 So, let me begin with the last remark that we did last time namely, given a surjective homomorphism from one group G to another group H, the first isomorphism says that the quotient group G by kernel of F is canonically isomorphic to H. Likewise, given a finite dimensional vector space V, you are told that its double dual V is double star is canonically isomorphic to V. Perhaps you are not told the true meaning of the adverb canonically, what is the meaning of canonically in these situations. So, I would like to explain this to some extent whatever nobody can explain fully after all using the category theory. So, for that I have to introduce another important notion in the category theory namely natural transformations and natural equivalences. The word natural is there or you can say canonical. So, that is what comes now. So, that is the topic today and later on we will even generalize this natural transformations itself. So, let F and G be any two functors from the same category C to D. There is a way of defining these if they are both contravenous functions also, but I am taking to begin with that they are covariate function. Exact similar definition is there for contravenous functions also. So, by a natural transformation of function that means a natural transformation from F to G, F and G are funcals from the same category C to same category D. So, this eta is a transformation of funcals which is also written with a twisted arrow, indication like this. We mean a head of data. So, what are they? So, that is what I am going to define. So, the entire thing you will be one single definition. For each object A of C, there is a morphism eta A from in the other category M D its morphism is in the other category from F A to G A. Remember A is an object in C then F A and G are objects inside D. Therefore, M D of that makes sense. So, for each A I am getting a morphism in this category. And this association is such that if F is a morphism in C itself from A to B then you must have the following diagrams commute. Namely, you have F A to F B, F A gives rise to eta A to G A and eta B F B to G B eta B. There is already F there is a map of A to B F morphism, F F is also there G F is also there F F and G F are related by this diagram. Remember F F and G F by the very definition they are morphisms from F A to F B in category D. Eta A and eta B are now they are in the category D. So, this entire map this morphism this diagram is in category D of morphism. So, eta B composite F F must be equal to this eta A composite G F. So, whichever way G F composite eta A you can write like that. If you reverse the arrows you will get for the same thing for contravenous functions also. So, this is the meaning of naturality. One more thing you have to one more definition I would like to give. Namely suppose each eta A is an equivalence in D. It is not just a morphism, but it is an equivalence invertible map these two are invertible. Then we say that eta is a natural equivalence of these two functions. Otherwise it is just a natural transformation. It is a transformations each transformation is invertible. Then it is called natural equivalence. When such equivalence exists that F and G are said to be naturally equivalent or naturally isomorphic. Sometimes when you are doing category theory you do not want to keep saying naturally canonically and so on just isomorphism. So, category theory is just the isomorphism it means they are natural isomorphism they are not just isomorphism here isomorphism there and so on. There will be commutative diagrams like this of isomorphisms. So, here is a quotation I had I do not know where I have read it it is not mine, but it is a quotation. I mean I do not remember where it is gone. So, for my colleagues etcetera it has become my statement now because I have been telling it since so many days. Now here is the example the second example which I wanted to explain which is easier vector spaces after all linear algebra you have learned. So, there is not much to explain here. So, consider the category k where k is a field you will define this one set of all, sorry the category of all vector spaces over k and linear maps. So, in that we can do something namely you can define v star as the dual space of linear maps from v to k this is just a notation I hope this is standard notation. Then the assignment v to v star itself is a contra variant from star. It is not just an assignment it is a contra variant from star on the same vector space to same vector space this category is the same. For example, you have defined what is a linear map and what is the corresponding association with linear map. So, take a linear map from v to w what is f star I have to do this star I am going to define the star is a construction here taking top star that is the functor ok. So, v will be v star w will become w star the arrow will be the other way around and that is some f star which must be linear map. So, you have to tell what is this f star is and then it should have the property that g composite f star is f star composite g star identity of star must be identity of corresponding star. So, these properties must be there. So, first of all I am going to define what is f star f star of v is nothing but p composite f ok. Now, both categories are vector cases c to d are both categories are f this capital F to be the identity functor and g to be the double dual double dual is what you have to compose this contravariant functor with itself that will be a coherent functor ok. So, you are taking the square of that instant contravariant functor the star again star. So, that is my g ok. So, g gives you v to v double star right. So, this is the contravariant functor actually I should write here just vector k to vector k ok instead of g v equal to v star I am just writing this one it is not a g of v is v star v double star that is the notation ok. Define a natural transformation first I want to say there is a natural transformation from identity map identity functor to this double star ok namely capital G this curly G. So, define a natural transformation ok what I have taken f to be the identity functor I just want to write it as f that is all eta from f to g as follows. Take a vector space over k we have to first define a linear map eta v from v to v is double star from v to v double star ok that is what you have to because this is f of v and this is g of v. So, between g of v and f of v I must have an eta v ok. So, this is done eta v operating on a vector v little vector v. So, this is capital v and this is small v must be a linear map on v star ok 2 k right. So, it acts on a function phi which is a linear map which is an element of v star v star itself is a map from v to v we just take phi of v I am sure that you have seen these things. So, you can verify the linearity part etcetera very easily v is in capital v phi is in v star and v star is what functions linear maps from v to k. So, it follows that f double star from v double star to v double star see this f star you have got what is f double star it is g of if under in under this notation g of f ok as the property that f double star of a is psi composite f star you have just verify this this combining with this one ok. So, it is just repeating this f composition once more once again ok f double star psi of f star ok. So, it is straightforward to verify that the following diagram is commutative v to w you have f then here is f star what is f star it is just g of f this f eta v eta w this is what you have to verify here in the definition remember this one. So, this f is identity here and g is taking double star. So, this commutative diagram I mean has to be verified that that is commutative ok which is very straightforward we have to take a vector v for f v what eta v makes. So, eta v eta of f v. So, you have to see that they are compatible ok. So, this is true in the category of all vector spaces so far I have not assumed anything about finite dimensionality. Now, I come to finite dimensions. So, there is a sub category of finite dimensional vector spaces this is a full sub category full means what all morphisms around vector k to vector k will be there in the smaller category also. Only the domains are now domains and co-domains are finite dimensional whose objects are finite dimensional vector spaces if v is finite dimension then we know that v double star is also finite dimensional. In fact, v star is also finite dimensional. So, the two functors star and again starting star namely g they are from f vector k f is finite dimension f vector k to f vector k the sub category to sub category they become funcals. Further we know that eta v is an isomorphism whenever v is finite dimensional. So, this is a linear algebra that you have learned I am not going to prove that I am going to explain the word canonical here this is an isomorphism we know right. So, what you might not have bothered about is that such a diagram is commutative. No matter what v you take you change v to w change v to w with any linear map the correspond diagrams are commutative. So, you do not have to worry about these transformations these isomorphisms. The same isomorphism you work for all f and all w that is the beauty of this way. So, we know that eta v is an isomorphism whenever v is finite. Therefore, eta is an equivalence of the two functors identity functor and the double dual functor. It is not just one vector space that you have got an isomorphism for all vector space together in a such a compatible way that is the meaning of the canonical isomorphism. So, this was just a technical just a verbose explanation of technically very precise statement namely there is a natural transformation which is an equivalence. So, category theory has achieved that. So, our next example is the word canonical occurring the first isomorphism theorem in group theory. This example is little more subtler little more difficult than the vector space case. There everything was built in very here I have to do some circus. Nevertheless, I appreciate this great thing it must be due to Neuthor. There are Neuthor isomorphism theorem and so on, but the first isomorphism theorem perhaps is not named after. Consider the category C whose objects are surjective homomorphisms from G to G prime of groups. What are the objects? They are not groups. I am making another category objects are morphisms which are surjective morphisms between one group to another group. These are the objects. You understand? How can you make it a category you are wondering? I am going to do that. So, I am going to do several such things you will see. So, this is only a beginning like this. So, objects are morphisms which are homomorphisms in this category need category of groups and they are surjective also I am assuming whose morphisms are morphisms must be what now we have to understand. One object is F from G to G prime. Another may be some little G from say here this what I am using F12 is G1 to G1 prime F1 G2 to G2 prime is F2. So, what is a morphism? Morphisms are pairs of homomorphisms alpha and beta such that the diagram is commutative. All of them are group homomorphisms. The top thing is one single object. The bottom thing is another single object. A morphism is a pair of homomorphisms not arbitrarily but they must have this diagram commutative. So, this is going to happen several times in later on we will see. The diagram commutativity is the essence of the whole thing. Note off because of commutativity of diagram if you take alpha of the kernel of F1 you take kernel of F1 here take alpha of that that will be contained in the kernel of F2. What is the kernel of F1? Some G1 which goes to triviality it will go triviality here also. Therefore, this way it will come to triviality that means just this kernel is contained in that that one under alpha. Therefore, alpha induces homomorphism alpha bar when you go modular G1 by kernel of F1 to G2 by kernel of F2. So, this is also theorem that you have studied in group theory. All right. So, there is a morphism for each alpha beta like that there is an alpha bar from the quotient to the quotient. This alpha bar will depend upon both alpha beta remember that but I do not need that notation usually people just write alpha bar here because it is given from G1 by kernel to G2 by kernel, G alpha is this way. On the category C, C of the first category whose morphisms are this one and I have given you what are the morphisms. So, we have two funtons now. What are the funtons? One I am writing it as Q, another I am writing it as I defined as follows. Q of any morphism like this F is G by kernel of F makes sense. Q of alpha beta for a morphism I have defined what is a morphism is alpha bar. Another functor I of F is the image of F take the image here and then I of alpha beta you take it as beta. Now beta has come into play. In this notation there was no beta. So, beta has come to play here. So, I have two funtons here. One is Q, another one is I. Q is a quotient, I is the image. This is quotient, this is going to be a functor and this is going to be another functor, both of them. This is a quotient functor, this is the image functor. The first isomorphism theorem tells you that there is a phi F phi depending upon F, there is this phi F kernel of G by kernel of F to G prime such that the following diagrams are commutative. The G prime is the image here. G by from G to G prime is surjective homomorphism. There is an isomorphism here. What is the meaning of that? G1 by kernel of F, G2 by kernel of F, kernel of F2, G prime to G is phi F prime, phi F2 prime are the isomorphisms making this diagram commutative. Look at now this fun tells you the quotient and this is the image. So, there are two, this is just one is the quotient functor, another one is the image functor. And phi F1 is the natural transformation which is an equivalence. I hope this explains the ordinary group here that you have studied. Let me now introduce one of the first deep results in category theory. So far whatever you have done are all easy parts. This is the first deep step you are taking in category theory. We are not going to do anything for other just a little bit of this one, just a jointness and then we stop there. Later on we will do something else. I would say that a jointness is the starting point of series category theory. So, let us just make a beginning here and then interested reader can pick it up from elsewhere. Like I have given that book, reference that book, that is a good book you can read from. There are many other books out there. So, a jointness. Once again I am going to define two things simultaneously, left a joint and right a joint. So, you can have a very vague picture of having a homomorphism then you have left inverse and right inverse. It is similar to that. Having said that the similarity end is there, this is much more subtle and much more stronger statement than that. So, F from C to D, G from D to C, these two are functor. Now once again I am taking only covariant functor. For contravariant functor, there is exactly same kind of definition, same kind of result, but you have to reverse the arrows. So, I am not going to do that. So, here they are both covariant functors. We say F is left a joint to G and simultaneously G is right a joint to F. When you say alpha composite beta is identity, beta is the right inverse and alpha is the left inverse of the corresponding functions. So, it is like that. Left a joint to G and G will be right a joint to F. How to remember? Namely look at this. There is a natural transformation, natural isomorphism, a natural transformation which is an isomorphism from one functor to another functor. These are actually bifunctor, double functor. What are they? There are two slots here, remember this here. There are two slots in each of them. This is C and that is D. So, this is a functor on the product category C cross D which I have not defined. All that definitions and etcetera, I will time consume. It is not necessary. So, you can I am doing it because I said I am not going to do it deeply, but if I have to do all that then I have to do all this systematically. So, home C, let us put a this slot here something namely some element from some object from category C. Some element from category D. G D will be inside C. So, both of them will be inside C. So, home C makes sense. Similarly, if this is C, the same C you have take, F C is an element, is an object in D. And this is D. So, these are objects inside D. A home D makes sense. These two are, both of them are sets. So, this must be a bijection from one set to another set. Eta must be bijection. Now, when I say natural transformation, automatically there are many other things built in here. See, first of all a bijection for every C and every D. So, that itself is a lot of data. But when I say it is a natural isomorphism, what is it mean? Let us recall. Actually, the definition is over, but now I have to explain what is happening. Just means that for every pair C D, with C and C and D and D, there is an isomorphism eta C D because now I have to write both C and D depends upon, if this eta will depend upon both of them. From home I have put C here and D here or home C G D to home D, F C D. So, G is the right. So, it is on the right side. F is on the left side. So, that is what it is. You could have written this bijection from here to here also. No problem. But that would be eta inverse. No problem. Such that I have not yet finished. It is just an isomorphism, such that what? Whenever you have a map from C to C prime and a map from D to D prime, morphisms inside C inside D, you must have a whole lot of commutative diagram. Once you have F here, home F G of psi will make sense. Except this is double functor. Home C F C psi will make sense. So, this entire diagram must commute. The most difficult thing here is this natural transformations like this. It has so much of data built in this one. So, unraveling this one is the difficulty. You have to practice it a little bit. Then you will see that you know you have told so many things in one single one single shot. So, I have put it here. We shall leave it to the reader to verify that any two left edge joint functions of F G are naturally equivalent. This is not a very difficult exercise. But the difficulty is that you may not be knowing what to do with this kind of things. You may be very well in computing you know 5022 equal to into 3092 equal to blah blah blah blah, immediately you may do. You are familiar with that kind of mathematics. You do not know this kind of mathematics. That may be one of the reasons why you have difficulty. Secondly, if I explain this, it will remain almost as difficult as it was. But maybe slightly less that is all because you have not spent much time on this one. And that is the reason I am leaving it as an exercise. Find out you know unravel this hypothesis. What it gives you in the special cases? What it gives you? You know we put special eyes something, special eyes something and so on. Try to do something. You cannot get something, come back and then read somewhere or maybe some other time I will explain. I can give it as a number of exercises to you later on assignment to you and so on. Right now, this is a remark which not even a well defined exercise. But I am imploring you try to work out this one. That whenever you have a natural transformation of two of them, two left adjoins for the same function will be naturally equivalent. Two right adjoins to the same function will be naturally equivalent. You try to prove that. So, what you have to do at least you must think. And then it will be easier for me to explain. Right now, let us give two examples of of these functors and then stop you know of natural transformation of adjoint functors and then stop. The first one is the set ENS to the app. We have function that assigns to each set S a free abelian group over it. So, what is the definition? FS is you must have noted this one. This is not a capital FS is usually got for free groups. But I am taking it as free abelian group now. I am assigning for each set a free abelian group over that. A right adjoint G from ab to ENS is got by just a forgetful functor namely what take the group and forget its group structure and look at the only the underlying set. So, that is a forgetful functor we have seen. So, I want to say that G is the right adjoint to F. Let us solve that is if you know what is an abelian group and what is a free abelian group you can work out this. The second one is little more subtle let us solve but here you have to know a little about commutative algebra tensor product and so on. Let B be an algebra over A like a polynomial algebra or tensor algebra and so on. An algebra over A makes sense. It is vector space or other kind of vector space but A itself is not a field it is just a commutative ring. Let A be a commutative ring and B is a A algebra. Let S from B mod, B mod is a category now. B is an algebra, B is also a ring. So, you can take the modules over that that is the category. You can think of this as a A module. So, typical example is R vector sorry a C vector space complex vector space can be thought of as a real vector space right it is like that. So, B mod to A mod with a forgetful function. You take a complex vector space forget this complex structure it becomes a real vector space. So, that is it is that is an example that is not exactly same here. So, B mod to A mod with a forgetful function that associates to each B module the underlying A module. Because B is larger ring than A if you scalar multiplication by B makes sense then scalar multiplication by A also makes sense that is the meaning of this. So, that is all. So, that is one functor. Then you consider the functor M going to M tensor B over A. Now, here B is thought of as a module over A as a left module and M is have to be right module matter is commutative. So, you can left and right you can interchange. So, you can make M tensor B over A. M is an A module becomes you know M tensor B becomes a B module. So, it is a left a joint this this morphism is a left a joint S and the functor N going to home B N. N going to home B N over A that means what all A homomorphisms from B to N. This is a right a joint to S. So, I have defined two morphisms one is B mod to A forgetting this one. Now, here N is an any module then it is home B N that becomes another vector. So, there are two of them the functor M going to M tensor B is a left a joint to S and the functor N going to home B N is a right a joint S is a forgetful functor. So, there are two of them here. Of course, the tensor product itself is left a joint S. So, S will be right a joint S that is not the what. So, this one is right a joint to S itself. So, this one will need you to understand what is a tensor product what is a algebra and so on. So, that is if you do not know that one I cannot explain this. So, it is only for those people who know enough algebra, but this one abelian group you must be able to do that. Verify these details. Next time we will define we will study some general topics of what are called as universal construction. This is one of the universal constructions by the way free abelian construct if free abelian group here that kind of thing in the categorical language. Thank you.