 So, in this chapter, we shall discuss the number of variants of singular homology. These include a simplicial and singular simplicial homology for simplicial complexes, CW homology and cellular CW homology for CW complexes and maybe one more also. So, we will see, we shall state how each of them is related with the standard singular homology that we have studied so far. But they are for special types of topological spaces, not for arbitrary topological spaces that you have to keep in mind. The proofs are all postponed to the last section. Each of these homology groups enhances our knowledge of singular homology and helps in computation in special cases. And it gives you some fantastic results about the topological spaces themselves finally. So, the first one is singular simplicial homology. The singular homology we understand. There we adjust continuous functions from delta n to at the topological space x starting point of the singular simplexes. But now we are going to replace it by simplicial. So, for that we need x to be not arbitrary x, but a simplicial complex. That is the whole idea. The singular chain complex of the topological space is too huge to carry out certain types of computations. At least in the case of a polyadron namely the underlying topological space of a simplicial complex, we can remedy this situation by introducing a subchain complex of S dot. So, the new thing that we are going to do is a subchain complex of the singular chain complex of Mod K which yields the same homology groups as S. When you go to homology, they will be the same. Same means what? Canonically isomorphic. There is a subchain inclusion map. When you pass on to homology, the inclusion induces an isomorphism. That is the final statement. So, we have seen earlier that simplicial approximation theorem remember that for a polyadron. Polyadron I mean a Mod K where K is a simplicial complex. The set of simplicial map is capable of capturing topological features of the space at least up to homotopy type. The simplicial approximation theorem just says that any continuous map can be replaced by a simplicial map up to homotopy. After that we have seen a lot of applications of this. So, this being so, it should be possible to have similar treatment while dealing with homology also. To be a little more precise, we can think of taking only a simplicial map sigma from delta n to K. See Mod delta n to arbitrary x, all continuous maps were taken. Now use the extra structure on x namely x is Mod K. So, instead of taking all continuous functions, you can take only those which are given by simplicial maps. So, you can think of sigma without the bar delta n to K, simplicial map and then pass on to the corresponding continuous function given by that. That is the meaning of simplicial maps. While forming the singular homology, you can just take that as a basis and take the group generated by that. That will be subgroup of s dot of x. So, if this turns out to be meaningful, then it would have same kind of advantage over the singular chain complex as simplicial maps over continuous functions. Continuous functions were too huge whereas, simplicial functions, simplicial maps could be actually write, you can write down and you can count them, you can unleash them. So, that is the whole idea. For instance, the chain groups become now combinatorial objects purely, you can put them in the computer and will be extremely handy compared to the ordinary singular chain group. This is what we would like to study now. Okay, so that is all pep talk or sales talk you can say. Now, we will come to brass text. So, just like we started with x comma a as a topological pair that K and L be a simplicial pair. K is a simplicial complex and L is a sub complex. Okay, now I am going to write down this Sn of x as with a, you know, double s you can say just to denote, just to distinguish this from ordinary Sn of x. Okay, so Sn of k be the subgroup of all, subgroup of Sn of mod k generated by all simplicial maps lambda from delta n to delta t. This is just taking instead of all functions, I can take only polynomial function or only linear functions here. So, that is what I am doing here. So, let Sn of k L be Sn of k modulo Sn of L, at least a sub complex of L. So, any simplicial map delta n to L would be also a simplicial map into delta n to k. So, this is a subgroup of that to take the coefficient just like we have defined Sn of xa as Sn of x modulo Sn of a. So, this similarly we have defined. Note that once again Sn of k L is a free abelian group. Okay, over what? Over all those simplicial maps from delta n to k which are not contained inside mod L. Okay, even if one of the vertex goes outside L, that will be inside Sn of k L. So, even though it is a coefficient, it is a free abelian group. Okay, if the base is consisting of those simplicial maps, that is that lambda delta n is not contained in L. Similarly, now take the same face maps. Remember, the face maps were linear maps. So, when you do boundary operator, which is the sum of these things, if you have started with sigma as a linear map, they will be also linear, linear in a sense, they will be also simplicial. So, daba denotes the boundary map of single chain complex. Then daba of Sn of k will be Sn minus 1 of k. So, this is the beauty of this boundary operator here, which is the sum of the face maps. The face maps are linear maps. So, this is also linear. Therefore, S dot of k, which is direct sum of Sn of k and Sn, S dot of k, which is direct sum of Sn of k form a sub chain complex. Okay, what are boundary operators here? The operator of degree minus 1, this is just restriction of daba, the same daba. That is there. And S dot of k L. So, this Sn of k is subgroup of this one, sub chain complex of this one. And the relative thing is sub chain complex of this one. So, what we have done is, only simplicial maps are allowed. Having, I mean, saying this one was easy, but the difficult task is to show that the inclusion induced map homomorphism in the homology is an isomorphism. That is not coming so easily. You have to have some patience. So, let us instead of doing that, let us get familiar with this new object first. So, compute a few things with this new chain complex. What is H naught of is curly S dot k for a connected complex k. Remember, a complex is a simplicial complex k is connected if and only if any two vertices inside k can be joined by an edge path. Edge path means what? Sequence of edges, one starting, one ending in where the second one starts. So, like that you have to take a finite sequence from one edge, one vertex to another vertex. If you can do that for any two vertices, then the k is connected. This part we have seen in part one. So, if you take that each one edge is a one simplex now. What kind of one simplex? It is simplicial map. So, you can take some of these that will be changed. What will be the boundary? Boundary of this minus boundary of that, boundary of this minus boundary of that and so on. All the middle vertices get cancelled. Only the starting vertex and the end vertex will remain. The starting vertex will come with a minus sign. The end vertex will come with a plus sign. So, what you get is boundary of sigma is V minus C where V and U are end points of this edge path. Okay. Therefore, you see that it is exactly same as in the case of the singular chain complex for a connected, for a path connected space. There also we had got the same thing. Now, you can use the augmentation map epsilon from S0 of k to Z, sending all the vertex maps. Vertex maps means what? Constant maps at vertices to 1. Then V minus U would have gone to 0 and that is in the kernel. So, all this if you do exactly the same way, what you get is H0 of computer H0 of H0 of S dot k of this chain complex which means for isomorphism to Z. Augmentation gives you an isomorphism. So, only thing here, there we took arbitrary pathes. But in a simple shell complex, you can choose a path to be simple shell. Some convenient notation now for going further. I need a little more conventions. A simple shell map delta x can be displayed. It is not like continuous function. Continuous functions, you have to write down formulas which may be very complicated. You do not know. That may be whenever you write a formula, it is more or less like you are writing polynomial function or trigonometry functions and so on. But here we have even more. We are linear functions. So, can be displayed by simply writing down the edges of the vertices EI of delta n under lambda in that order. Delta n is E0 even En. Look at the image of these. They will be vertices. Write down them in that sequence. Completely we will understand what lambda is over. This is what we are going to do. Thus an ordered n plus 1 tuple of vertices of k define a simple shell map if and only if all those vertices belong to a single simplex k. If it is a simple shell map, lambda is simple shell means what? Lambda of the entire delta n must be a simplex. The only thing is it may not mean n simplex. It may be 0 simplex also when all the vertices have mapped to the same vertex. It may be 1 simplex or it may be a 2 simplex and many other possibilities are there. So, entire thing must be a simplex. From n simplex down all the way to 0 simplex, all these possibilities are there. So, what happens is this map may not be injective. That means the V0, V1, Vn, these are the images of E0, E1 and En. They may not be distinct. So, this is what is the difference. We are not writing down in this notation, we are not writing down the subset simplex whatever wherein we did not repeat the vertices. Here we are repeating and this does not represent the image as such but if you cut down all the repetition then it is the image. Otherwise it will tell you exactly what the map lambda is. That is very important for us. So, let us keep this, understand this notation correctly. Where VIs are in some simplex F, F is a simplex of k and VIs image of lambda EI. It is equal to lambda EI. So, if you take any simplex F, it may have a number of vertices. Take a sequence in that which is a not necessarily distinct elements. Just a sequence V0, Vn, Vn, we like that. That itself will give you a singular simpler, singular simplicial simplex. Singular is the word for functions which are not injective in this sense. In particular, take k itself is delta n. Remember, this is like studying Dn. We have studied S of Dn and S of a sphere to begin with. So, I would like to study when k is delta n, what is this simplification? The standard n simplex. Then all the face maps Fi from delta n minus 1 delta n are simplicial and can be represented in the form what happens under Fi. What happens to E0? E0 will be E0. Up to EI minus 1, it is EI minus 1. As soon as you come to EI, it is shifting to EI plus 1 and En. So, this is your Fi. I mean as a function. Now, this denotes the function from delta n minus 1 to delta n. We introduce some convenient notation here. Instead of writing which one, so you can write, people write all of them except EI is there. So, put a hat on EI. It is like hiding EI. So, hiding EI or dropping EI, rest of them are there. This is a notation. So, where hat indicates that the corresponding entry is deleted from the sequence. That is all. In this notation, it is very convenient notation, boundary of V0, V1, Vn, this denotes the inclusion map of delta n inside delta n, inside some fn. In some particular order, which order I do not know, V0, V1, Vn I have chosen. Given any simplex, you can enlist the vertices in several ways. What are several ways? Different order, choosing order of the vertices. That is all. So, the boundary of V0, Vn will be nothing but I ring from 1 to 0 to n minus 1 to the I, V0, etc. Vi hat fn. It is the same thing as taking the map and then applying Fi. When you apply Fi, Vi will drop off. So, this is summation this one. This is the new interpretation when everything is simplexial map instead of Fi themselves can be included like this. This is like Fi summation Fi's because you are composing with identity map. Once you have this notation, I can now write down the chain complex of delta n itself. How many q simplex, q plus 1 simplex? How many q simplexes will be there? Tell me. Singular q simplexes. q simplex means now E0, E1, E0, E1, Eq mapped onto some into delta n. Delta n has n plus 1 element. E0, even Eq has q plus 1 element and there is no restriction, all maps you have to take. So, when you take that, you get n plus 1 raise to q plus 1 singular simplexial q simplexes in delta n. This may look huge, but compared to all continuous functions, this is God's grace. All continuous functions are uncountable. So, this is just finite set. So, these many simplexial ones correspond to the sequences of length q plus 1, no other condition with n plus 1 element. Just the chain complex of delta n will look like z raise to n plus 1, raise to q plus 1, z raise to n plus 1, raise to q and so on up to z raise to n plus 1, this corresponds to S0. This corresponds to Sq, Sq minus 1, etc. This is S1 and this is S0. And you know what are the boundary operators. So, given a singular q simplex sigma in delta n and any vertex v in delta n, we shall use. Now, I am doing this one we have done like this in getting the generators for SN, the cone construction we have used. So, I am using the same thing v sigma extension of sigma by this vertex, etc. v sigma denote this was a q simplex. So, this must be a q plus 1 simplex defined by v sigma i when i equal to 0 is just v starting point. Then v sigma i for i greater or equal to 1, it is v sigma of v i minus 1. In other words, v sigma 0 will be v, v sigma 1 will be sigma v 0, sigma of e 0 and so on. You have to shift sigma because the first 0th plot has been taken by v. Afterwards, you just repeat what is sigma ii. So, this must be sigma ii minus 1. You can then extend this notation. Once you have got v sigma where sigma is simplex, you can take summation. Summation sigma i v of that is nothing but summation ni v sigma i. So, that is the notation v sigma where sigma is a summation is v of summation ni sigma i is ni v sigma i. So, this is what we meant by cone construct here. It follows that boundary of v sigma exactly the way we have done in the case of that. See, we first v naught will drop out minus 1 raise to 0 into v will drop out and the sigma is left out. After that minus plus minus plus will occur, but what happens is the first v naught is there and then you are operating sigma on the rest of it. So, boundary of sigma. So, this is the formula for the boundary on v sigma. I am not putting another formula, I am just deriding this one. Dabba is already defined all the time. So, this allows us a simple way to construct a chain homotopy of the identity map of delta n to delta n with the chain map which is a constant. This theta s dot of delta n to s dot of delta n. So, what is it? Fix any vertex v naught belong to delta n. Take xi naught of v equal to v naught. See, I have defined this xi from for or each n and it must be chain map. So, xi naught of v equal to v naught and for all vertices v and delta n take xi k identically 0. See, this is a map not on delta n to delta n, but at the chain level. So, 0 means say makes sense. So, send all the other vertices to 0 for k positive. All vertices and all simplices everything to 0. Only one of the vertex at delta naught is selected to v naught, v naught. All the vertices here are going to become for all the vertices in delta n. S naught is collapsing to one limit. One simplex, two simplex, all of them are mapped to 0. Check that xi is indeed a chain map. In all other dimension it is 0. There is nothing to check. Only delta, only the at the 0th level there is one single constant map there too. So, now for each k positive, non-non-negative, we can take p, I want to define a chain homotopy between these two. p of s k to s k plus 1 of delta n to be the map p sigma equal to v naught of sigma is cone construction. Then what is p composite boundary plus boundary composite p? It will be identity minus xi. That is precisely what I want. p of boundary operating upon something and boundary of p operating on this one, some total is identity minus xi. You have to verify this. Look at p sigma is v naught of sigma. We have already got boundary of p is given by sigma minus v of boundary of sigma. So, this one is sigma minus v of boundary of sigma. And then you have to operate p of, add p of boundary of sigma, p of boundary sigma v naught of sigma. So, that will cancel out what we will have to order the identity minus xi. So, immediate consequence of this is that on homology the two induced homomorphisms are the same. The identity map induces the identity homomorphism on the homology. I have to look at what happens to this xi not xi star. So, then you have computed something. So, why they are equal on the homology? They are because they are chain homotopy. We have constructed chain homotopy here. So, since it is easily seen that xi star is actually zero map on the homology. On positive dimension the map itself is zero. The kernel is the whole thing. The image is the whole thing. So, the quotient will be zero. What embossing does it not level? What is the kernel? Kernel is any v minus delta, v minus any difference between v minus, sorry, v minus u, some total of each v minus u is connected by an edge. They are in the boundary. So, kernel is in the boundary. So, h naught of that one is also zero. Therefore, since it is easily seen that xi naught is zero on h cube for all we conclude that all the homologies h cube of s dot of delta n is zero for cube positive. Okay. But h naught of s naught of delta n, there will be just infinite cycling as usual which we have already computed. If identity map is zero, what is the meaning of that? The group must be zero. So, that is what I am using here. So, we shall stop here today and tomorrow or next time whatever whenever you need me, we will continue our study of these things and make it a better, make it even better than this one. Thank you.