 So, continuing the study of chain complexes, given a chain map between two chain complexes which are themselves exact, we need to dig a deeper as to know how the nearby components are related, okay. How the chain maps influence the nearby these graded components. This will prepare us with a powerful technique known as diagram chasing. What this means, we will see when we actually study this. And this is going to be an essential tool in what is known as homological algebra, okay. And one important, one very important result there, namely snake lemma, okay. Before stating the snake lemma, I will give you a number of simple shorter lemmas which are all proved by this technique namely diagram chasing. Consider the following commutative diagram of r modules and r linear maps in which the two rows are exact. If f1 and f3 are isomorphism, so is f2. This is the statement. What is the picture? Picture is this one. So, 0 to m1 to m2 to m3 is one short answer sequence. 0 to n1 and 2 to n3 is another short answer sequence, okay. So, that is what it is. Following commutative diagram of r modules and r linear maps, the two rows are exact. Column wise, we have three maps here f1, f2, f3. Of course, this map I do not have to write. This is a 0 map. This is also 0 map. So, I do not have to write it. The whole idea is that they are commutative means if you take f3, alpha 2, it is same thing as beta 2 composite f2. Similarly, f2, alpha 1 is same thing as beta 1 composite f1. So, this is the data, okay. Now, what is the conclusion? What is the further thing? Suppose further that these two are isomorphisms, f1 and f3 are isomorphisms, then f2 is an isomorphism, okay. So, this is one way relation. It does not say if and only if, okay. So, f1 and f3 are isomorphisms, then f2 is an isomorphism. That is the conclusion. So, it is f2. So, that they are already homomorphisms of modules. All that you have to say that this is a bijection, assuming that f1 and f3 are bijections, okay. So, let us prove this point wise. Namely, pick up a point in m2, show that there is a point in m2, you know, an element in m2 which comes to that. f2 of some m2 is equivalent. That is all, okay. So, starting with an element here, somehow I have to use these two facts, that f1 and f2 are bijective, in particular they are surjective. Start with an element here, there is no way to go here, right, but you can go here. So, pick up the element beta2 of that element. Now, you see something nice happening. This is surjective. Therefore, I can pick up an element here, okay, which lands on to that one, right. At least if I cannot pick up an element here itself, but after going here, I can come up, okay. So, there is an element here, such that f3 of that is equal to beta2 of this n2. Now, this map is surjective. So, there is an element here m2 which goes to this element. So, what we have got is an element here. If you come all the way here, it is the same thing as this element. So, an element here, if you come this way, the same thing as this element just means that if you come this way also it is same element. So, I have an element f2, m2 here, f2 of m2 will be an element here, n2, both of them are going to same element under beta2. So, now I am in a good shape. Now, I can use the fact that this is an exact sequence. Two elements are going to same element under beta2, just means that their difference is going to 0 under beta2. So, that is in the kernel of beta2, okay. The kernel of beta2 is equal to image of beta2, beta1, right. Therefore, the difference element f2 of m2 minus n2 or n2 minus f2 of m2 you can take, that is beta1 of some element here and that is anyway unique element because this is injective map. So, that is some element here, that is all. Now, this is surjective. So, I can pick up an element m1. So, f1 of m1, beta1 of that is the difference here, okay. Now, I can complete the nice thing. If you look at the image of that here, okay. So, what does that element does? Under f2, it will come to the same element here namely, this m1, f2 of alpha1 is the difference of these two, okay. So, you add that element to m2 that will become f2 of m2. So, modify this m2 by the image of alpha under alpha1, alpha1 of m1, okay. That will map under f2 to the element n2 that I have taken. So, that shows that f2 is surjective. You see starting with here, we went there and then observed that something is happening and then we went this way and modified this element. So, this kind of technique is called diagram chasing. This is the diagram. So, you can just glare at it, do this way, do this, do this. So, I have got the proof. So, without the diagram, writing all these homomorphisms etcetera, image etcetera is a bit difficult. In any case, you have to have this picture in your mind if not the picture on the paper. So, better to draw the diagram and then everything will be clear, okay. That is the point of this diagram chasing. The same technique can be used. Light modification according to the whatever you have to prove a lot of theorems. The next one, let us prove that next one also. Namely, why f2 is injective. So, what should I do? I take an m2 such that f2 of that element is 0, okay. I want to show that m2 itself is 0, right. That is the injectivity. So, come here. It will be 0 here also because this diagram is negative. That means m2 when come this way it is 0. So, alpha 2 of m2 in m3 has the property that under f3 it is 0. But f3 is injective. Therefore, alpha 2 of m2 is 0. So, the first thing is I wanted to prove m2 is 0. At least I have proved that alpha 2 of m2 is 0, okay. Now, use this part. If alpha 2 of m2 is 0, it is in the kernel of alpha 2 is equal to image of alpha 1. So, it comes from an element here. So, this m2 is alpha 1 of some m1 here, okay. But if you go this way and come here, it is the same thing as f2 of alpha 1 that is 0 to begin with. That means f1 of m1, when you take beta 1 it is 0. Now, beta 1 is injective. Therefore, f1 of m1 is 0. But f1 is also injective. Therefore, m1 is 0. Therefore, alpha 1 of m1 which is m2 is 0. So, I think first you have seen the surjectivity which might have taken a little more time. Now, you see the injectivities take less time. Once again what you have done is you are using both these diagrams or the neighborhood things whatever happening is used to conclude something is here. So, that was the theme. The isomorphism f3, isomorphism f2 induce, I mean influence this one. This has to be an isomorphism. There is no uniqueness here. There may be several isomorphism but it has to be an isomorphism, okay. So, based on this kind of techniques, we will prove several results. So, that is what you have to train yourself now, okay. So, I have written down the proof here which is a carbon copy of whatever I have told you. In your formulas there is no need to write. You practice writing this one. There is no nothing to worry about that, okay. So, I have proved first injectivity here and then the surjectivity. So, you have to correctly take the difference whether you take n2 minus m2, f2 minus m2 it does not matter but you have to correctly write these signs that is all. After modification by alpha 1 m1, this m2 will exactly map onto n2. So, that was the surjectivity part, okay. So, let us leave the next one as a exercise, as an assignment to you to work out. But let me state this. What is called as 4 lemma which is very, very important. Similarly, another one 5 lemma, okay. So, in the first case, there are only 3 important ones. The two other important ones were actually 0. So, this is a simpler case. The next one named 4 lemma is a one step ahead and it is a very powerful lemma. So, what does it say? I will not read this one at all now because it is easy to explain with the picture. So, there are 4 terms now, okay. May be non-zero, may be 0 and so on. There is no assumption on last thing being 0 and so on. And the whole thing is exact. This is exact and the diagrams are commutative. They are all R-module homomorphisms, okay. So, the statement is if, okay, f1 and f2, there are assumptions on f1 and f2. f1 is surjective and, sorry, f1 and f4 and things, f1 and f4. f1 is surjective, f4 is injective. This is the first standing assumption. Now, there are 2 parts. If f2 is injective, then f3 is injective. If f3 is surjective, then f2 is surjective, okay. So, you may get confused which one is what and so on. But for that, you may have to practice a little bit, okay. So, proof is exactly similar, but unless you try, you won't learn. So, you have to try this one, okay. So, the third one is, each statement is easier, okay. This is ready made thing. It is like a lab tech. That is like a tech. This is a powerful thing that will give you this one, okay. So, this is a ready made thing and this is very useful also. Here, there are 5 terms and the n terms are arbitrary. They are not necessarily 0, okay. And this whole thing is exact and what you have is commutative diagram R-module homomorphisms. If f1 and f2 are both isomorphisms, f4 and f5 are isomorphisms, then f3 is an isomorphism. So, this is a direct generalization, okay, one step generalization of the first lemma that we had in which the end things were 0. So, automatically this f5 and f0 here, they are automatically, sorry f4 and f5 here whatever, they are automatically isomorphisms. So, now I have made these ones arbitrary and put an isomorphism here and isomorphism here, okay. So, if f1, f2 and f4, f5 are isomorphisms, the center one f3 will be an isomorphism. So, this you can prove directly by chasing or you can use this lemma, okay. Now, I come to the central result of this section namely what is called a snake lemma, which is going to be a strong input in the homological algebra. So, this time you have a exact sequence like this and bottom it is the other way around, it starts with 0, here it ends with 0, okay, four-term sequences. Given a commutative diagram of homomorphisms, r-module homomorphisms wherein the two rows are exact, okay. Suppose you have such a thing, then there are two rows of then there exists r-module homomorphism delta, so this is a statement, there is a delta from the kernel of this last f double prime here, the kernel of this one is a sub-module of m double prime to the co-kernel of this map f prime. The co-kernel remember kernel is just set of all points wherein f double prime is 0, co-kernel is n prime divided by the image of f prime, that is by definition the quotient here, okay. Kernel to co-kernel of f prime that will be a connecting homomorphism delta, what does it connect? That is important thing here such that look like this one, kernel of f double prime to all this co-kernel of f prime I have put here, but there are obvious morphisms here kernel of f prime to kernel of f double prime, co-kernel of f prime to co-kernel of co-kernel of f f prime, these are just alpha and beta alpha prime beta prime restricted to the sub-modules. So you take kernel of f prime here, kernel of f here automatically because these are diagram commutative beta of a kernel of f will be contained inside kernel of f prime. So take the restriction, similarly take the restriction of alpha, okay. So these maps are alpha and beta here restricted, here they are the quotient maps of the corresponding alpha prime and beta prime. When you go modulo the image here that will map into the modulo of n modulo the image of f and when you go n modulo image of f that will give you a morphism from n double prime modulo image of f double prime, okay. So these are obvious thing whatever you want to recall first isomorphism, second isomorphism or whatever various isomorphism theorems in Abelian groups. So there are morphisms like that, okay. So these morphisms that I have not written they are obvious morphisms, they are here it is alpha and beta you can write, here you can write alpha prime bar beta prime bar induced morphisms. So the point is that you have a morphism like this and you connect them you get a six-term exact sequence. The entire thing is a exact sequence 1, 2, 3, 4, 5, 6 terms. Moreover the assignment the connecting homomorphism what is a naturality property. What is the meaning of that? Of course now we know how to express naturality property. So you must make a category and another category the assignment must be a functor. It is a functorial isomorphism. So what are the categories here I will explain later on first let us just this more this categorical part we will explain later, okay. Why there is such a map that is fantastic and then it is this part is exact namely kernel of alpha beta here equal to kernel of alpha etc those things are easy, okay because because they are there here and this is just a restriction this is an intersection and co-kernel is also easy, okay but why the image of this equal to the kernel of delta and the image of delta is equal to kernel of this alpha bar those are somewhat mysterious. So this is a fantastic theorem even the definition of this is fantastic, okay. But having seen the diagram changing as a technique the surprise will be a bit less now you will see how it is done, okay. So first look at this one the central one that was the big one thing here m prime to m develop prime n prime to n to n develop prime, okay. I am taking the kernel of f develop prime that is one of the important things if I put kernel appear, co-kernel appear this vertical arrow these will this will become a short exact sequence or a slightly wrong or exact sequence not a short exact sequence, okay this is an exact sequence kernel is is precisely equal to the image here, right because this is inclusion map after all and f develop prime image kernel is precisely this one is by definition I have taken here, okay when you quotient out n develop prime to the image of that that is obviously a surjective map because it is quotient map, okay. What is the thing that has been quotient delta precisely the image so its kernel will be precisely equal to the image of f develop prime the same thing here here, okay. So these two term homomorphisms I have made into a you know there are these things are now vertically exact sequences, okay. So the statement of the theorem says kernel of f prime to kernel of f to kernel of f develop prime then all the way come here co-kernel of f prime co-kernel of f to that is exact this six terms first three here the last three here exact sequences what else what we need to define is the homomorphism delta from kernel of f develop prime to this n prime, okay. So here you have to do some kind of diagram chasing but this time very systematically go down go left go down go left go down down back down back down, okay. So that I have indicated by double arrow what is the meaning of go down. So I have defined this map right from here to here I have defined so take an element go down here it is an element of m develop prime after all and this is map is inclusion this is surjective therefore I can pick up an element which goes on to that. So here I have to pick up something I do not have a homomorphism here that is why I have double arrows here the single arrows are homomorphism they are linear maps. So I have to pick up some element here which goes to that one, okay we have picked up now push it down here come down to n if you look at the image of that it is the image of this but that is zero because it is coming from kernel of f develop prime therefore this element when you start with the pick up something and push it down that element comes to n double prime zero therefore it is in the image of alpha prime so pick up an element here this time there is no ambiguity because this alpha prime is injective so I get a unique element here but ambiguity is here already to push it down that is in the image of alpha prime so I get an element here now go model of the image of m prime to this one namely go to the quotient go kernel of f so you have landed from an element here you have landed the element here there is no ambiguity in this one in the sense that there is an element here you can get an element here the point is there is a choice here so you do not know whether this will be always the same element so somebody else picks up will we get the same element that is the first surprise that yes you will get the same element no matter who picks up what element here so there is an ambiguity here only from here to here there is no ambiguity when you pick up some series of beta so what is the ambiguity ambiguity is precisely take the difference of you have picked up two elements right I mean one person picks up one element another another element or you must have picked up two different elements at two different time the difference would be 0 under beta the difference beta of that difference will be 0 because both of them are coming to same element therefore the difference element comes from m prime so it is alpha of m prime now if you come to n prime here alpha prime of that element will be the difference image of under f the difference of these two you understand what is going on there are two different elements come here and you have picked up two elements here okay this element goes to the difference here so it comes to the difference of those two elements here so the two different elements here are going to same two different elements here alpha this whatever element m prime here f prime of that must be the difference if f the difference is coming from here they will have the same image here in core kernel of f prime they have the same image because their difference is coming from m prime okay so whatever element you get here there is a unique element so call that delta of this element so function delta is already defined okay so you see with the picture it is easy to explain then writing alpha beta and then you do not know what what you have written down but you have to write down also okay now the same thing I will use to show that this delta is a homomorphism this time I will use the fact that picking up element which goes to the same element here okay you have the freedom you can pick up which element you like okay so to show that this is a homomorphism what I have to show so some alpha times this goes to alpha times that under a modular homomorphism and some goes to the sum right these two things I have to show so multiply by a scalar okay start with an element here multiply by a scalar pick up the original thing multiply by a scalar that will go to zero that will go to the same element so each time you can multiply by a scalar multiply a scalar and so on so that part is easy okay let us look at the sum suppose you have x and y two elements are here in the kernel of f prime you come here okay in m double prime and pick up say m1 and m2 going on to that one don't do that okay okay do that m1 and m2 are going what you will pick up for the sum you just pick up the sum m1 plus m2 would have gone to the sum of these two okay when you come here it will be again f of m1 plus m2 would have gone to the sum of those now these two elements are are alpha prime of something alpha prime is a unique I mean is an injective mapping there are two elements which go to correspond two elements the sum would go to the sum so here also some would go to the sum okay so here instead of picking up some element which goes to the sum since I have picked up elements for each of them I will take the sum as the one which goes to the sum so I use that liberty and then everything becomes okay so delta is a homomorphism of our modules all right so you have got this picture so all the proof I have written down there may be typos or one that doesn't matter okay so verification is now of exactness of the six terms so so let us go go back to this picture there is a picture this picture the verification that this is exact you have just take the intersection of these three three sum because because we have know that this is exact here similarly beta bottom if this is exact when you are taking quotient the exactness is easier so the most important thing is exactness at the kernel of f prime from here to here the exactness at co kernel of f prime at these two things you have to verify the kernel of delta is equal to image of beta the kernel of the image of delta is equal to co kernel of alpha bar these two things you have to verify again the proofs are identical I will prove only one of them whatever you want to okay whatever you want to they are all similar so let us say let us say kernel is in the image here okay let us prove the kernel is in the image of beta restricted to this kernel of f so go back here how we have defined this one so delta when you come here it is zero some element is here delta of that in co kernel is zero what is the meaning of some element is in the co kernel is zero in the co kernel it is in the image of m prime because this is quotient by m prime so this element delta delta is different image of that under this one that is zero means it is f prime of some element here okay so take that element if you come here okay what happens to that if you come here these two it is same thing as image of this one which means that the element here f of alpha of that is the element chosen here okay there is an element alpha of that one and the the one which I have chosen here both of them are going to the same element here that is the meaning so the difference of these two would be going to zero here that means it is in the kernel of f this f is kernel so that is that element is here okay so it is you can pick up an element here which goes to the difference here okay now look at beta of that I want to say that that is this element of course you have to correct sign okay so beta of that when you come here the same thing as this element right whatever I have started with because this composed beta of that is this element okay by the by the commutative of this diagram but that means what beta this inclusion map so beta of that is that element means beta here itself is that element that's all identity inclusion map followed by beta is that element means beta of that is that element because this is what we have start with say x here okay x here beta of that is x so this is beta this is because this is just electrician okay so this way every time what I am doing is going across this wherever diagrams and showing that whatever I wanted so other three things namely if you start from here that would be zero is what you have to show if you take an element here come here okay and you have picked up some element here we can take this itself image of this itself that would go on there but when you go to n it is zero because it is a kernel so once this is zero here it is zero here it will be zero here also therefore delta composite beta is zero that beta is taken from kernel of f so this completes the proof of that kernel of f double prime the sequence is exact so similarly I have to prove that co kernel of f prime is also exact okay so that is exactly similar again left to you as an exercise some part you have to practice now while there is one more thing I have to explain namely what is the meaning of the naturality of this map okay so look at this picture the data okay I want to make a category out of this every object of that category looks like this where m m prime etc are modules over r okay so such a thing I would give you a name it let me call it a snake so set of all snakes not set the collection of all snakes that is the object of this category so I am going to define a category which is called snake what are morphisms you know how to do all that you take another picture like this then m prime to say m m prime 1 m prime 2 and so on so I have done that one here so whole lot of commutative diagrams that is a one single morphism in this category this is where so m prime m m double prime n prime n n double prime is one single object the other object is p prime p p double prime q prime q q double prime is another object from one object to another object okay one is backward and forward like this a morphism will be set of morphisms like this okay phi prime phi phi develop prime psi prime psi psi develop prime so six morphisms will make one single morphism in the category of snakes the important thing is that all these squares these square these squares must be commutative okay so entire box here okay there are two boxes here so they are all commutative so from wherever you start if you come here there are two different ways of going from here to here and here to here from here to here here to here so they are all here to here this way this way so all of them must be commutative whenever there are two different ways of from one to the other both of them must be the same that is the meaning of commutative so this itself forms a category okay from this category we get a from to what category category of six terms this so this another category there are six terms here and the whole thing is exact connected by by one two three four five morphisms the whole thing is exact the exact six term exact sequences there is a so there is an assignment as soon as you have one snake you get a six term exact that's assignment if you have homomorphism from one to the other that will give you a one map kernel of f prime to kernel of whatever i have let us see in this picture the kernel of f f prime to kernel of g prime kernel of f to kernel of g kernel of f divided by kernel of g divided by and then delta and delta prime okay then co kernel correspond co kernel okay so that is a morphism and proof will be obvious from from what you are doing is just you know once you have this diagram commutative is kernel to kernel co kernel to kernel there are morphisms is obvious okay that doesn't need extra proof this is elementary module theory that's all okay so the new delta that you have defined is what is the connecting homomorphism also commutes that is the important one is delta another one is say let us call this as say delta p this is delta m m m p p p so delta p okay those things are also commutative is the important thing here okay so this is a picture okay correspond kernel of g is here there is a delta is a delta here i have just written delta instead of delta m and delta p and so on this diagram is also commutative is what you have to do so one single thing is important other things are obvious so you can verify that the if you the stunning thing somewhat a strange thing you may say that there are choices involved and yet this gave you a canonical map you have factorial thing so that is very important but the point is that the choice was choice was in one single place okay and it disappears as soon as you go here okay the as soon as you come here so though there is no map map will be final defined only here okay because of commutativity of diagrams so final thing will be the same under two different ways you are coming okay so all that is not the justification justification is just to verify this that is all down toward you verify that this diagram is commutative okay the checking is straightforward because we have done enough of diagram chasing something which you have to do on your own so i am leaving it back to you okay so the student is advised to go through the definition of connecting homomorphism delta in the above lemma carefully and memorize it why because despite the so much of developments in theories and so on while dealing with the connecting homomorphisms quoting theorems and lemmas may not help so you have some particular element the other element the image of that whether it is plus or plus of this or minus of that you will never be able to determine unless you chase the diagram i have got this one while doing my own research work okay so by telling you that sometimes you have to actually see the definition of that to prove a certain thing to prove a certain formula so better to know how this is defined so it comes in many other things so at this level it is the simplest so better clarify this completely at this time okay thank you