 So, today we should take up another theoretical topic classification of covering projections. Like classification of spaces, now we are classifying the functions which are all covering projections, obviously what we are going to do is we are going to fix the base space space is fixed up to homeomorphism. So, the covering projections will be, will be classified up to homeomorphism which take one projection to another projection. It is not just the classification of the top spaces up to homeomorphism. The homeomorphism should preserve the covering projection. So, this leads to what are known as covering transformations and so on. So, we shall go on introducing some of these. And finally, we will see that they again use group theory namely they are closely related to subgroups of pi 1 of n, ok. So, let us go through this topic. So, here is the definition. Take two covering projections p 1 and p 2 over x. We say they are equivalent if there is a homeomorphism from x 1 bar to x 2 bar such that when you compose it with p 2 it gives you p 1. So, the projection map p 1 is taken to p 2 by f. So, that is the meaning of this one, ok. Is the definition clear? Now, this is easily seen that it is an equivalence relation first of all. If there is a map like this you take f inverse that will give you f p 1 composed f inverse will be equal to p 2. So, this means that the p 1 is equal to p 2 implies p 2 equivalent to p 1. Every p 1 is equivalent to itself no problem. Again, if there is a p 3 and a homeomorphism g from x 2 to x 3 such that p 3 composite g equal to p 2 then you take g composite f of p 3 that will be equal to p 1. So, that will show that g composite f is a covering projection is a equivalence between x 1 and p 1 and p 3. So, this is a transitive relation. So, equivalence under this one is actually equivalent equivalent while you are calling it equivalent. So, it is the equivalence relation. So, it is justified, ok. So, equivalence classes of covering projections is what we want to study. Given a covering projection p from y to x cross i remember there is an exercise you given to you earlier restricted to x cross t you get various covering projections. And exercise was that these are all equivalent without I define that one now in terms of this definition what you have to show that is that all these think of x cross t as if x it is x for getting t ok. Then under that homeomorphism x cross single t going to x under that homeomorphism all these covering projections restrictions x cross t they are all equivalent to each other that was the exercise ok. So, that is an example provided you have solved it you will see some more things like that here. Two covering projections over x ok are equivalent if and only if they define the same subgroup pi 1 of x of two conjugation. Therefore, if two subgroups corresponding subgroups are not conjugate to each other then you will know that the covering projections are not equivalent. So, this is a starting point of relating the equivalence classes of covering projections to subgroups of pi 1 of x. So, let us see how this one comes this is one straight line and one line proof p 1 and p 2 are covering projections f from x 1 to x 1 bar to x 2 bar is an equivalence. Let us pick up points x 1 bar and x 2 bar over inside x i bar over the same point pi 1 in x say x base point such that f 1 of x 2 is equal to x 2 ok. Take one point here f 1 f 2 there take that as x 2 there. Then since f is a homeomorphism we have f check of pi 1 of x 1 bar x 1 bar is equal to pi 1 of x 2 bar x 2 bar this we have seen because f is a homeomorphism. Therefore, when you apply p 1 check of pi 1 of x 1 bar little x 1 bar is equal to p 2 composite f check of pi 1 of x 1 bar x 1 bar which is p 2 check of pi 1 of x 2 bar x 2 bar. So, now suppose these two are conjugate to each other now this time I do not know what point I have to take I have taken p 1 check of f 1 and this is pi 1 of x 2 x 2 at some point some other point suppose these are conjugate to each other ok for some element tau in pi 1 of x x where this x is p 1 of x 1 bar which is equal to p 2 of x 2 bar. I do not know what map we have but I have chosen x 1 and x 2 x 1 bar and x 2 bar sitting over the same point. Then only I can talk about pi 1 of x x when I talk about this one there are two subgroups and I am assuming that these two subgroups are conjugate to each other that means this one is conjugate of that one all right. Now I will produce a map which takes x 1 bar to x 2 bar and this map will be a homomorphism such that when you compose with f composite p 2 will be equal to p 1 so that is what I move ok. So, let lambda be a path in x 2 bar such that at 0 I am starting point x 2 and when you compose p 2 composite lambda it is this tau. So, I have lifted tau that loop tau into x 2 bar ok that is what it is end point of this lambda will be x 2 bar ok x 2 hat some other point. Then we have lambda inverse pi 1 of x 2 bar x 2 bar lambda this is we know right lambda is a loop from lambda is a path from x 1 bar x 2 bar to x 2 hat. So, x 2 hat loops at x 2 hat ok looped as x 2 will get converted into loops at x 2 hat. So, this is the isomorphism ok. Therefore, it follows that p 2 check if you apply of pi 1 of x 2 x 2 hat on this side it is p 2 check of this thing p 2 check of lambda inverse is tau inverse then I have p 2 check of pi 1 of x 2 bar x bar that is tau on this side ok. But we have assumed that this tau is nothing but p 1 of x 1 bar. So, what I have what is now for some other point you see x 2 hat some other point at x 2 bar the fundamental group maps to the same group under p 2 check ok. So, equality was given in the in the earlier case forward case in the backward case we had to get this one after starting with a conjugate. The conjugate is there when you fix your points yourself. Now, the conjugate will tell you what point you should take so that the groups become equal that is the idea. So, now, p 2 check is equal to p 1 check because I have changed the base point in x 2. Equivalently I could have changed the base point in x 1 by reversing the arguments here ok these are symmetric argument. So, now what now you apply the lifting criteria you see I have told you that we keep on applying the lifting criteria to map f from x 1 bar to x 2 bar ok or to get these maps either to p 1 or to p 2. So, you have p 2 from x 2 bar to x 1 bar take p 1 from x 1 bar to x take x 1 bar as y then you get a map from this y to x 2 bar right but y is nothing but x 1 bar. So, you can do the other way around also. So, both ways you will get maps f and g such that p 2 composite f is p 1 and p 1 composite g is p 2. But what happens x 1 bar goes to x 2 hat and g of x 2 hat comes back to x 1 bar. So, this is the starting point assumption whenever you are lifting you are specifying the one point the lift of one single point you have to specify ok. Now, if you take p 2 composite g composite f or f composite g ok see p 2 of f is what p 2 of f is p 1 then p 1 of g is p 2 ok. So, p 2 f composite g will be equal to p 2. So, g composite f or f composite g here is the lift of p 2 itself through p 2 but identity map is also a lift and f composite g of x 2 hat will come back to x 2 hat. So, identity map and f composite g agree at one point and x 2 bar is connected and so on that we have to assume right we have assumed connect connected spaces. Therefore, f composite g must be identity map by the uniqueness of the lifts. Similarly, g composite f is also identity map of x 1 square. So, f and g are inverses of each other automatically ok. So, that gives you that lifting problem which we have solved is used here to show that if the fundamental group level p 1 check and p 2 check give you conjugate subgroups then the coverings are equivalent. What are the assumptions? Assumptions that they are connected coverings. So, what we have seen is equivalence classes of a connected covering space corresponds to a unique conjugacy class of subgroups of fundamental group. In fact, take a covering projection x bar to x you have freedom to choose the base point above inside the fiber as you keep changing the base point the subgroups will become different but they are all conjugate. So, within this space if you fix the space if you change the base point what happens is you are getting conjugates of groups nothing else happens if you change the whole space but keep the group as conjugate to itself then it has to be a homomorphic state. The points get base points may get shuffled that is all ok. If you fix the base points also one going to other even the same conjugacy subgroup we will get no other conjugate subgroup ok. The conjugacy number of conjugacy classes will also depend upon this number of these things up to something of course several of them may give you the same conjugacy class ok. So, to complete the picture given any subgroup k of pi 1 we should also be able to tell whether there or not there is a existing covering. So, this is not done yet. So, what we have got is each conjugacy class goes to a subgroup conjugate subgroup each each isomorphism class goes to a conjugate subgroup. If you have a subgroup will there be a covering which comes to that that is the fundamental problem that we have namely ontoness of this one ok. If two coverings go to the same conjugacy class of a subgroup then they are equivalent. So, that is a equivalence classes going to the conjugate subgroup is a injective mapping. To complete the picture we ask whether it is subjective subjective means what given any subgroup can you construct a covering such that under P check it will come to one of the conjugates. Of course, once it comes to one of the conjugates you can actually make it a subgroup also by changing the base point that much you know already ok. In particular suppose I take the trivial group k equal trivial group is also allowed you have to answer this question to all the subgroups of pi 1 of x. So, take trivial group what is the corresponding question does there exist a covering projection for which the total space is simply connected. Remember P check is injective mapping if P check of pi 1 of x bar is equal to trivial group it follows that pi 1 of x bar itself is trivial. So, given any space do we have a covering that has that the covering is simply connected the top space is simply connected. So, this is a special question I mean a particular case of the question, but it turns out that the method of attacking this problem is first to solve this problem the simplest problem namely very k equal to trivial group. Once you solve this one it will tell you how to get other groups ok. So, that is the way it is done I mean in theory in how to solve this problem you solve the simple problem first. Our next goal is to ask this you know solve this problem alright. So, I repeat instead of going to next next thing I repeat a repeat this one what we want is now starting with a space which is locally path connected and path connected at a certain time. Take any point does not matter and look at the fundamental group of that space at that point. Look at subgroups will there be a covering projection corresponding to these subgroups means covering projection must have isomorphic to fundamental group isomorphic to the subgroup under P check. P check is already injective mapping right inject monomorphism so on to the image will be isomorphism ok. So, this is our goal. So, let us make a few definitions starting with a covering projection P from x bar to x a covering transformation what do I mean of covering transformation let us understand this one ok from x bar to x bar itself let us say homomorphism such that P composite F equal to P. So, in other words it is an equivalence of X itself with X X itself there may be other equivalence is right that is what it is now homomorphism from X bar to X bar such that P composite F is P itself P 1 composite F equal to P 2 defined the two being equal. So, these are the equivalences of X bar with itself ok. So, these equivalences will form a group now under composition because if F is an equivalence what we are calling covering transformation then G is a covering transformation composite will be also covering transformation inverse will be also covering transformation. Therefore, under composite this forms a group which will denote it by GP ok. GP is a set of all covering transformations forms a subgroup of all homomorphism because each of them is homomorphism under the usual composition of maps. This group is called the DEC transformation group it is also called the Galois group of P. The DEC transformation group of P or the Galois group of P of course heavily depends upon P it is not all homomorphisms of X bar ok. So, this is the definition of DEC transformation Galois group. The basic lemma again group theoretic here is the following. There is an injective mapping free it is just a function theoretic set theoretic map of GP to the right cosets of K in pi 1 of X. Remember the right cosets of K have flowed some good role here right earlier also they had come I will recall it again. That K is the is the image of pi 1 of X bar under pH ok. So, the right cosets of K ok. You take a sub of that that will be GP. GP there is an injective mapping ok. Moreover K is a normal subgroup you filled only if this field is an isomorphism that means not only this map field which we are going to define is a bijection it will be homomorphism it is an isomorphism. In general it is just a set theoretic map ok. Moreover K is a normal subgroup we filled only if this is an isomorphism of groups. In any case the cardinality of GP is always less than or equal to the number of sheets of P which is precisely equal to F inverse of of any point or P inverse of any point number of sheets. Why suddenly number of sheets come? Look at we have we have a bijection of the right cosets with the fiber earlier. So, this is nothing but the right cosets as I said is the fiber of P number of sheets. And GP have an injective mapping to this one. So, the cardinality of this one will less than cardinality of the fibers ok. So, this is the lemma. So, non-trivial part is to define this mapping and and what show that K is show that it is injective. Then if K is normal then this becomes surjective and the map becomes a homomorphism all right. So, this thing we will do next time not today all right that is the goal ok. Thank you.