 We have started classification of covering projection, let us continue that work now. The first lemma we had last time already stated it, let me recall it, there is an injective mapping P from the group of covering transformation, this is Galois group of the covering transformation P to the set of right cosets of K in pi 1 of xx where K is P check of pi 1 of x bar x bar. In particular, the cardinality of the group of covering transmissions is less than or equal to number of sheets of P. Why this part follows the latter part? Because we have already established a bijection from the right cosets of K with the fiber of P on x bar, on x, on x. The number of points in the fiber is the cardinality of the sheet, number equal number of sheets. Since there is an injection, so cardinality P will be less than your cardinality P. So, this part is easy once we define an injection map. The second part says that if K is a normal subgroup, in fact, if and only if the map phi, which is a synthetic map, is an isomorphism of groups. So, this is possible if K is normal, then the right cosets form a group first of all. So, map phi is a map from a one group into another group and the claim is that it is actually isomorphism. Not only it is injective monomorphism automatically, but it is actually isomorphism. That means you are on to also. So, several things you have to verify. So, let us do them one by one. First, the definition of the map phi. Given a covering transformation, look at the image of x bar under this covering transformation. Since P of f is f, it follows that sorry P of f is P, it follows that P of x bar is equal to P of fx bar. That means fx bar is in the same fiber. It is the same fiber. So, you take phi of f, I am going to phi of f to be K of P composite gamma, where gamma is a path from x to fx bar. When you take a path from x to fx bar, its image will be a loop in x because both x bar and fx bar are mapped at the same point, namely here in this particular case it maps on to x. Therefore, P composite gamma is a loop. Take its class and take its right coset K, K of that. The right coset of K with that element. This is the definition. Now, the only thing that there is some ambiguity at all, it is the path joining x bar to fx bar. To take another path, then what happens? Gamma suppose omega is another path, gamma composite omega P of gamma, composite K of P gamma, sorry, so gamma composite omega is a different path. That means it will be a loop. Therefore, by our lemma 7.12 or whatever, endpoint of gamma must be equal to endpoint of omega. Therefore, what happens is the point from x to x bar, fx bar, this coset class is independent of the different path that you are taking. See, you are taking both of fx and x bar same point and two different paths. That will be a loop. That is a loop means below it is in the image of K, it is inside K, it is in the image of feature. So, if you composite by another element here, the right coset does not change. Therefore, the right coset is independent of what gamma you take provided you choose the same point fx bar with x bar. So, that is the path how phi is very different. To show that phi is injective, it is similar, proof is similar now. Take another g, whatever g belong to g, another element such that phi f equal to phi g. What is the meaning of that? The two cosets are the same. Suppose the path joining x bar to gx bar is tau, then what it means is k of p composite gamma is equal to k of p composite tau. Okay, phi f equal to phi g means this one. But then p gamma p tau inverse will be inside k. It just means that p of gamma star tau inverse is inside k. It is a loop. As soon as you can lift to a loop, 7.2 will tell you that gamma and tau has the same end points. So, fx bar must be equal to gx bar. Once fx bar equal to gx bar, remember what are f and g? They are covering transformations. That means pf is equal to p, pg equal to p, which means they are lifts of p. At one point they agree, then they must agree everywhere by the unique lifting property. Okay, so the first part a is complete. Well-dividedness of phi and injectivity of phi is established. Therefore, the immediate consequence there, that is also complete. Namely, in particular the cardinality of gp is less than the number of sheets. Okay, now let us come to this case. Normally if we do not live, p is an isomorphism. I will just assume that this phi is surjective. Okay, instead of isomorphism, let us just assume surjectivity and see what happens. All right, suppose phi is surjective. Okay. First of all, we have already had bijection. Again I am using this bijection between the right cosets of k and the fiber of p. From this it follows that for each z belonging to p inverse of x, this is surjection means there exists a g belonging to gp such that g of x bar goes to z. Okay, because the map from the g, when you take g, you take g of, you take a path from x bar to gx bar and then you are taking the right coset, then you are taking whatever element gives that, when you are again lifting it, which means you are taking the same path, then you are taking the end point. So this is what the process is going here. To come to from g to the fiber p inverse of x. This just means that this map, direct map composition leads to isomorphism. Now it becomes a surjective map. Okay, is taking a path and taking the end point. So g of x bar is z. Starting g, g of x bar is z. Okay, for every point it has to be, there must be a g. Now take any point, any element of pi1 of x, lift it at x bar. The lift of omega x z, okay, see z is also a point over x, I can lift it at any point in p inverse of x as a starting point. Okay, look at g operating upon omega bar, g of omega bar. G is a transformation from x to x, x bar to x bar. You can compose omega bar with g, g omega bar operating upon 0 is g of x bar, that is z by definition. That means the starting point of g composite omega bar is nothing but z. Okay, but g composite omega bar, if p of that is nothing but the same omega bar because p composite g is nothing but p itself. Okay, start with a loop at x bar. Okay, then if you lift it at g, for arbitrary g, then they are all loops. Okay, if it is not a loop, then g of that will never be a loop because g is a homeomorphism. You take a path omega, okay, lift it at x bar, it may be a loop or it may not be a loop. Okay, if it is a loop, then g of that is a loop for all g and that will give you all for all elements because of a surjectivity. That means that the loop, the lifts of omega at all the points that in the fiber, they are all loops or if it is one of them is not a loop, then none of them is a loop. So that is the meaning, if we do not lift omega bar is a loop, this is a loop, if we do not have the omega bar is a loop. Okay, this one we prove that, this proves that the covering is normal, but then we have already established covering is normal is equivalent to subgroup k is normal as a subgroup in pi 1 of x. Okay, so surjectivity of phi gives you normality. Okay, so in particular if phi is an isomorphism, it is surjective also, k must be normal. Okay, now we can do the converse. Suppose k is normal, then I have to prove that phi is surjective, phi is a homomorphism. Okay, while proving normality, I did not have to worry about whether it is homomorphism or not, just bijection was enough, surjection was enough. Now I have to do the other way around also, namely once k is normal in the converse part, suppose k is normal subgroup of pi 1, then I have to show that phi is a homomorphism and it is surjective, by injectivity we have already done. So what do they mean by k is normal? The right cosets of k forms a quotient group, right, which we denoted by pi 1 mod k. Okay, otherwise it is only a coset space. Now it is a group, given an element in this group, how do you represent it by a coset k omega, lift this omega to a path at x bar? Look at the end point z with the end point, p check of pi 1 x bar z, look at the fundamental group at z and look at its image. This thing we have seen is conjugate to k, okay, but same element, omega, omega bar, you come back and so on. And hence, by normality, conjugate of k is equal to k. So what you have is, you have x bar, x bar and there are maps p and p, but the base point you have changed, okay, one is x bar and another one is z. But you apply lifting criteria to p x bar to x, okay, its image is in the image of the other group, therefore it can be lifted, okay. So apply the lifting criteria to one of the maps, to the map p itself. So both of them are p, right, but the base points are different. So you can demand x going to z and get a function f, okay, so that p composite f is p, which means it is a lift of p through p itself, so p composite f is p and p of x bar is z. Now reverse the roles. Now lift the same map from z bar to x, okay, then what you get? You will get g from x bar to x bar, which takes z bar to x, x bar, z to x bar. But then g composite f and f composite g would have fixed x bar to x bar or z bar to z bar, but identity also fixes by the uniqueness of the liftings, f composite g and g composite f must be both identity of x bar, that means f and g are homeomorphisms, which means it is a covering transformation because both of them commute with p, pf is equal to p and gf equal to pg, p, pg equal to p, okay, okay. So what I have done is I have done something more than this, so here what I have first shown that the normality implies the map fees is subjective first, so it is a bijection, okay, so how do you do that? Lifting, lifting right here up to f bar, such that f of x bar equal to z, we have now used this lift of omega at x bar to join x bar to z, okay, this z bar that is the meaning of f of f bar equal to z, so lift of this whatever path you take that will go there, that means what we have seen is phi of f equal to okay omega, for each omega I have got an f, okay, so this shows that phi is onto, actually this is not correct, what you have done is take omega, lift it, it goes to some z, that will give you f, f of, by the definition f of phi, phi of f is nothing but I have to take a path, we join, but that is the whole time, so it is z and then take omega 1, right, so that is that, that is what is happening, so what is written down is fine, but explanation is needed, okay. So now you have to show that this is a homomorphism, so activity is done, so this is a homomorphism short option, so here is the workout, take omega and tau, path is from x to f x bar and x to g x bar respectively, then if you take g composite omega, okay, now these paths are inside x bar itself, g is a path from g x bar to g of f x bar, because I am applying g, initial point is x bar for omega and end point is f x bar, so g of that will be an initial point for g omega and end point is the g of f of x bar, okay, therefore if you take tau star g composite omega, this is one path, this is another path, tau starts at x bar, ends at f x bar, okay, tau sorry, tau starts x and ends at g x bar, this one starts at g x bar and ends at g of f x bar, therefore this map is from x bar to this path is from x bar to g composite f x bar, therefore in the definition of phi it will be the end point of phi composite g f, it will be the end point of that one, okay, so now you look at what we have to do to get the cosets, but the very definition phi of g composite f is k of phi of this path, but what is phi of this path, it is phi tau star phi of g composite omega, phi composite g is phi itself, so phi composite omega, so this is phi tau star phi composite omega, which is k of this slope into k of this slope, this is the group law in the quotient space, how to multiply right cosets, right, that is a group theory, okay, k of some x1 x2 is kx1 kx2, right, but this is by definition phi g, this is by definition phi f, okay, so you see if you took left cosets then you won't get a group homomorphism, but you would have got an anti-group homomorphism, the way we write the maps g composite f is the reason for that, so here you get the correct law here, so you should take the right cosets, so this comes with the proof of the lemma, so let's go ahead, this is only a beginning as it's as the name indicates lemma, so immediate corollary you see those things are stunning, this is something which you can remember very easily, so what it says start with a connected normal covering blah blah blah all those basic assumptions are there, you have to assume x is locally path connected, path connected, damn that, okay, take a point x bar in the covering space such that it comes to px bar in the bottom, x in the bottom, then we have an exact sequence of groups and homomorphisms, what is it, trivial group to pi 1 of x bar x bar p check, this part is exact just means that this kernel of p check is equal to image of this is 1, that means p is injective, pi 1 of xx, this part we have already, we have already established, now we have another group here, another group homomorphism psi from pi 1 of xx, the group of covering transformations and the last one is trivial means it's psi, it's a surjective homomorphism, kernel of psi is precisely equal to p check of is just k, okay, so this follows easily by the first isomorphism theorem of group theory, so here it is, all that I know have this notation k equal to p check of pi 1 of xx bar, then look at q which is the quotient map, quotient homomorphism pi 1 of xx bar to the right cos x, okay, this is a quotient map, put psi equal to phi inverse of phi inverse composite q, we have established phi is an isomorphism from what, from where to where, this lemma, earlier lemma what it says, phi is an isomorphism from, of course I should not expect you to tell this way, what we have seen this lemma I am just recalling group, this group of covering transformations to the right cos x, right, so what you have is a first isomorphism in this picture, you have an isomorphism this way, this is an isomorphism map here, okay, these right cos x, okay, the right cos x to gp is an isomorphism, but this map is, this is a, this is herjective mapping in the kernel of q is same thing as p check by the very definition of right cos x and the quotient group, k is nothing but image of this and this is injective, okay, so by the image if you go, go to the quotient map, this is an exact sequence, so change this one by this map and go here and what is the definition of psi, you just try as, you just take this phi inverse q composite phi inverse, go back here, q composite phi inverse, that is psi, okay, so it is actually first isomorphism, if you have homomorphism such that right cos x have gone to the same point, that means its kernel of this one is k, therefore modulo k, it is isomorphism, so whichever way you like, so I have clearly to write down, if you don't know the first isomorphism there also then this will follow, all that you have to know is the quotient, quotient map as kernel precisely k and that is all and quotient map is a surjective mapping, so this quotient gets identified with gp, okay, so instead of writing this group here, I am writing gp now, which is more descriptive in terms of our covering transformation, you have fundamental group here, you have some subgroup k and then you are going up here as a fundamental group of this one is upstairs, the covering transformations of that is a quotient group, so this is the picture, okay, so this is now a theorem which we keep preferent to, all right, corollary is again interesting corollary, namely you start with a connected covering space etc, suppose the covering space, the top space is simply connected, which means in the earlier theorem, earlier lemma etc, k is trivial, then the quotient is same thing as fundamental group itself, gp will be isomorphic to pi1 of x itself, okay, so this p, p from gp to pi1 of x itself, because there is no right cosets, right cosets are elements of x itself, right, therefore this itself is an isomorphism of covalent transformation, okay, the covering projection p has the following universal property, this is the extra thing that you have, that you have to digest, which is not at all difficult, what we said, given any covering projection q from z to x, there exists a map f from x part to z such that q composite f will be p, this p, the covering from simply connected space to x factors into f, some map and q, q is the covering projection, given covering projection or you can see that this big space, this big space, it sits over all coverings of x, only thing you have to assume that it is connected, okay, all coverings of x sits over that, that is why it is called universal covering, okay, we are going to, this is universal property first, we are going to make a definition, because of this property, okay, so here is what I have drawn the picture, x is here, p is the covering projection and this is connected, simply connected, by the lifting criteria, its image in pi 1, pi, image of this pi 1 is trivial, therefore the lifting criteria says that this can be lifted, once trivial group, it is inside the image of q checker, right, that was the lifting criteria, so it can be lifted up, in fact there will be several lifts depending upon the first base point, one base point where you want to go, you have to select, then it is here, okay, so this p is factored into two maps, q bar is given, this is your choice, there is some choice here, not so much of choice, in fact this f itself becomes a covering projection, right now we are not doing that, okay, so only this much, that there is a map like this and this x bar is a covering for x, it will sit over salt, all covering, what we will prove that this f itself is a covering, so universal covering is a covering for everything, so that is why we are making a definition here, namely universal covering, so let us look at the definition now, start with a covering projection, let us assume everything is connected, fix a base point x and a base point above x bar, okay, p x bar is x bar, we say p is a universal covering, universal covering projection, you can just say just universal, if for any given connected covering projection, q from z to x, okay and an element z belong to z, such that q z equal to x, see x bar and z are both sitting over x, that is a necessity, okay, as soon as it is mentioned that there is a unique map f from x bar to z, such that p becomes q composite f, f of x bar equal to z, the uniqueness follows by the unique lifting property, lifting property, lifting criteria, the image of p check is trivial because x bar itself, pi 1 of x bar is trivial, so it is contained inside image of q check, that is all you need, okay, image of q check may not be trivial, but image of p check is trivial, so this is contained in here, so we can factor p like this, so we have just defined the definition, we have just given the definition of universal covering, let us see just a little bit of property of universal covering, given two universal coverings, p i, x i bar and x other, both of them connected of the same spaces, okay, you do not have much choice, namely they are all homomorphic features, actually as covering transformations they are the same, namely there is a homomorphic side from x bar to x 1 bar to x 2 bar, such that p to composite size p 1, remember such a thing is we have defined equivalence of covering projections, therefore a universal covering space if it exists is unique in the equivalence class, there is only one equivalence class of universal covering space, maybe there is none, we do not know that yet, okay, in that it happens that universal covering space is may not exist, okay, so how to show that this uniqueness, very easy, you use the universal covering both ways, you will get maps from the other one way, other way and so on, just now I have shown that F is a homomorphism and so on, just here, same thing you have to do, apply universal property p 1 to obtain the map psi from x 1 bar to x 2 bar and the universal property p 2 to obtain a map psi from x 2 bar to x 2 bar, what is the property of this psi and psi, p 2 composite size p 1, p 1 composite size p 2 and psi of x 1 is x 2 bar, psi of x 2 bar is x 1 bar, now you look at psi composite psi, psi composite psi, they will take x 1 bar to x 1 bar and x 2 bar to x 2 bar, therefore they must be identity maps, which means one is the inverse of the other, okay, so this is why universal covering space if it exists, it is unique, so this is the remark, connected universal covering space over x, if it exists, it is unique up to equivalence of covering projections, in particular the previous theorem, previous Corolli says that simply connected coverings are universal coverings, okay, simply connected coverings are universal coverings, but we still do not know your simply connected coverings, we can now complete the answer to the question of existence of covering projection corresponding to other problems, assuming that simply connected covering exists, so that is the next project, you will stop here.