 But today we shall first complete the proof of pushouts using the existence of proof of the free products and then discuss one or two special cases of pushouts. First let us complete the existence of pushouts. So this is the statement of the theorem which will also recall what is the definition of pushout and so on okay. Let eta ij from gi ij to gi be a collection of homomorphism of groups. Let hv the free product of these domains gi's, the free product of collection of gi's together with homomorphisms, a free product always comes with the homomorphisms of each factor group gi to h. Let n be the normal subgroup of h, normal subgroup in h generated by the set alpha i eta ij, eta ij is from gi ij to gi then alpha i, so they are elements of h, alpha i eta ij coming from h but I may come from the other direction and alpha j eta ji of h inverse, these elements I would like to cancel out, so I take the normal subgroup generated by them that is my n for various i not equal to j, select all of them h varies of gi's okay and i and j are different take all such pairs in lambda, take the normal subgroup generated by that not just a subgroup, go modulo that call that as g, h by n is g and q is the quotient map quotient homomorphism okay and put beta i equal to q composite alpha i, so then I have this collection g, g is this one, gi's are the given groups, beta i's are from where from gi beta is from alpha i composite q, alpha is from gi to h to g, so this will be from gi to capital G okay, gi's are s as before, eta ij's are also s before okay, so this collection is a push out diagram, namely if you have another such collection right, then there will be a say g prime, gi, gi, j, beta ij's and so on, beta i prime and eta ij's prime, then there will be a unique homomorphism from gi to g prime making all these diagrams communicative okay, roughly speaking what you want to say is the following, take the free product ignoring these eta ij's, take the free product of gi's, go modulo this normal subgroup that will give you the push out, so that is the essence of this theorem, let us prove this, so this is the picture we have, h is the given free product ignoring these things, h has the property that if you have gi, gi's these are monomorphisms, if you have any g prime and beta ij's, beta ij's, etc., there is a unique map gamma here, that is the definition of h, I mean universal property of h, similar property for g we want to prove, what is g, g sitting again you see these are alphas by definition, these are the inclusion maps here okay, eta ij's are given maps okay, these beta ij's are defined as alpha j composite q, this beta ij, alpha ij composite q, it is the definition of this beta ij's, these beta ij's this fact, this diagram becomes the push out diagram is what we want to prove, namely start with a arbitrary g prime and homomorphism beta j's, beta ij's and so on, beta ij's, such that if you start from here, go like this or start from here and go like this, they must be the same, so these diagrams are commutative, then I must produce a unique homomorphism, which is I have denoted by dot, dot, dot, this is what I have to produce okay, ignore this part, apply the same property for h, I get a gamma here okay, it is a unique homomorphism, making these diagrams commutative, forget about this part, this g part wherever g is involved, forget about that, it is a commutative diagram, now look at this gamma, what is the property, if you take an element here, come here, an element here, come here they are same, but what I have done, this composite this inverse okay, I have killed them, I have killed them here okay, because they are going to same element here, that is the meaning of this one, if g i and beta ij are same, this composite this is same, so when you take this one, this all of them come here, this come here they are same, therefore in the normal subgroup the generators here, they are mapped to identity element by gamma, therefore the entire normal subgroup is mapped to identity element under gamma, in other words n is contained inside the kernel of gamma, therefore gamma factors down by first isomorphism theta, where is the diagram here, to a map gamma bar from g to g prime, because g is nothing but h by n by definition, so there is a map here, there is homomorphism here, automatically I mean it is defined by the commutativity of this one, this gamma bar is defined by gamma, gamma bar is q composite gamma bar is this gamma right, therefore once this diagram is commutative all the other diagrams will be automatically commutative because starting with that those things are commutative, if you come from here to here the same thing is going here, therefore coming from here to here, here is same thing as this one is same thing as this one and so on ok, therefore it is one line proof essentially, this gamma bar make this diagram commutative, why this is unique, the uniqueness follows because if it is fitting here then this composite this must be this one, but there is only one such map here and this is surjective map, then this is a surjective map, if there are two maps fitting here such as their composite is gamma, this is unique ok, so composite is gamma then they must be gamma bar must be equal to gamma prime bar, because this is surject, so that completes the proof of push up diagrams very easily by once you know for the free product ok, so we have taken a middle course namely free products it is a special case which immediately give you the general case, but there are other more special cases and so on, all those things are also quite easy once we have proved this middle thing, so let us do those things now ok, so we come to a notion called free groups, so first let us first make a definition which is similar to definition of this push outs and so on, but much simpler once again it is given by a universal property, start with a non-empty set ok, by a free group F which we denote by F S because it depends upon this S with S as a basis, so free group always comes with a basis ok, it is a part of its definition with a basis S, it is like a space X comes with a topology that underlying topology has to be there right, so with a basis S, we mean a group F which contains S as a subset and which has the following universal property, what is the property? Take any group H and a set theoretic function from S to H that function has a unique extension F hat which is a homomorphism from the whole group F to H, F hat is a homomorphism, F is just a set theoretic map and F hat is unique, there is no two homomorphisms this property, F hat of S equal to F S means F hat is an extension of S, such a thing is called a free group, the simplest case is one element, suppose S contains one element, then the infinite cyclic group is the free group over that one element, namely the generator there, infinite cyclic group has precisely two generators, I mean one of them, if T is a generator, T inverse is also generator, like in the additive group of integers which written additively one is a generator minus one is also generator, to write it multiplicatively T is a generator, T inverse is also generator. So, that T or T inverse one of them you take, then the finite, then this infinite cyclic group becomes a free group over that one single element S, why? Because, because of this property, I have just checked this property, namely take any group and send the generator S to any given element, that is the function, function on the singleton set S, that extends to a unique homomorphism from the entire what infinite cyclic group, how? All elements of this cyclic group are T power n, T goes to G or T goes to little h in h, then where do you send T power n to h power n, over, that is the only way you can extend it and you can extend it. So, this is the first simple case, it has this property that property has been extracted to define the free group over any set. So, I repeat a few things, the simple and enjoyable exercise to convert this into categorical definition, namely that F is an initial object in an appropriate category, let not do that one. It then follows here that a free group on a given set is indeed unique, the uniqueness as usual up to isomorphism of the set so that the set goes to the set projectively. That is a consequence of this definition of the uniqueness, there is a unique map as usual I will not go through that one, as usual homomorphism is unique, this part will take care of F itself is unique. Now, one can go through the construction of the free product and make appropriate changes and then get the definition of the construction of free group. Instead of that we will deduce it in a very simple step from the construction of free groups. The key is the infinite cyclic group generated by one single element there. So, this is what it is, that Sb asset, Sb wrong trace let us denote the infinite cyclic group generated by Gs. What is Gs? It corresponds to all elements S power n where n is an integer. So, that is a group. So, take Gs to be that group, you just have to notation. Then the free product of all these Gs is indexed over the given set S. This is a free group FS, each the free group FS. We have the definition of FS, we have the construction and properties of this free product, we will verify that it satisfies the properties of FS. That is all we have to do. Notice that each Gs is contained inside this free group, free product. That we know. Therefore, each singleton S is there. That means the set S itself is a subset of this group, the free product. Now take any function from S to H. Just now we have seen that each element S belonging to H, S mapped to H, F of S, there is a unique homomorphism FS from this infinite cyclic group Gs to H. This is the first stage, such that FS of S is this FS, not S. This is a type organ. This FS, F is the given map. That means the element of H. Now by the universal property of this free product, all these FSs get extended to unique effect from free product FS H. Such that when you restrict it to restriction means this inclusion map FS from Gs to H as well. It will be equal to this map, FS. So FS of S will be FS, wherever S is belongs to S. The uniqueness follows because free product has this uniqueness. So this proves that the free product over an arbitrary set free product of what? Of infinite cyclic groups over an arbitrary set S is the free group over that set S. This is one thing. The second thing of importance is what is called as amalgamated free products. Even if you do not use it, if you understand this one your understanding of pushouts and also of free products and free groups will be better. Therefore I am introducing this. Amalgamated free product itself is useful in three-dimensional topology. Right now we are not doing that kind of topology here. Okay. So this special case arises when all the Gijs in the definition of their pushout, they are the same group A, one single group A. Okay. All Gijs are equal to A. Okay. And all the homomorphisms Gij to Gij. Okay. These are all monomorphisms. Now there is no need to have double suffix here because they are all same A, A to Gij. Okay. You can put an A to I or you can even ignore that and identify A with subgroups of some particular subgroup of Gij for each I. You can think of collection of Gij subgroups and A is a common subgroup to all of them. So notation becomes very simple. All right. Now you have to do pushouts. This is not a free product now. But pushout makes sense. Right. Take the pushout for this special case that is called amalgamated free product. So free product is there but there is some identification. The A part is common to all of them. In the free product only identity element was common. All the identities were collapsed to a single identity empty sector, empty word. So there is only one element after all in any group there is only one element which is identity element. But we have so many groups each of them has an identity element to begin with. So they were all collapsed to a single empty word. Okay. Here you have to do a lot of collapsing because the A part is not collapsed. Okay. A will be a subgroup of the whole thing. So let us see how the structure of this amalgamated free product looks like. This is not difficult because we have understood the free product completely. Okay. So let us do that. So a notation for amalgamated free product is you use the free product notation but instead of writing all Gij and so on now I index inside you write that A there. That shows that A is common to all of them. This is the notation for amalgamated. If you have only two of them for example G1 and G2 then you write G1 star suffix A, G2. So that is the notation for amalgamated free product. Okay. So this diagram is the simple diagram now. Of course G i and G a there may be many of them dot, dot, dot here. Inclusion maps alpha i G i and this is a push out diagram. Means what? If we have another of G prime here and similar diagrams here then there is a unique map from G to that G prime. So that is the definition of push out the special case being this A being one single group and all these things are inclusion maps. Okay. So let us state a similar similar to the reduced rewards inside a free group. What is the equivalent of reduced rewards inside amalgamated free product? Let us state A and go through only a portion of this. I will not repeat all those things which you have done for free groups again. Okay. Then I will say only this is similar. Okay. So this is the statement. Start with a right concept representative for A inside each G i. So let T i be a subset of G i. Okay. Each element of T i represents a right concept of A and I would like to include the right concept A itself represented by the identity element. Do not take any other element of A. Any element of A would have been representative but there you do not have freedom. You have to take identity element and one belong into T i. So one is an element of T i in this. This is a choice. You are free to choose except that for A itself you have to choose one as representative. Okay. Suppose you have chosen such a representative. Then every element of G which is the amalgamated free product of G i's can be represented uniquely in the form and this form is called normal form in the following way. How? Take an element G. There will be some element of A. This may be trivial but I have to write A. This A is inside A. Then it is a product T 1, T 2, T k reduced to the words inside this T i's because I want it reduced. They are inside this T j i's but not one. The element one should be thrown away. Okay. That is why we have taken one here so that we can throw it away. If it is an arbitrary element of A we could not throw it away. That is the whole idea. The T i's are strictly non zero non identity elements of capital T ij's. These are T ij's are representatives of right cosets. Remember that. So take the reduced word in T 1, T 2, T k reduced words in that set. Every element G can be written like this. Take a reduced word and multiply it on the left by A but A is an element of A. Okay. Include this empty word. Then it will give you an element of A. After all elements of A are inside G. A itself is contained inside G as a group. So that is also allowed. So this is all. Okay. So here k is greater than or equal to 2 then consecutive elements here coming from different groups, different T i's. Okay. J i is not equal to J i plus means the ith and T i and T i plus 1 are in different T j i's and therefore they are in different j i's and therefore this one thing will be reduced. Okay. So that is the meaning of this one. The first part itself is not reduced to representation because if T 1 is already inside anything because A into T 1 can be combined, right? But that is the only way to get all the elements of G i. Suppose T 1 is in G 1. Then all the elements of G 1 I want to get there. Not just the representatives of the coset. So I have to multiply by A. Alright. So how to get this one? This very straightforward start with a G as a reduced reward. Okay. Take a reduced reward in G i's that we know. Okay. If two of them are in the same group you combine them and so on. You start with the reduced reward like that. In the free group ignore A. Just take the free group free product of G i's and take a reduced reward like that. Okay. Now what you do? Look at G k. It must be A k times T k for some T k. Yes or no? Because this T i j's are T a k's are coset representatives. So write G k last one as A k T k. Where A k is in A. T k is in one of any wherever it is, wherever G k belongs to. But should not be this one. If it is one, if it is one what happens? It is just A k. That T k is absent. Okay. Then I go to the next day. Now I take this A k, multiply it on the right by G k minus 1. What I have done is G k I have broken it into cosets. I represent a coset and an element of A. This A i combine with G k minus 1. Rewrite it as a coset. Right coset of A k minus 1 times G k minus 1. Like this you keep shifting this portion of A all the way back to this side. On the left side. Leftmost of this. Automatically these G i's will be, since G i's are in different G i's, what you get is T 1, T k, they will be in different case. So automatically this is where normal form. So once you know how to get reduced word, you know how to make it normal form. Okay. Question remains why this is unique? People say it is obvious because there is only one way you have to get. If you are satisfied with it, it is okay. Otherwise the same proof like what you have to do, take normal forms and define these allies from each G i into the permutations of all these elements. Do the same thing for each other. Then you will get the uniqueness of this one. Actually once you have done the uniqueness of free groups, this part, this algorithmic. Okay. So I am accepting this also as a proof. There is no choice here. Choice is what T i's are, you have to choose. After that there is no choice. All G k is given, T k and A k are well defined. There is no, there is no ambiguity in that. Okay. Getting the reduced word itself, you didn't know what it is. But now we have proved that one. So this is a proof. So final remark is that you can use the universal property of the pushouts to obtain many results. But often what may happen is you may get stuck up. You don't know what to do. Then this normal form will tell you. It's like a mother, it will help you out. Just like the reduced words will help you out. Okay. The same, they play the same word, the same role as reduced words in the case of amalgamated free groups. They are called normal forms. For example, using this one, you can immediately see that each G i is a subgroup of G. Go back here. If this starting with this element is already inside one of G i, then what will you have done? Look at the normal form. It is just the way you write inside G i A times some concept representative. Nothing else. And once you have uniqueness there, it follows that inside G i going inside this one is injective. Okay. Let us stop here. The next thing is putting all these preparations to produce. Emily do some topology. So you have done a lot of algebra and preparation of these things. Pushouts, various things and you know classification of G coverings and so on. Now we will reduce, reduce a lot of things about Aldermable group itself. That is our next talk. Thank you.