 So, so far we have introduced various notions such as graded modules, chain complexes, short exact sequences, long exact sequences, diagram chasing technique, then using that 4 lemma, 5 lemma and then the big result namely the snake lemma, ok. The snake lemma is going to be a good feedback, going to give you good application, very soon we will see that. So, guided by the experience in complex analysis or in differential equations and so on studying differential forms and so on. So, we are led to measure the deviation of a chain complex being exact, from being exact and that is precisely the role of this homology modules, the homology groups, ok, associated to a chain complex. By the very definition, a chain complex has an operator daba, self operator of degree minus 1 such that the d square is 0, which just means that the kernel of d or daba whatever you want to say, kernel of daba contains the image of daba. So, how far the difference between the image and the kernel is what we want to measure, one being a subgroup of the other or one being a submodule of the other. So, a nice way to measure this thing is to take the quotient and that leads to the definition of homology groups as a graded module now, ok. So, let us make formal definitions, start with a chain complex C star, the homology groups of C star to be a graded R module, ok. So, denoted by h star of C, which is the direct sum of h n of C's. So, each h n is a module and I am taking direct sum over that. Where C star itself is graded over that, C star itself is direct sum of C n's, right. So, what are these h n's I have to define? After that, h star is defined as a direct sum, ok. So, h star of C star, h h n of C star is defined as kernel of daba n by the image of daba n plus 1. Remember daba n plus 1 starts from C n plus 1 to C n and daba n starts from C n to C n minus 1. So, the kernel and image are both submodules of C n, ok. So, the image of daba n is contained inside kernel of daba n. Therefore, the quotient makes sense and this will be a again an R module. Ok, define this one for every n like this and take the direct sum that is the entire homology group of C star. So, what we have done is the homology groups associated with chain complex. The beauty of this is that you can make it into a functor. So, that is our next object here. If you have a morphism from C star to C star prime another chain complex, remember a morphism of chain complexes consists of a sequence of module homomorphisms C n to C n prime such that they commute with the corresponding daba operators here, daba here and daba prime here. They are compatible with the structure, ok. Therefore, what happens is a chain map has a property that kernel of daba n will be contained inside the kernel of daba n prime. Image of daba n plus 1 will be contained in middle daba n prime. This is the consequence of daba composite f is equal to f composite daba, ok. And hence by the isomorphism theorems of modules or just for homology just for abelian groups, what we get is this f very f induces a graded homomorphism which I will denote by h star of f. h star consists of various homomorphisms now. You can write it as h n of f if you like from h n of C star to h n of C prime star. So, the entire thing I am writing h star in the obvious way namely, ok restrict f to kernel of daba n prime, kernel of daba n it will go inside that. Then quotient it by this one. So, in the image also you have to you can quotient by this one, ok. So, this gives you the corresponding homomorphism. The very nature of this is such that if you have another map from say G from C prime to C develop prime, ok. Then G composite f will have similar properties, ok. So, what happens is h composite h star of G composite f will be h star of G composite f star of h. Similarly, h star of identity is identity. So, this is what we get h star of G composite f f G I have taken g is this way. So, h star f and h star f. Thus h star will be a covariant functor from where from the category of chain complexes to the category of graded modules. When you pass on to the homology there is no chain complex now. We have lost the tabas here, ok. So, this this functor is called the homology. Homology associated to a chain complex only. Later on we will associate it to different things, ok. The simplest thing now to prove is the so called weak additivity of the homology functor. Namely homology commutes under direct sums. The homology of a direct sum of chain complexes is isomorphic to the direct sum of homology of chain complexes. It is easy to verify this one, you know, routine for direct sum of two chain complexes, ok. But then you can realize it that there is nothing is very special about two. The same argument will go through any number of components, infinite many components, direct sum, ok. Such verifications you will leave to you because these are just practicing with homology of with you know homomorphisms and abelian groups and so on that is all, ok. So, above property of the homology namely direct sum of chain complexes will you rise to direct sum of the corresponding homologies, ok. So, this is you can call it as additivity of the homology, ok. So, what we would like to do is let me let me go back to this chain exact sequences of where was the exact sequences of modules and so on of chain complexes like this, ok, of chain complexes like this C prime to C to C term, exact sequence of chain complexes, short existence of chain complex. Where in the middle one may not be the direct sum of these two so how is this related to the homology of this one and homology of this. So, this is what we would like to, can you expect that the homology will be also deleted by the same kind of exact sequences like this. In other words I am asking suppose you have this one then you apply homology here, then you get 0 to h star of C prime, h star of C, h star of C double prime 0, 0 will go to 0 anyway. So, would that be exact, what is easy to see that g composite f is 0 map of the chain complex here. Therefore, h star of that will be also 0, so composite of this to be 0, this composite this is 0, right, this composite is 0 and so on. But if this is exact will that one be exact that is not clear at all, right. This is a chain complex composites are it is when that will give you the corresponding composites are also 0. So, if this is exact means kernel of g is equal to image of f f f then it is not at all clear whether the homology will have that property. Indeed, let us go back here now. That is the kind of thing we want to study. Let us go back here. Okay, so this is the question one we want to ask namely do we have h star of C dot isomorphic to the direct sum of h star of C prime and h star of C double prime. That is too much at least do we have a corresponding short exact sequence of homology modules. Okay, the answer to both of them are negative in that strong sense. But will you go a little relax a little bit then you get a beautiful correct answer. Well, correct answers are always beautiful and that is given by application as an application of the snake lemma. Okay, so this is the kind of at you will be interested in however the answer in general is no. So, we are forced to refer to the proper homology groups property of homology groups as in theorem 3.1. Okay, weak additivity, the best thing that we can say about this activity of homology from follows from theorem 3.2. The theorem 3.2 is the snake lemma. So, here is the sorry, see here is the theorem 3.2 that we are going to present here now. This is the best thing that we can solve. So, what is the meaning? What is the statement? Given a short exact sequence, now I am writing down the maps also here alpha and beta, maybe earlier I wrote it as alpha f and g. There is a short exact sequence of chain complexes. Remember, this means that there is a whole lot of chain, these short exact sequences of modules here, each c prime n, cn and c double prime n and there is a commutative diagrams is alpha and beta, alpha and beta and etcetera will commute with the boundary operators there. So, that is the meaning of this. Once you have such a thing, there is a factorial long exact sequence of homology curves and it is denoted like this namely dot, dot, dot it starts from somewhere because this n is going from minus infinity to plus infinity, right. So, at nth level what happens? hn of c prime, the corresponding hn of alpha here to hn of c then hn of beta, this is, see this is exact means this is exact, okay. But from here where do you go? It is not exact here, this is the point, okay. So, hn of c prime you have landed but next thing will not be 0. So, what you will get? hn minus 1 of c prime, okay. So, hn minus 1 of c prime. Then the three terms again repeat namely hn minus 1 of c tau, hn minus 1 of c prime, c double prime. Again this will go to hn minus 2. So, that is the meaning of this dot, dot, dot. That is the meaning of this dot, dot, dot on the previous. So, this long exact sequence, this long sequence is exact. That is the meaning of long exact sequence of homology curves and then there is this factorial one. Now, is it easy to guess what the meaning of this factorial means? If you make this one as a category which we have already done, category of short exact sequences of chain complexes, then there is a functor to the long exact sequences of modules and module homologisms. What is that? H, hn, hn, it is a way it is done, okay. And that is exact and association is a functor, okay. So, this is a consequence of the snake lemma now. So, the snake is appearing here, you see. So, this is going to be our delta. The corresponding we had delta there also. We are not going to distinguish and introduce too many notations. So, delta was there to remind you that this is also something to do with that we are putting the same delta here. So, just to recall what was the in the snake lemma, let us see what we had got here. From m prime, m to m, this was a snake and then we had thernal of f prime to kernel of f to kernel of beta and then the snake delta to co-cernal of f prime, co-cernal of f and co-cernal of factorial prime. This is 6th term, no 6th term exact sequence. So, this was the snake there and the association was functorial, this is functorially. So, one sequence another sequence there is a morphism there. So, this is what we had verified. Now, we will use that to prove this theorem, which is not at all difficult, but you have to guess how to get these terms out of this kernel and co-cernal. So, that is an elementary module theory or elementary obedient group theory. So, chain maps homomorphisms there, remember alpha and beta respectively induced chain maps of quotient modules. I am not directly going to the homologies here. So, take c prime star, I mean c prime n, if you want to put n everywhere you can go and putting it, c prime star divided by the image of daba prime. So, this is the co-cernal of daba prime. c star by the image of daba, similarly beta bar from c star of image of daba, c star by image of daba to c prime, c double prime by image of daba prime. These are co-cernal of correspond image of daba prime here, image of daba here, image of daba prime, image of daba here. Also upon restricting to kernel of alpha kernel of daba prime kernel of daba you can just restrict alpha, you can just write alpha itself or alpha prime I have written alpha prime and beta prime with the restrictions of kernel to kernel. For each n we then have a snake as follows. Now, out of this one single sequence I am writing a snake now. From the snake I will get a six-term sequence. Putting all the six-term sequences together I will get the long homology except sequence. So, these are the two steps there I have to do. So, how to get a snake first step? Okay, so this is a snake. Remember snake had one zero on the top and another zero at the bottom coming from here. So, this is m prime, m double prime, n prime, n n double prime in the snake, c prime by image of daba prime, c prime by image of daba n plus 1, c develop prime by image of daba n plus. These were the co-cernals of some other other steps. These were the kernels of some steps. Okay, now we apply the snake lemma to that. Take the kernel here, kernel here, kernel here, co-cernel here, co-cernel here, co-cernel here and see what are those six groups. They happen to be the six terms which I am interested in namely these six terms and they will be connected by this delta. Okay, so the snake always gives you the six terms sequence. So, notice that the kernel of this map daba n prime, this map, this is induced by daba prime. Okay, this is induced by daba prime, daba prime is cn prime 2, cn minus 1 but it has gone into kernel. But this part goes to zero in any case under daba. When you compose with this one, this goes to zero. That is why this map is defined. Remember that. It is a kernel, the same thing as kernel of daba prime divided by the image of daba n prime and that is why it is the homology corresponding. This is nothing but kernel of daba prime divided by my, it is a hn of c prime. Okay, there must be a prime here, sorry. Okay, so it is isomorph to kernel of daba prime by the image of daba n plus 1 prime. Okay, this daba n plus 1 prime here. Likewise, this will be hn of c daba n prime. And what are the co kernels of this? Okay, the co kernels are also, you can just look at, what is the meaning of co kernel? You have to take this one. Okay, image of this, it quotient out by the image of that. The image of this here is same thing as image of cn prime, which is nothing but image of daba n prime and divided by the kernel of daba n minus 1. This is the hn minus 1 of, hn minus 1 of prime, hn minus 1 of c prime, hn minus 1 of c, hn minus 1 of this. Okay, so put them together. What you have got is this part, but this is true for every n. And what happens just behind this one that is precisely the same delta at one stable over. So all the hard work is done by the snake lemma. So here we are just reaping the harvest. Let us now introduce an important concept here, which is guided by the homotopy in the topological spaces. But how you got this one? You may wonder, but to understand this one, you have to wait or you will understand it if you have done something like Poincaré lemma in your differential topology. Okay, so these things were cooked up from experience Poincaré lemma, for example, tells you what you have to do. So chain homotopy is purely algebraically defined here. So if you just see it like this without knowing all these things, you will lose the motivation why it is done. So you have to just cook up, you have to just mug it up till you see a little more and then you will see oh, these two things are fine. Okay, given two maps module graded homomorphisms of degree zero between two chain complexes, we said they are chain homotopy. If there exists a graded homomorphism D star from C star to C star prime of degree one, okay, these are degree zero maps, this is degree one map such that so capital D composite DABA plus DABA prime composite capital D is F minus G, okay, it is easily verified. When I say easily verified, there is work for you to do, okay. The chain homotopy, D star is a chain homotopy. If there is such a thing exists, then F and GR chain homotopy, okay, so this chain homotopy is an equivalence relation. The important thing to see that is F is homotopic to G, G homotopic to H implies F is homotopic to H, okay. So this is not at all difficult, you have to just see that it is an equivalence relation. The idea behind this definition would be clear when we consider the topological situation from which it has emerged. So that is why word homotopy is there, okay. The important property of chain homotopy that we are interested in why we have brought it here is the following bearing of this one in the homology. Chain homotopic maps induce the same homomorphism of homology groups. Once you have said this one verification is totally easy, okay. So you have F and G, you have to take the nth level what happens Hn of F. An element in Hn of C star is represented by an element in the kernel, represented by modulo the image. So take F of that, okay, modulo the image corresponding image there in the C prime star, right. You have to show that G is also like that, G of that element is also same. That is what you have to show, okay. So if you look at this one F minus G for any cycle here, cycle means element of something which is daba of something, okay. If some element is daba of something, when you put that element here, okay, this you don't know what happens but this part will be 0, d of daba will be 0. So it's daba of d of something. But if it's daba prime of something, therefore the difference this element is in the image of something so it gets killed. Therefore Fx minus Gx will be 0 when applied on an element X which is in the kernel of daba. So I have proved it for you anyway. So you write down the details now, that's all. So exercise has been taken away already. You just write it down. Let us stop here. We will pick it up from here next time.