplex can be thought of as a map from delta n to x, okay. So that takes the place of alpha here and then what we get is we would get h of sigma h, okay, h composite sigma, sigma is same thing as h of alpha star, you have alpha sigma, sigma star would be equal to alpha composite identity, alpha sigma, sigma cross identity star of h, okay, sigma cross identity star of h operating on the identity map here. We take the identity element in delta s dot of delta n, okay, in fact s n of delta n, the identity map from delta n to delta n is one of the generators there. On that it should be, it should have this property. So this gives us idea that it is enough to define this h sigma, h of sigma, not for arbitrary sigma but for just psi n. Suppose you have defined it for psi n, then for arbitrary sigma, h of sigma must be given by this formula, okay, by the functoriality. So therefore our task of defining h sigma, see, h has to be defined on the whole of s dot but then it has to be linear and so it is enough to define it on the generators. Generators are arbitrary, simple shell, arbitrary continuous maps from delta n to x, even that is not necessary, just define it for each delta n, for each n, you define it only on psi n, okay, where psi n is the identity map of mod delta n to delta n, it is a continuous map. So if you define for each n, h of psi n, then we are done, then I will take this as a formula for h sigma, then extend it linearly, that will be there. Of course we have to verify that it is a chain homotopy. So what is the property that will make it chain homotopy, so that I have to, I can define it just by h of psi n, so that is our next goal, analyzing this one. So the demand for h is that eta 1 star sigma, eta naught star sigma must be equal to h of daba sigma plus daba of h sigma, so this is the meaning of that h is chain homotopy between eta 1 and eta naught. For every sigma if it is true, then it will true for every chain, okay, so this is what we want, but then by functoriality if we can manage to have it only on psi n, then we are done, because if once we, once 33 is true, I can take sigma star of these things, for sigma, if you put say xy n is sigma, sigma can be thought of as sigma star of xy n, okay, and then use this property for xy n and then you will get the property for sigma. So we can apply simply sigma cross identity star to both sides of certify to get 32. So our task is to define xy n, h of xy n so that it satisfies this 33, alright. Now I have to recall the prism construction, after all it is going to be the prism operator. This was elaborately introduced in part 1, but now I have to do it the way I want it to right now here, okay. So in delta n cross i or i cross delta n, whichever one you want to use, you can use that. So here I have just taken delta n cross i, let b top i denote e i cross 0. Remember even e naught even e to e n are the vertices of delta n. So for delta n cross i, you have e i comma 0 at the 0th level and c top i is e i comma 1, the same thing at the level 1, okay. So similarly you can take the bare center of delta n, place it at the level half, let us have a notation for that gamma n. For each n I am going to do this, okay, the bare center of delta n, okay, take at the level half inside delta n cross i and put it at gamma n, okay. So just for, suppose this is delta naught, then delta naught cross i is this one, okay. This will be your b naught and this will be c naught and this will be gamma naught. Now here in this one, this is delta 1, this is delta 1 cross i, so this is b naught and this is b1, c naught and c1. This will be gamma naught and this will be gamma 1. No, this point is not given a name in this one because it is the other way. This, you can think of delta naught, this other delta naught, this gamma copy of delta naught. Gamma 1 will be here, so that is the, that is the charm here. So similarly for delta 3 also and delta 2 also, you can do that. What you have to do is, inductively, you first, first inductively define for each point cross i and I mean vertex cross i and then then edge cross i, okay, and then delta 2 cross i and so on. Edge cross i when you define, remember you are not going to divide the bottom space and the top space. You are only going to divide the vertical things, cross i at the centers, at all the centers like this and then you take another center here and then join them, cone construction. Okay, so this will be the prism construction, all right. For any simplicial complex k, if sigma from delta n to k is a simplicial map, sigma equal to E i equal to V i, then we shall denote sigma by just V naught, V 1, V n. So this notation also we have introduced earlier, okay. So this is for a simplicial map. If x belongs to mod s, namely s is a simplex where s contains this entire sigma of delta n, okay, this may be a sub complex. Then it may not be a sub complex, sigma could be any simplex also, okay. Sigma is a simplicial map, then we have this notation, but later on we will have this, but this notation is not available, no. So then x sigma is just the linear extension of V naught, V 1, V n with x, namely the, I have to describe what this means, x sigma on E naught, the first one will be x now, then V naught with the image of E 1, V 1 with the image of E 2, V i minus 1 with the image of E i, okay, by shifting. So we have this notation, all right, and extend it linearly over all the simplicial chains. So if rho is summation nj sigma j is a singular nj with support of sigma contained inside mod s, okay. Then for any x belonging to mod s, it makes sense to talk about x rho namely summation nj x sigma j. So x sigma j is a singular n plus 1 chain, this makes sense, all that I have to do is send x to E naught and the rest of them shift it, take the delta 2 phase, delta naught phase of that and there you define sigma j, so that is the meaning of this, okay, omit delta naught, then you have delta 1, sorry omit E naught, then you have E 1, E 2, E n, that will be a copy of your delta n, one lower dimension, there you take sigma j. So this will be, if x is a, sigma is an n simplex, this one will be n plus 1 simplex, singular simplex, okay. So now the definition of h is completed by induction again. The first thing is xi of, this h of xi naught is gamma naught extended c naught minus b naught. So this is what I am doing here, c naught minus b naught, then extend it by gamma naught. So c naught minus b naught is a zero chain whereas h of xi naught will be a one chain, what is that one chain? It is this to this one, sorry this to, this to this one up, gamma naught to c naught minus gamma 1 to c 1, okay. So if it is like this, gamma 1 to c 1 will be like this, sorry, gamma 1 to b naught will be like this, minus of that will be like this. So it will be this, this, all the way if you put an arrow here upward, arrow here upward it will be going up like that, okay. So that is the meaning of this definition here, gamma naught of c naught, gamma naught is a center, c naught is there, so you go up, gamma naught is center, b naught you are coming down, but there is minus sign, so you are going up, okay. So that is your one chain h of xi naught. Check that it satisfies 33 on s naught, you have to, you have to look at this far, right. H of daba, daba xi n, what is daba xi n? There is nothing there, xi naught there is nothing, but daba of h of xi n is there, what is that? The boundary of h of xi n will be c naught minus b naught, right. C naught minus b naught when you omit gamma naught, but when you keep gamma naught there is nothing, so it is c naught minus b naught which is just restriction of eta 1, the zero cell taking eta 1, it is b naught or c naught and zero cell at eta naught level is is b naught, this is a binary definition, so c naught minus b naught is c naught minus b naught. This part is zero, the boundary of h of xi n, boundary of this one, you omit this one to get the first c naught minus b naught, then you keep this one and omit this one, there is empty set, so there those things are zero or c naught again gamma naught minus gamma naught you get, you have to cancel out, okay. So, this is satisfied. Now, inductively assume that you have defined h of xi r for r less than or equal to minus 1, so that it satisfies 33, then actually defined on s r, r less than or equal to minus 1 by only defining h of xi r, sorry, h of xi n now, okay, r else we have defined, now we want to define it for h n. The moment you have defined h of xi r, h is defined on all of s r, you have to remember that, once it is defined xi r, it defined for all sigma in there, so it is linearly defined for all the entire chains there. So, we have already defined it for s r, r less than or equal to minus 1, now we want to define it for s n, okay. What do we do? We have defined for xi n, so h of xi n you define it as gamma n of c naught, c 1, c n at the top, okay and minus b naught, b n at the bottom, so this is precisely eta 1 minus eta 2 of xi n, okay, that is why we have taken that, minus h of boundary of xi n, boundary of xi n is an n minus 1 chain, so there h is already defined, so I can take h of that, you delete that part, okay, so that is a trouble making thing, you delete that one. So, by induction again, first of all boundary of h of boundary of xi n is, boundary of h of boundary of xi n, I am thinking, okay, is eta 1 star of boundary of xi n minus eta 1 star of boundary of xi n minus h of h component xi star boundary square of xi n, okay. So, that is equal to boundary of this whole thing namely eta 1 star of xi n and eta naught star of xi n, so that is boundary of c naught, c 1, c 1 minus b naught, b n, okay, therefore boundary of h of xi n, I am writing the same thing again, which is, yeah, this is boundary of h of boundary of xi n, you remember there was a, in the definition of this one there is a h of boundary of xi n, the boundary of this I have found out, now I am finding boundary of h of xi n, okay, boundary of h of xi n by definition it is gamma naught of the whole thing, so first you delete gamma naught whatever is in the bracket will be written down there, c naught, c 1, c n minus b naught, b n, b n, minus h of boundary of xi n and then minus gamma n you keep and then take the boundary of the rest of the, boundary of whatever is inside that is boundary of c naught, c 1, c n, minus boundary of b naught, b 1, c n minus boundary of h of boundary of xi n, okay, the boundary of h of boundary of xi n we have found out. Boundary of h of boundary xi n is boundary of this minus this which is just this factor, So, whatever is in the bracket, it is just 0. So, you get only the first line, the first one is eta 1 of xi n, second one is eta naught of xi n minus h of boundary of xi n, okay. You take this last term on the left hand side, then what you get is this formula, the formula for eta, this is the problem in who you want it. This will, but when you take this one, namely this last term here, take it on this side. So, the construction of this prism operator is over, okay. Remember, once this is done, the homotopy axiom is completely proved. So, let us go to the excision now, okay. Here we shall present a proof of the excision theorem. Recall that we have defined a subdivision chain map SD from C dot k to C dot of SD of k. SD of k was a subdivision, the baricentric subdivision of k, where k is a simplicial complex, okay. This idea will be now completely generalized here, okay. This was not all that great thing. This was used to prove Brouwer's left hand fixed point theorem, but the idea is very good. So, we are going to extend it to the entire singular chain complex now, okay. So, subdivision chain map at the singular chain complex level itself, and a chain homotopy D from S sec to S sec, okay of this SD with the identity map, okay. Simultaneously, these two things will be defined, okay. So, again this will be done inductively. First of all, if tau is a zero-chain, SD of tau, you take it as tau. That means each point is taken to itself. This was the case with SD also in the case of, in the case of simplicial complex, okay. All the vertices were taken to vertices. The extra vertices were somewhere taken. So, that, we do not have that trouble at all now. So, it is in some way, it is much simpler here. And D tau to be zero, okay. You start with this. This is the starting hypothesis. And S naught level. Having defined SD and D on n minus one chains, we shall now define them on n chains. First, let Xi n be the identity singular and simplex, just like we did it for the prism operator, okay. On delta n, let us first define SD of Xi n. SD of Xi n. Now, you see, this is a game. So, again, again it is some kind of repeating it. Beta delta n. So, there we took the center gamma, gamma of this one, gamma n. Gamma n was what? It is beta delta n shifted to half level. Okay, because there we are doing it x cross i. So, here we are doing within x itself. So, it is beta delta n extending SD of the boundary of Xi n. So, it is a cone construction over this one. Look at whatever it is doing on the boundary of Xi n, okay. Say SD of that one is defined. You take beta n extended by beta n of this. This will be the definition of SD of Xi n, okay. And D of Xi n is going to be beta of delta n of SD of boundary of Xi n minus Xi n minus D of boundary of Xi n, okay. See, this SD is common. This bracket, this bracket, SD of boundary of Xi n minus Xi n minus D of boundary of Xi n. So, take SD of that, okay. So, beta of delta n is going to be that. I am not very sure whether brackets are correct here. This bracket is correct. Let us see. D has to be n plus 1 chain. D of boundary of Xi n, okay. So, let us go ahead with it. Now for any n chain, n simplex, SD of sigma will be sigma dot of that. Sigma dot is the chain map induced at the chain level, okay. So, sigma dot of SD of Xi n, okay. So, this idea is similar to what we did in the homotopic theory. After defining for Xi n, we take, for arbitrary sigma, you take sigma dot of that. D sigma equal to sigma dot of D of Xi n. So, sigma dot depends on how defines how denotes the homomorphism induced by sigma on the chain complexes, okay. So, then extend them linearly over all the chains by taking the summation. Summation ni sigma i SD of this one is summation of ni of SD of sigma i. And SD of sigma i should be sigma dot of this. So, this also you can say it is sigma dot of this, okay. Notice that once we define these maps on the universal singular simplex namely Xi n, okay. Then the rest of the definition is forced on us by the functoriality. By functoriality, it has to be like that. So, that helps us to define the whole thing. Just like in the proof of homotopic invariance, this is a typical example of what Vajnani does in such situations and work nothing down. So, once again, I keep telling that if you know, if you have to do some existence theorems, construction and so on, you better understand how far it is free, how far what is the free choice and all that. Once you are, once you see for example, like in the existence of different solution of differential equation, there is a uniqueness part. The uniqueness part actually tells you what could be the map, what could be the solution, okay. Here is a case wherein because it has to be like this, okay. Because of functoriality, that helps us to define, okay. Cut it down and define it for only for Xi n, then you have done, okay, okay. To see that S t is a chain map, chain map means what you have to commute with, Dabba. Dabba composite S d of tau must be S d of tau of Dabba of tau. This is what you have to do. Again, we can do it by induction, okay. Induction on homogenous elements of degree n, then namely tau is a chain, n chain. If n is 0, there is nothing to prove. There is no commutative diagram. So, you have to start with n equal to 1 to n 2 0, okay. So, so you can assume that n equal to 0 it is done because for minus 1 there are obviously 0 0s here. Having proved this for n minus 1, we note that it is enough to consider the case again instead of arbitrary tau here, you just take for Xi n, boundary of S d of Xi n should be equal to S d of boundary of Xi n. If you prove that, then again by taking tau star of that, this formula will follow. So, then we have boundary of S d of Xi n is S d of Xi n, S d of boundary of Xi n, okay, minus beta delta n of boundary of S d of boundary of Xi n, okay. So, first take boundary of Xi n, then S d of that, okay. Then beta n you have to take and then the whole thing minus this boundary of Xi n has to be subtracted and then you have S d of that. S d of boundary of Xi n is one first term here minus beta delta n of S d of boundary of boundary of Xi n, okay. Why? Because I have already verified S d is commutes with Daba on lower dimension chains. So, here I am using that one. So, S d comes out and Daba goes in. Daba Daba is 0. So, this part is 0. So, what I left it S d of boundary of Xi n, okay, boundary of S d of boundary of Xi n, okay. So, Daba Daba is 0 is always true and this Daba comes out and S d goes inside, S d comes out and Daba goes inside because S d we have by induction hypothesis, okay. Similarly, now I have to show finally that D Daba plus Daba D is equal to S d minus identity. So, again you do it for by induction. A degree 0 there is no statement. A degree 1 this follows from the fact that D naught has been chosen to be 0. So, suppose we have proved the statement for n minus 2 chains, then you want to prove n minus 1 chain as a consequence boundary of D of Daba Xi n, okay. You could boundary D plus D boundary of boundary of Xi n because boundary of this part is 0. So, I can add and subtract here, okay. So, but this is now S d minus identity of boundary of Xi n, okay. So, that is boundary of S d of Xi n minus boundary of Xi n. See here S d and boundary I have interchanged, okay. So, this is boundary of S d of Xi n minus Xi n, okay. Thus boundary of S d of Xi n minus Xi n minus D of boundary of Xi n is 0. So, if you take it on this side boundary of boundary of something here, the boundary of S d of Xi n minus Xi n and this term comes here negative boundary of the whole thing is 0, okay. Therefore, this was an elementary computation. Therefore, now D Daba plus Daba D of Xi n is D of Daba Xi n first part. Then boundary of D Xi n I have written down here by the definition. So, first term is D of boundary Xi n. This term when you boundary of this one first beta n, beta delta n goes away what we have left it with the thing inside the bracket, okay. Then minus beta Xi n is there but then you have to take the boundary of the inside thing, okay, right. So, this will leave you just these two terms. The rest of the terms just cancel as you can see. The second and third term will be S d of Xi n minus Xi n will remain D of boundary of Xi n, D of boundary of Xi n cancels out, okay. This whole thing will be 0. So, S d of Xi n minus Xi n is left out. So, having proved this one, let us just write down whatever you have done so that we can use it as a quoting. A subdivision chain map S d and a chain homotopy d have the following property. What are the properties? First of all, if you have singular simplicial k, it will go inside S of S d of k. This S d will automatically take care that we are doing it inside the barricade subdivision of k if k is simplicial complex. If it is not, then we do not have to bother about it. So, it is generalized the old construction. D of S d of k is contained as S of S d of k. So, our new chain complex, our new S d, the subdivision chain map has the property that it preserves the simplicial maps. Simplicial singular subcomplex are preserved. It defines a chain map and a chain homotopy at the subcomplex levels also. And further, second part is S d induce a chain map on the quotient space also because of because of on the subcomplex it is there on S dot of mod k to S d of mod k namely which is S dot of mod of S d of k. This is the subcomplex chain. This is the full chain complex. This is a simplicial one. Then you quotient out, you get the just a simplicial chain complex. This is singular simplicial, this is simplicial chain complex. So, all these things are now generalized and one single S d gives you all of them when you respectively take the quotient and the sub. Okay. If tau from S d of k to k is a simplicial approximation to the identity map, then on the chain complex level, we have tau star composite equal identity. We proved it much harder in the earlier case. Now this follows by because of the chain map. So, let us stop here and next time we shall prove the excision, basically excision and a few things which are left out. Thank you.