 Last time we introduced the simplicial homology of a simplicial complex as a sub-question of the singular chain complex. We stated the theorem namely the inclusion map from singular simplicial chain complex to the entire chain complex of singular chain complex induces isomorphism on the homology. Similarly, the quotient map from the singular simplex is to the ordinary simplicial homology that will also induce isomorphism on the homology. Together if you take P inverse composite I star that will be an isomorphism from the homology of the simplicial homology whatever you call to the ordinary homology of the singular homology of the underlying topological pairs. So this has many many consequences now so we will continue the study of this one today. First of all the comment that I made last time I would like to repeat it namely when you take a triangulation the simplicial homology will depend upon triangulation as such. However once we have established the above theorem we see that the simplicial homology is functorially same as the singular homology of the underlying space. Therefore it is independent of what triangulation you have taken. The only thing that we have to ensure that an arbitrary topological space may not have a triangulation if it has a triangulation then we can choose any triangulation and then compute the homology with the simplicial homology so that will give you the singular homology as well. So this has a tremendous application now so one by one we will try to recapitulate some of this. The first thing is c dot of k l is defined as a quotient complex of a free chain complex namely all singular simplicial maps however even the quotient is also a free chain complex. So one begins with a choice of an arbitrary order of vertices put an order on the vertices then each n simplex in k one can choose the non degenerate singular n simplex given by an order preserving map from delta n to the vertices of of k. So that one single thing there should be only one order preserving map after all. So that will represent all such collection call collection of all such things will represent a free basis for c n k. So that way we get that c n k is also a free abelian row if k is finite then all c n k will be of finite rank also. In particular if k is compact mod k is compact then we know k is finite so these are the few things which you have to keep in mind. Moreover what happens the boundary homomorphisms are given by just the incident matrices that is matrices centuries are either 0 or 1. You see so what you have to say is if a matrix is defined from if you take a set of basis say for n simplex is as alpha 1, alpha 2, alpha n and for n minus 1 simplex is as say beta 1, beta 2, beta m then whether alpha 1 is beta 1 is incident at alpha 1 or not depending upon that at the alpha 1, beta 1 entry or 1 1 entry we will put a 1. If it is not incident it is not a sub of this one you put 0. So that 0 1 matrix you know you can put a matrix for information for the boundary operator. So once you put that matrix boundary operator is completely determined. So this way you can computerize the entire information on the simplexial chain complex in the computer and if programmed properly in a single minute in less than 5 seconds it will give you the homologies. So this is the way computers have you know this is the only way actually computers have entered into topology to begin with sometimes that was done in 80s but now it has become a huge industry. The process of choosing the order on a K you have to be cautious about it because as soon as you choose an order it is your choice somebody else may choose something else and so on. Functoriality will be lost, computation you will arrive at the same thing no problem but the factoriality will be lost so in applying the theories you have to be careful ok. Process of choosing an order on a vertex says taking strictly increasing sequence of vertex maps defines a right inverse to the homomorphism phi star from s k s dot of k l to c k l this is a quotient map and thereby c k l and a sub complex of s k l ok. However one should take care to note that this identification is not factorial. You can identify it as sub complex but somebody else may identify with some other sub complex and so on. So there are problems of that nature so you have to be careful. We also note that s dot of k l plays an intermediary role from going from c k to s k we cannot go directly. So, s k l to s k I can go curly s to ordinary s I can go by inclusion from curly s to c I can come by come by quotient map ok that is why it is called sub quotient of this one. Suppose k 1 and k 2 are sub complexes of a simple shell complex k such that the union is is the whole of k since any simplex singular or not of k is a simplex in k 1 or a simplex in k 2 because k 1 union k 2 is the whole of k and they are sub complexes it follows that this is a curly s k 1 plus curly f k 2 is the whole of s k similarly c k 1 plus c k 2 is the whole of c k a non-degenerate simplex in k will be either inside k to k 2 or inside k 1 otherwise you do not get k equal to k 1 union k 2 as sub complexes. So this is much much stronger than saying that mod k is the union of mod k 1 and mod k 2 of course they it will be but if I just say that and do not say that k 1 and k 2 are sub complexes then it is not working ok for a sub complex which is stronger than saying that the total space is union of the subspace is here. Therefore in a very strong sense there is a accession here you see we needed open subsets and so on and accession theorem there you do not need anything you can this was the sub group of this one in general but it is actually equal in this case. So we are we wanted h star of this one to h star of that inclusion map in isomorphism but this is actually equal so h star would be identity isomorphism here. So the beauty is that immediately without any verifying whether something is accession and so on it is there already. We can apply mariovetry sequence we can apply all the consequences of mariovetry sequence to whenever you have two sub complexes spanning the whole of the whole of the k. So this will be very useful now both of them for curly s as well as the c. So as a striking and very useful information increases result of equivalence of singular and simplification homology in the following. Suppose you start with a k comma l a polyadral pair it just means that k is a simplification complex and l is a sub complex. If dimension of k is less than or equal to n ok so this is the only condition we are not assuming that it is finite dimension finite or finite dimension dimension of k less than n then the homology of this pair beyond n namely bigger ok they are all 0. So the homology survives only up to the dimension n and the nth homology is always a free abelian group ok why it is a free abelian group because remember h n is defined as some kernel divided by the image. The kernel is a subgroup of a free abelian group so it is free abelian. The image is 0 here because h n plus 1 c n plus 1 is is 0 there is no c n plus 1 here at all. There are no non-degenerate n plus 1 synthesis in a dimension which is less than equal to n ok. So this is the beauty of c n k rather than s n k ok h n of k l will be a free abelian group and h n of k this is 0 because the c n c n plus 1 c n plus 2 etc they are themselves 0 further if k is a compact polyhedron that is what I already told you h star of k l is finitely generated because the whole thing is finite number of cells you know the chain complex itself is finitely generated free abelian groups all of them ok. So this information though it is very easy now I mean this is not difficult theorem this will be very important in this context let us study what the role what is the role of the barycentric subdivision. Remember barycentric subdivision was used in the simple shell complex used to get the one of the very important theorem namely simple shell approximation. So let us understand whether barycentric subdivision has any role to play in homology ok. So start with a simple shell complex k and let s d of k denote its barycentric subdivision this little s d is denotes the subdivision now this capital s d I am going to define a chain map from c dot to c dot ok this is done inductively as follows at c naught level c naught of k is what free abelian group over vertices of k this is a free abelian group over vertices of s d of k what are the vertices of s d of k there are some additional vertices they are all barycenters of various simplexes in k. So they will all be vertices there so there are extra vertices here. So from vertex c here to vertex here there is an inclusion map extend it linearly to get c naught of k to c naught of s d of k. So this is just the inclusion of vertices ok the vertex set of k is a subset of vertex set of s d of k. So on c naught we define s d to be the extension of the inclusion map linear extension because they are abelian groups over this free this you know the basis as these sets. Now suppose we have defined s d on c n minus 1 ok then I want to define it on c n. Now s d is a chain s d is a abelian group homomorphism ok and we would like to have it as chain map or first let us define it as a group homomorphism ok. So any n simplex sigma in k if I define what is s d of sigma then again I can extend it linearly because c n is a free abelian group. So define s d of sigma to be beta sigma this extension you know you have done this kind of extension own construction s d of the boundary of sigma boundary of sigma is a chain which is in c n minus 1 n minus 1 chain. So boundary of sigma is defined by induction hypothesis ok s d of boundary of sigma is defined this is a chain now you extend it by beta sigma to make it a n chain this is an n minus 1 chain to make it an n chain so that is the definition of s d of sigma ok. So beta sigma denotes the barycenter of sigma here ok inductively we will have to verify whether this is a chain map it is a homomorphism that is what by definition but I have to verify that it commutes with the boundary operators ok. So at c naught level there is nothing to bother because beyond that there are zeros only ok so at c one level you have to verify starting with one. So let us just see how it looks like ok at n equal to 1 let us look at one of the edges e equal to u comma v ok that is a one simplex. We have s d of e by definition is beta e to boundary of e s d of boundary of e but s d of boundary of e remember it is just the identity inclusion map at 0th level it is just the identity extension of inclusion map so it is will be boundary of e one way beta of e boundary of e is v beta of e boundary of the other part is minus c minus of u. So this is what the definition of s d of e gives us on the other end boundary of s d of e after taking this one I have to take boundary boundary of this by definition look at this one you know you drop beta e and write v then drop v and write beta e but with a minus sign right that is the way boundary operator. Similarly boundary of this one is u minus beta e that is a minus sign so beta e beta e cancels away what you are going to be minus c ok. So what we have what is boundary of s d of e we have computed we have to compute s d of the boundary of e ok on the other end s d of the boundary of e is by definition s d of boundary of e is v minus u but at the 0th level s d is identity map it is v minus c. So these two agree so we have verified that it is a chain map from c one on c one to c not that part is a is fine that it commutes with the boundary operators. Now suppose you have done it for n minus one up to n minus one suppose you have done this job then I can verify it for n ok let now sigma be any n simplex then boundary of s d of sigma is s d of boundary of sigma minus by this is by definition and first line is definition ok minus beta sigma of boundary of s d of boundary of sigma s d of sigma was beta sigma times this one so we first drop beta sigma what you get is this one now you keep this one and then take the boundary of that one ok. So now this is same thing as s d of boundary of sigma but this one is beta sigma i is it is boundary of s d we have verified is s d of the boundary in the in the dimension one less therefore it becomes s d of boundary square of sigma the boundary square of anything is 0 therefore this drops out what you have left with s d of boundary sigma ok so that verifies for the n chain now. So what we have what is capital s d from c dot to c dot k to c dot s d of k is a chain map ok all this we could have done on s dot k as well ok there is no problem but however important thing to notice here is that s d is not functorial ok because what because of the barycenter taking barycenter operation itself is not functorial ok. So what is the functorial essence here if we have a simple shell map free from k to l ok we have a commutative diagram when you pass to c dot k to c dot l to c dot of s d k to c dot of c dot this is introduced by phi dot phi so the any simple shell map induce is say functor chain map at c dot of s d of k to c dot of s d of l whereas it will fail to do that if I take instead of c dot if I take s dot so this is the problem but here it is functorial ok so checking this one will be left to the reader as an exercise for time being so that you will carefully go through what is happening here now let us continue what is this good for this the barycentric chain map s d ok let tau from s d k to k be a simple shell approximation to the identity map see s d of modulus of s d of k is k is same thing as modulus of k is same topological space x ok so there is an identity map identity map can be simply approximated by just taking any subdivision any subdivision you can do all that you have to do is the extra vertex is there you should send it to the one of the old vertices within that simple shell complex within that simplex so such a tau exists all that you have to do is to take tau to be identity on the vertices of k but extra vertices become no that is what are extra vertices of s d of k they are barycenters of some sigma a barycenter of some sigma should be sent to some vertex v inside sigma that is all there are many choices but do not go out of a simplex the barycenter is in the center after all just move it on the on the bound to the boundary one of the vertices on the boundary that is it ok such a thing will be a simple shell approximation take any simple shell approximation by the very theorem we have the mod of tau is homotopic to the given map to which it is simple shell approximation therefore in this case mod tau from mod s d of k to k which is this is is to special same this map is homotopic to the identity map because it is a simple shell approximation to the homotopic in particular what we get is when you pass to the homology homology of this and homology of this are the same and tau star will be actually the identity map ok so this information has something to play with the chain map s d which is from this here to here ok on c k c k to this one so what if the relation that is what we have let tau from c s d k to c k with a chain map induced by tau ok after all tau is from s d k to k if you take c dot to c dot here you will get a chain map the chain map is ok we actually claim that tau dot composite s d is identity map ok so this barycentric subdivision is the right inverse to tau dot or barycentric subdivision chain map has lots of left inverses at the chain level the tau dot is one choice any simple shell approximation to identity would have given you that there are many choices for tau dot ok so this also we will do by induction on c naught look at c naught to c naught what s d did s d just was the inclusion map and what tau did tau was a retraction ok all the new things were sent to some old things on the old things there was no changes identity map therefore on s naught is clear ok or c naught it is clear all right suppose this inequality holds on c n minus k that sigma equal to v naught v n v n be a non degenerate simplex in k I want to verify tau of s d is equal to identity that is what I want to verify now what is beta sigma beta sigma is this barycenter where does it go under tau to one of these vertices I am I am putting it as vi I do not know which vertex ok because that depends upon the choice of tau then tau dot s d of sigma this is the right side here left side here tau dot component s d of sigma is tau dot of beta sigma boundary of s d of sigma by definition but tau dot is a linear extension of a simplexial map therefore it is tau dot of this into tau dot of that right the tau dot of beta sigma we just wrote down it is a vi the rest of this tau dot of s d of boundary of sigma ok why because here we have verified that tau dot of s d tau sorry yeah it is daba of s d is s d of daba ok that is what we have verified by induction hypothesis ok this is equal to vi of daba vi this is induction hypothesis next induction hypothesis this is by chain map so here it is induction hypothesis with tau dot of s d is identity all right so vi into boundary of sigma I am writing the formula for boundary of sigma it is minus 1 to the j sum summed over j v naught vj omitted vn ok now look at what happens if j is not equal to i ok then if j is not equal to i then that vi will be there twice so that will be a degenerate simplex so we can throw away all of them in c dot they are all 0 so the only term that survives is when j equal to i outside is a vi inside vi is omitted so they may be all distinct in fact they are distinct ok so but what is this one this vi omitted thing is in the i th place you can put this one back in here by a permutation but that will make minus 1 raised to i come again I will sign so minus 1 raised to i into minus 1 raised to i will make it sigma ok in the first place it is just vi if I put back into vi in this place then it is sigma so tau dot of s d of sigma is sigma so thus we have arrived at one of the important results here namely the subdivision chain map induces an isomorphism on h star ok how do you get this one how do you get this one tell me because s d ok is the right inverse of tau here when you pass on to the homology tau star composite s d star is the identity but we know tau star is an isomorphism so any right inverse of an isomorphism must be an isomorphism ok at this level there is no isomorphism as such this right inverse from the tau tau tau dot is not an isomorphism but when you pass to the homology tau dot star is an isomorphism being homotopic to the identity map it is actually the identity isomorphism therefore in any case the Paris centric subdivision see this is different chain complex this different chain complex but and the homology we have seen that they are the same homology they are homology of mod k right so s d star is actually the identity map in that sense but right now to see that this identity map requires a lot of identifications here what you know is this is an isomorphism ok and it is a canonical isomorphism because s d s d itself we have seen that it is a it is a chain level map which is canonical which is factorial ok so we have some exercises here one of this exercise is to verify that for the last one ok these are not very difficult exercises I hope you will try them before we proceed further ok so let us stop here today