 Today, let us first begin with the statement of the classification theorem. We begin with a group G and a connected locally path connected semi locally simply connected space B with a base point B naught and it is simply connected covering P from E to B. Of course that also has a base point E naught mapped on to B naught. Let of course G be any group I have already told G curly G be denote the set of all equivalence classes of G covering something then assignment alpha going to E alpha where alpha is an element of home pi 1 B B naught G. There is a homomorphism from the fundamental group into the given group G to construct the group extension G covering E alpha starting with the G covering E then take its equivalence class that is a map alpha from E alpha going to E alpha which we have denoted by mu. This is a canonical bijection notice that on the right hand side you have equivalence classes on the left hand side they are just the set of all homomorphisms. In particular if G is abelian this set becomes a group otherwise may not be a group. So, the equivalence classes of G curries they themselves form a group in a strange way. This is the consequence of this bijection. In fact what we want to do is we have already defined the inverse map the so called inverse map we have not verified it mu from GB to home JG and we want to show that this mu and nu are inverses of each other. So, the first thing is to show that mu composes nu is identity and nu composes mu is identity and then there is a part which says canonical part. So, these three main parts are there. To start with the equivalence class of a G covering which I denote by zeta after all total space is some E prime P prime B. And then because E is the universal covering you will have a map P bar from E to E prime which is actually lift of P through P prime. If you make it base point from base point going to base point then this is unique map. So, take a homomorphic alpha from G to G given by the equation 19 whatever the equation was let us have a look at equation 19 and come back here yeah. So, you start with a loop representing the given element of the pi 1 lift it look at the end point the P bar of the end point is in the same fiber as E naught prime inside E prime. Therefore, it is related to E naught prime by an element of G that G is alpha of omega this is the definition defining the relation for alpha ok. So, you have to remember this I will be using it again perhaps. So, where are we now? So, we have this defining relation. Now, consider the assignment G comma E going to G of P bar E ok. So, I want to say that this assignment respects the equivalence classes see first I want to define finally what I want define a map from E alpha to E prime ok. So, that this becomes a G map. So, that these two are equivalent. So, before that to define a map like this I will define it a from G cross E it is G comma E G cross E into E prime. Then I will show that it respect to the relation there the defining relation for E alpha ok. So, what is that? An element G E is equivalent to G times alpha H and H inverse of E for any element H of the fundamental group J ok. So, this is equivalent to this one ok. This one is mapped to by the definition G times alpha H the first part and then P bar of H inverse E ok. But this is same thing as G times alpha H as it is P bar of H inverse is alpha of H inverse times P bar of E this is just what what the definition of alpha tells you just now what we equation 19. So, alpha H and alpha H inverse cancel away what you are left with is G G of P bar of E which is the definition of this map G E. G E goes to G G of P bar of E. So, all the elements which are equivalent to this one also going to same number. Therefore, we get a well defined map from the quotient and that time calling it as F ok. So, if commutes with the projection maps is obvious because of because of the definition here ok. So, it is a covering transformation from E alpha to E prime which is the zeta that is the G covering. Moreover, let us look at F of instead of just one element G 2 times E multiplied by G 1. So, what happens to this by definition it is a F of G 1 G 2 E but F of G 1 G 2 E is G 1 G 2 P bar E but that is same thing as G 1 of G 2 P bar E is nothing but F of G 2 E. So, G 1 has come out it just means that F is a G map between G coverings any G map is an isomorphism is what we have seen. So, therefore, see starting with E prime we looked at alpha and then we perform this E of alpha which is which is mu and now we are showing that this equivalent to the ordinary. So, mu of mu mu first and then mu is identity this is what we have what mu of mu is identity. Now, let us look at the other way around. Now, I start with a homomorphism instead of starting with a equivalence class of G G covering start with the homomorphism G to G then you perform the extension E beta by the very definition E beta is nothing but for any omega beyond G we have 1 comma P omega E is beta omega E. Therefore, if omega twiddle is a lift of omega in E at the point E naught P bar omega tilde is a path from the base point 1 comma E naught to 1 comma omega twiddle of 1 this is a path applying omega which is nothing but this omega twiddle of 1 by dimension is 1 comma phi omega phi E naught that is the definition of phi omega, but that is that is same thing as beta omega phi E naught why because this action of omega comes on this side equivalence class under beta. So, it is beta beta omega comes out beta omega 1 E naught. Therefore, from the definition again equation 19 for alpha this means that this should have given you alpha alpha must be equal to beta. So, starting with a beta you take E beta that is mu then the corresponding homomorphism is back to alpha itself whatever alpha. So, this alpha must be this beta so mu of mu of mu of its identity. Remember that one way we get equality for the homomorphisms other way here we are only showing that the two G coverings are equivalent under F they are equal they are not may not be the same, but they are equivalent under the G maps. So, keep that in mind. So, this shows that mu is a bijection. The next thing is we have to verify that it is a canonical map canonical bijection. So, what is the meaning of the canonical I have to explain you and then prove that one this is the meaning. Take any function F from B prime to B let us choose a base point B naught prime going to B naught that will induce a homomorphism from pi 1 of B prime B naught I am going to call it as J prime to pi 1 of this I am which I am going to get say J. There is a homomorphism from here to here this F check. When you compose any homomorphism from pi 1 this J to G with F check what you get is a homomorphism from J prime to G. So, that is F check star. So, this is the map from home to home which are under mu correspond to the equivalence classes of G coverings over B and here equivalence rise of G coverings over B prime. So, whether you first take F check and then go to mu or first take mu where this way we have to go this way first you take mu and then take F check F star. F star is what the pullback the pullback of the G covering over B to a G covering over B prime or you first take F check here convert the map alpha into a homomorphism a homomorphism alpha into homomorphism from J prime to that and then take the extension from E prime remember whenever we do this mu it has to be you have to use corresponding universal cover corresponding simply connected cover and then extend the group structure to G you do that you get a G covering over B prime. So, this diagram commutes So, this is totally independent of what F is you see the same F star you have taken go here homomorphisms at pi 1 level are induced by F this one is induced by again F it is a pullback of the bundles pullback of the coverings. You can put mu from G B to here by reversing the arrow here because that is just the inverse of the if you reverse arrows here reverse arrows here and put mu here that will be also commutated provided you prove this one alternately if you prove that then this will this will get over because mu and mu are inverse of each other. So, the whole point is we have defined a mu not by one single set home we have defined it for all possible different here you know if you change B if you change G the domain of mu will change but the change if it corresponds under F under a home under a continuous function then there is a commutativity of the diagrams this is the whole idea. So, once again the important thing here is to understand what is happening at the universal cover level namely simply connected cover level under this F check you have a map from F from B prime to B it induces a homomorphism at the pi 1 level it also gives you a map from E prime to E the lift of F because E prime is simply connected I am taking simply connected covering E prime over B prime that F check has F bar whatever the lift of F ok we will have something to say about the group GP GP prime on one side and GP on the other side that is the thing that you have to understand once you understand that the canonical property falls out very quickly ok. So, once again you have to identify pi 1 with the group of core transformations to understand this again and again ok. So, under this identification ok we must figure out what the homomorphism F check from pi 1 of E prime to pi 1 of B B not corresponds to ok. So, as usual P from E to B P prime you have simply connected covering places with base point preserving etcetera. Note that there is a unique map F bar E bar B such that F composite P prime equal P composite F bar I am just taking this composite this this is simply connected. So, it is a lift if you specify the base point is unique lift that is all that has been used. So, F bar of E prime is E not ok. Now, a covering transformation psi from E prime to E prime corresponds to an element omega prime of pi 1 of B prime B not. What is the rule? The rule is the end point of the lift of omega twiddle omega twiddle is a lift is equal to psi of E not prime of the base point. Same story inside E to E also which you have which you have been using ok. Now, the point is if you take an omega prime in in pi 1 of B B prime F F of if you can have at F check of that F composite that will be a loop in pi 1 of B at B not that is the way F check is defined at pi 1 level by just take composition with F right. So, it follows that F bar of the lift of omega in say omega prime inside E prime will be the lift of omega inside E F bar if you come come on this side from take a loop here lift it here push it via F bar is same thing as first you push it and then lift it by the uniqueness of the lifts ok. So, that will tell you what is the relation between this covering transformations under F bar ok. So, it follows that we have F check of psi psi is an element of the covering transformation or think of this as now an element of pi 1 operating at E not this why I am taking as a as a group of we have a covering transformation it is equal to F bar of omega prime to little 1 by definition, but there is same thing as F bar of psi E not prime ok. So, this covering is defined by this this is also another lift. So, once you once you fix the effect of base point it is uniquely defined that is what I am telling ok. Since both psi F composite F bar F bar will be psi are lifts of same F composite B bar and agree at one point T psi of its one must be equal to F bar of psi. So, this is the commutative diagram that I am telling you here covering transformation here there will be covering transformation there psi F of F check of psi if this corresponds to the fundamental loop element F check of that will be the this element that is why I have written F check of psi ok that the covering transformation is precisely that. So, the F check can be thought of as a map from the covering transformation GP prime to GP under these identifications. So, that is also canonical property ok. Start with any any continuous function here base point preserving then this identifications of universal covering and there are the covering transformations in the fundamental group that itself is canonical ok. So, this picture tells you that now now we can complete the proof of canonical property the commutative diagram of this commutative square from home home J home J prime G to home J G etcetera. Start with the homomorphism from J to G composite with F check to get a homomorphism J prime to G remember F check is J prime to J then again follow by F you get a J prime to G. So, let us let us call it as beta. Now, consider the pullback covering F star of E alpha F star of E alpha by definition is the is the equivalence class of this one is is F star of mu of alpha. On the other hand you can take F check of alpha that is beta and then then perform the extension that is mu of that. This is E prime of beta what you have to show that these two are in the same class namely these equivalence classes also. So, in order to show that F star composite mu is mu composite F check we shall define a map here which is a G map that would mean that the equivalence class of this as the G covering is you could equivalence class of this one as a G covering this lambda should be a G G map that is all we want all right a continuous function which is a G which is respect the G actions then you are done. So, to define a map from here to here I will use the universal property of the pullback F star of E alpha remember what F star of E alpha was E alpha itself is a G covering right it is covering projection there is a projection map from E alpha to B we will denote by P prime this is our notation anyway F is a map from B prime to B pullback let us call the total space as z we know that it is a subgroup it is subspace of E alpha cross B prime and so on that we know already there are pi 1 and pi 2 projection maps here. So, this map pi 2 is the pullback cover G cover we want to show that this cover and E prime E prime beta these two are equivalent that means I must find a lambda here which is research at this triangle is commutative and this lambda must be what lambda must be a G map is what we want to do to define this there is a unique way there is an easy way of doing the whole thing using the universal covering of z what I have to do define a map from here to E alpha and one here to B prime such that when you compose with F and P prime when you come up to B they are the same this compose this one is same thing as this compose this one then there will be a unique map here this map is already given to you so you want this one to fit here so you choose this one so that this compose this one this one that is all so what is this map well if you just look at what are those elements this this map comes automatically okay so that this has to be commutative right so let pi 2 prime be the covering projection this pi 2 prime this is the the covering projection defining E prime beta so what is it it is by very definition pi 2 prime of G E prime is P prime of E prime okay there is a somewhat overlapping this may be I should denote this one by different notation here you saw P prime was used for covering projection from E prime to B prime okay so so this is this notation I maybe I should change it to Q the mother thing okay there is a typo here but this pre-prime is canonical notation E to E to E P to E B and the other one is E prime P prime B prime so that is a nice notation let us not change that one to get a map lambda we need to map pi 1 prime from E prime this one such that when you compose with this Q pi 1 prime of Q composite pi 1 prime must be equal to F composite pi 2 prime okay that is what we want to we want to define so where this is the covering projection this this is this this is Q and what is the definition of that that is G E equal to G E goes to P E this obtained by the original E to E P from E to B the universal covering there this is the definition of E prime okay so I just take pi 1 prime of G E prime equal to do not disturb G at all G F part of E prime okay why this is well defined if if there is a class here equivalence classes here why they go to the same equivalence classes here that is what you have to do so that is what I have to check under the equivalence relation on G cross E prime defining what is this one remember this is E prime of beta extension of E prime to B prime by the by the homomorphism beta right so it is a quotient of G cross E prime the defining E prime of B prime what is the relation relation is G times G comma H times E prime is this H is transferred to the other side with a beta beta H G of beta H time E prime okay so I have to show that these two elements go to the same element under pi 1 prime okay but what is the defining relations here on in this is inside E alpha we have the first G of F bar of H E prime I am touching the image of this one G of F bar of H E prime this is by definition F check of H F bar of E prime right we have we have just seen that F bar of H F bar of Psi is same thing as Psi F check of Psi that is what we have seen that F bar of Psi F check of Psi I am referring to this this picture F bar composite Psi is F check of this one F bar of that okay so F check of that F bar of E prime that is equivalent to now this is an element in the group right so it will be it can be pushed to the other side G times alpha are taking alpha of F check of H then it is just F bar F bar of E prime but alpha of F check is by definition beta so this is G of beta H F bar of E prime now look at this element under pi 1 prime precisely it comes to this one so if these two are equivalent their image are equal therefore pi 1 prime is well-defined okay it is easy to say that if you compose with Q okay further so I have written P prime here again Q of pi 1 prime by definition P of F bar of E prime but that is F of P prime of E prime but that is F of pi 2 prime of GE prime this is the relation between the E and E alphas and E prime and E prime beta the quotient maps are defined using the original P or P prime okay this is this you just change it as Q here that is all therefore we get a map lambda from E prime B what we have verified here is this composite this is equal to this composite therefore this is a unique map lambda here and you know how it is defined it is defined as pi 1 prime comma pi 2 prime of an element here is order to pull off the first coordinate and second coordinate so that is lambda okay so lambda of GE prime is the first coordinate is yes GE comma F bar of E prime this is this whole thing is first coordinate second coordinate is P prime okay that this is a G map is very obvious because if G1 G2 if you put here okay so that is G1 G2 will come here nothing happens here that G1 G2 G1 will come out here the same thing as you can come put it all the way outside by the definition of this action action on the pullback so it is a G map okay so this completes the proof of the entire theorem as promised we shall use this one in the final result of proving various forms of fan compass theorem thank you