 So, today we begin the classification of G Coverings. It will take some time before we complete it. Complete thing will not come today. The guiding principle here, the guiding or our guide here is the simply connected covering over the base space B. That just means that we are first of all assuming that simply connected covering exists. So, B is a path connected space. P from E to B is a simply connected covering. So, once for all this notation will be fixed. We are going to fix a base point B0 in B and E0 inside E above that. That means P of E0 is B0. Then this covering becomes a GP covering where GP is the Galois group of P, which is identified with the fundamental group Pi1 of BB0. Let us just recall how this identification is done. I am going to use a simpler notation J for Pi1 of BB0. Then this isomorphism phi from Pi1 of BB0 to GP. GP is the Galois group. Group of all covering transformations of P. How it is done? How it is done? So, let me recall this. Namely, starting with a loop at B0, lift it to a path in E at the point E0. Look at the end point. End point is in the same fiber. Therefore, there is a G covering, there is a G map, there is element of G at current transformation, which takes E0 to the end point of this path. And that G is unique to omega is sent to, the class of omega is sent to that covering transformation. This assignment is an isomorphism. This is what we have proved earlier. And we know that under covering transformations, this is the action of covering transformation becomes an even action. The quotient space is precisely B and quotient map is precisely B. So, these things we have already seen while studying the universal covering or simply connect the covering and so on. If we change the base point E0 to some other point E0 prime belonging to the same fiber, then the J action of E gets changed because this time you are lifting the loops not at E0 but E0 prime. So, corresponding change is just got by take the path from E0 to E0 prime inside E. So, that is P of omega will become a loop in B at B0. So, the entire action gets conjugated by this element and inner conjugation is an automorphism. So, the new action is nothing but got by an automorphic change that is what we have studied last time. On the other hand, if this P omega represents a central element, this omega is a path now inside E, P omega is a loop in phi 1 inside B at B0. So, if this is an element or central element in J, central means commute, then the conjugation is trivial. In that case, the covering, G covering structure does not change, otherwise the covering, G covering structure will change. All right. Suppose now instead of just changing the base point at the top level, we change it at the bottom level, the B0 to B1 inside B. Then the corresponding expression for J itself will be different. It is no longer J, but some other subgroup, some other group, which is again isomorphic to J. It is phi 1 now B1. And what is the automorphism here or isomorphism given by H tau, where this tau is the path from B0 to B1. So, we are conjugating by this H tau. It is a path. It is not a loop. So, it is not an inner conjugation. So, under this again, there is some automorphism of this group. So, the action will be, the G action will be different. Once again corresponds to some automorphism. That is what we have seen. So, there are all these change of base points, etc., all these things are taken care in the concept of G covering. They may give you different G covering and theorem that we studied two modules behind, that tells you what exactly the changes can be. Namely, once the covering projection is the same, then the G actions are different only by an automorphism. So, this is the theorem that we have already studied. Now, in the classification, this becomes our central idea and this itself gets extended. This idea gets extended. Now, let us go back to another construction of G coverings. Long back we introduced what is called as extension of G actions. So, we use that one. Starting with a homomorphism from this fundamental group to any other group G. Take some homomorphism, group homomorphism. Extend the E action, the J action on E is the, you know, E is the simply connected covering, that is fixed. To a G action by looking at this E alpha, what is the definition of E alpha? I am recalling it. It is G cross A, the total space, but now we are going to have a quotient here, namely equivalence relation. G comma T times E, where T is an element of J, okay? E is equivalent to, is identified with G times alpha T, remember alpha is homomorphism, alpha T is an element of G. So, this G and this alpha T are combined inside G, that is a group operation inside G. So, this one element of G comma E. This is then equivalence relation and you go modulo this equivalence relation, that is the space E alpha, okay? So, this was the definition of E alpha. On E alpha, G will act from the first factor here. So, it becomes a G space. There is a G action on this quotient space now, okay? So, we call the, have notation E prime free alpha. Because the action of J on E is E1, it will follow easily that the G action on E prime is E1, okay? Therefore, we get a G covering from, from E prime, again you take E prime itself is a quotient in this way. But now you take the quotient by G action, then you call what you get is the same space B, the original space B, P prime from E prime to B given by P prime of G E equal to P E. The first factor here, the second factor here, sorry. The first factor totally disappears. Why? Because first you quotient it out by the, by whatever overflow here, then G action will, will further take away all this. Everything is identified, okay? So, this is a G covering starting with a J covering, we have converted it into a G covering. Why? This map alpha, okay? If alpha or isomorphism, then you can treat this as a J covering itself, okay? Both of them are J covering. The usual map here, E going to 1 comma E, that is a canonical map. That will be a J map. Therefore, it is a J isomorphism, okay? So, you understand this covering, this construction, I have just recorded. Once we have extension of G action, etc., we have studied there itself, all these things are rigorously done under exercises. Now, let this curly G build, you know, set of all equivalence classes of G coverings over B. G is a group. This curly G denotes all the equivalent classes of G coverings. It has nothing to do with the fundamental to Kruppah B. That would have been a J coverings. So, that I would have denoted by curly J B, okay? Alright. Now, suppose you define a homomorphism, you define a satiatic map from homomorphisms of J into G into equivalence classes of G coverings by taking a homomorphism alpha to this extension fibers, extension of actions, E alpha, and take this equivalent class. So, we get a satiatic map here, okay? The aim is to prove that this itself is a bijection. Starting with a homomorphism, we get a G covering. Starting with a G covering, you must get a homomorphism such that if you use that homomorphism and come back, you must get the same equivalence class and vice versa. So, this must be a bijection of equivalence classes here. Here, just J homomorphism, there is no equivalence classes here, okay? Every homomorphism stands on its own, alright? So, how do you do this bijection? By precisely constructing its universe, directly constructing universe. Therefore, this proof is going to be completely constructive. We have constructed the map view instead of saying there is a bijection. We are giving you a bijection, okay? Of course, we are giving the map and then showing that it is a bijection. So, it is a constructive proof. Constructing the universe of new. Let us do that, okay? Yeah. Given a G covering, T prime to B representing one class, I have to construct a homomorphism, right? Let us fix a base point E naught prime inside E prime such that P prime of E naught prime is the given base point B naught. This you can do, alright? So, EPB is the standard simply connected covering with E naught, B naught as usual. Once E is a simply connected covering, this is a covering by the lifting criteria because it is simply connected. This P can be lifted. So, there will be a P bar such that P prime composed P bar is P, okay? If I specify where the initial point E naught goes, then this P bar is uniquely defined. So, what I do is, I demand P bar of E naught is equal to E naught prime. So, this P bar is completely determined by this property. Namely, it is a covering, it is a lift of P and it takes E naught to E naught prime, okay? So, these notations I am going to fix up now. Remember, this is some arbitrary G covering. This is the simply connected covering over B. I told you this is going to be our guide, okay? So, out of this, we will now construct a homomorphism from J, namely G P of this one to G P of G P prime of that. This is J is G P, group of current transformations of P and G is group of current transformations P prime. That is what we are going to construct now, okay? So, construct is obvious now more or less. Starting with a class here which we can represent by a loop, okay? Lift this loop to a path at E naught through P in A, okay? Take P bar of that lifted path. That will be a path at E naught prime and if you take P prime of that, that will give you back omega. Therefore, you can think of this as a lift off. There is a loop here. So, instead of, I do not know whether I can lift it here, all right? But I can lift it here and then I take the image there. That will be lifted. That is all I am doing. So, P bar composite omega twiddle is a lift off omega through E naught prime, through P at E naught prime through P. P prime of this endpoint is B naught, right? So, it follows that the P bar composite omega twiddle is in the same fiber. Therefore, there is a unique G such that G of E naught prime is equal to this endpoint. So, starting with omega, we have got an element G. This G is an element of capital G. The relation is precisely given by this equation 19. I am calling this G as alpha omega. Alpha omega operating upon E naught prime gives you the endpoint of the lift off omega through P prime, which I read, write it as P bar of omega twiddle. So, this alpha is the function from G to G. There is no ambiguity in the definition because the end points depend only on the homotopic class. End point is always the same, no matter which loop you take to represent the homotopic class, okay? So, I am recalling the isomorphism between G P and J here, okay? I just recalled it just now, but let us denote phi as the inverse of this. Let us see what is the inverse of this one, inverse of J to this one. Starting with phi, starting with the omega, lift it, okay? At E naught, look at the end point and that end point corresponds to the image of E naught under a covering transformation and that covering transformation is phi omega. That is a unique such thing. So, phi omega of E naught is the end point of homotopic. This is the correspondence for the inverse of that map, all right? So, if this is the case, we apply P bar, P bar of phi omega E naught, okay? We will give you this P bar phi omega, omega prime. Therefore, it is alpha omega of E naught prime, okay? Alpha omega P bar of E naught. E naught prime is P bar of E naught. So, this equation is equivalent to that. So, this will tell you what is the relation between alpha and this phi, okay? Phi of omega, P bar of that is alpha omega of P bar of E. This you can take it as definition of alpha, okay? Either this one or this one or this one, okay? Now, you take P prime on both sides, P prime of P bar of phi omega, okay? What is it? P prime P bar is always P. It is P of phi omega, but phi omega is a covering transformation. So, it is P, okay? So, this is just P. Same thing, if you take P prime of alpha omega of P bar, P prime of alpha omega, alpha omega is an element of G. P prime is a covering transformation, is a covering projection. So, it is same thing as P prime. P prime of P bar is P. That means, these two are both lifts of P and they agree to one point because of this definition, okay? At E naught, they are equal. What does it mean? They are equal everywhere. So, what I get is this is to lift the steps and equation tells you that they agree at P naught. They are equal everywhere. Alpha omega composite P bar is the same thing as P bar composite phi omega, okay? So, this is the equation we have got. This is equation for the covering transformations and P bar and so on. This is valid for all points of E, okay? Now, given any another element tau, let us say, tau is an element of J, bracket tau, you can compose it on the right, okay? P bar, P omega, phi of tau, start phi tau, phi bar omega, phi tau. In this equation, you operate on the right where phi omega, another covering transformation. So, phi omega, phi tau taken on both sides. So, I get this one, okay? You evaluate it at E naught, you get this equation, okay? But phi is a homomorphism. Phi omega composite phi tau is nothing but phi of omega star tau, right? Omega star tau. So, this is nothing but P bar of, you lift omega star tau as, you know, this is loop composition. Lift it and take the endpoint, okay? So, that is, by the definition in 19, this is nothing but alpha of this element now, alpha of omega star tau operating on P bar of E naught or E naught prime. So, this is the LHS, RHS, what is the RHS? This part, this part is alpha tau, alpha tau of this one by very definition. So, I will do it. On the other hand, again by 19, P bar of phi tau is alpha tau of P bar of E naught. Then I have left multiplication alpha omega. So, the whole thing I get alpha omega composite alpha tau. Alpha is a homomorphism. That is what we want to prove. I do not know that. But this component is equal to P bar of E naught. On one hand, it is equal to this one here and on the other hand, it is equal to this. So, these two are equal. Means, alpha omega star tau operating on P bar E naught is equal to alpha omega into alpha tau operating on P bar E naught. Now, you use the fact that the actions are fixed point free. Therefore, if they agree at one point, these elements must be the same. Therefore, alpha is a homomorphism. The proof is somewhat similar to what we have done for earlier, right? Due to fixed point finesse, these two. So, construction of the homomorphism is corresponding to any G covering, we have got a homomorphism. But what we want is corresponding to equivalence class here, there must be homomorphism. So, what we have to show? We must show that this alpha is independent of the class. For whole class, whatever you choose, you should get the same alpha. It is same for the entire class. So, that remains to be shown. Let us show that one. So, let us verify that alpha is independent of the G equivalence class. Suppose P double prime, P double prime to B is another G covering. And F from E prime to E double prime is a G equivalent. G equivalence means it is G map. That is the meaning of two coverings are equivalent, right? So, you have G map F from E prime to E double prime. I can assume that E naught double prime is equal to F of E naught prime. This is the base point I am choosing. Base point I have to choose. So, that is the base point for E double prime. Then what happens? P bar is there. You compose it with F that will role play the role of P bar for the P double prime. Remember, we are use P bar for all these from E to E prime. Now, you compose with F, you get E to E double prime. This F compose P bar will play the role of E prime for E double prime. And it maps E naught to the base point. Base point is B M, I put a base point. Now, let us assume that whatever you have constructed for E double prime is beta. Beta from J to J is corresponding to homomorphism for E double prime. I have to show that alpha is equal to beta. What is the equation for beta? By definition, this 19, what I have? What is beta omega of E naught double prime is F composite P bar of omega twiddle F. Because this is the lift of omega in E double prime. So, its end point defines beta omega. For alpha inside E prime, we have the old equation alpha omega of E naught prime is equal to P bar of omega twiddle. So, this is for E prime and this is for E double prime. Therefore, beta omega of E double prime which is F of P bar of omega twiddle which is nothing but F of P bar omega twiddle is I am substituting this one alpha omega of E naught prime. Now, use the fact that F is a G map and alpha is an element of G. So, this alpha omega comes out F of E naught prime. But what is F of E naught prime? It is E naught double prime. So, you have beta omega E naught prime E naught double prime is alpha omega. Again by G action which is even action so fixed point free. Therefore, this alpha omega must be equal to beta. But this is true for any omega. Therefore, alpha equal to beta. So, what we have done is we have a map from here. We have a map from home JG to JB mu. Now, we have a map from JB to home JG mu. Right? So, we have constructed a map like this. What we want to show that mu and mu are inverses of each other. So, we have constructed this map. We want to say that these are inverses of each other. That we will do next time for this for today this is enough. Thank you.