 We shall study the relation with the fundamental group of the covering space theory. Having established the homotopy lifting property for covering projections, you will have to continue the study of lifting maps through a covering projection. So throughout this lecture, one or two lectures, I fixed the notation. P from X bar to X is a covering projection. X naught is a fixed point in X which will be called as base point for X. We put F equal to P inverse of X naught. This F stands for fiber over X naught. Inside the fiber, we take a point X naught bar as a base point for X bar. So this X naught bar and X bar, X are fixed and X naught bar is mapped onto X naught under P. So to begin with, we know that every path at X naught inside X can be lifted to a path in X bar at X naught bar. The question is, suppose I take a loop at X naught as a path, you can lift it. But the lifted path may not be a loop. If it is not a loop, there is nothing you can do about it because of the unique lifting property. Of course, if you take the base point different, there may be a lift. But at that base point X naught bar, there is no other way. So we have to investigate when a lift of a loop will be a loop. So suppose the lift of a loop is a loop in X bar. Then that loop will represent an element in pi 1 of X bar at X naught bar and its image under P check. P introduces a homomorphism at the group level. That will be an element of pi 1 of Xx which is precisely the loop that we started with, represented by the loop that we started with. So this is how the question of lifting a loop to a loop is related to something happening at the fundamental group level. So the algebra enters here. So for studying this problem, let us make two fundamental lemmas. Let us first state and prove fundamental lemmas and then keep using that to derive several algebraic properties. So the notations are as before. P is a covering projection, X naught is a base point at X and X naught bar is a base point in X bar sitting over it. Take a path in X starting at X naught. Let omega bar be the lift of omega at X naught bar. The first part of the lemma is this omega bar, if you look at the end point of omega bar, this will not depend on the path itself but only on the homotopy class of omega. Whatever lift you take in the representative, in the representative of omega if the path may be different but if homotopy class they are same, the end point of the lift will be always the same. So this is the first statement. The second statement is omega bar, the lifted path is a loop if and only if omega is a loop in X such that the class represented by omega is in the image of P check. In this case lift of any member of omega that means take this class, this is an element of pi1. So take any path which represents that one up to path homotopy. All of them the lifts will be loops. So this second part we have already observed we have already observed and the second part or second part will be an easy consequence of the first part. One loop, look at the end point, end point is the same thing as the starting point. So all other things will have to be loop because they should also have the same end points. So this part follows easily. This we have seen, this is the meaning of loop. Once the more loop is there omega has to be P check of omega bar. That means this omega bar is in the image of this P check under this one. This will be some group of pi1 of X bar, a pi1 of X. So second part is easy. The first part needs explanation why the end point is independent of the homotopy, independent homotopy class. This thing we have seen already in the case of exponential function. Remember how the degree of a loop inside S1 was defined. You lifted it and then you look at the end point. If you take another class and another represented in the same class, lift it, the other end point will be again the same. So that end point there happened to be an integer because it is the exponential map. Here there is no integer or anything but we want to say that the end points is independent of the homotopy, the representative of the homotopy class. It only depends on the homotopy class. So proof is more or less the same but we happen as if we use the integers and exponential and so on. Fundamental property of exponential function was covering and that is what we are going to use here. Namely for any covering the fiber, fiber is a discrete space, discrete subset of the top space X bar. So that is what we are going to use. Okay, once I have said that one, I have said it. That is the proof of the whole thing. However, let us go through it slowly and see once again why this is true. So here is a full picture of this one. You can keep referring to this all the time. So here is your X and here is your X bar and this is P. This omega is a path here from this point to this point. These dot, dots, dots here are sitting over this point. You can call this as X naught and these are all in the fiber of X naught. So here this is X1 and these are all fibers above X1, sitting over X1. So this omega is lifted to a path here, omega bar. Omega prime is homotopy to omega. This is path homotopy. When you lift omega prime, this ends at the same point. That is the statement. We have to get proof that one. Okay. So this is I cross I. This path goes to omega and that path goes to omega prime. And when you lift it, this is H bar actually. It just looks like H. H bar is from here to X bar. That is the homotopy defines the homotopy of omega bar and omega bar prime. Okay. So what we do? We take this homotopy H with the first path, this path being fixed. This bottom being fixed. We lift up this one. We just lift it. Automatically, the H bar restricted to this T I cross 1 will be lift up omega prime. And what we want to show is that the end point is the same. Okay. Why the end point is the same? Look at this part. What is this part? Under H, all this namely 0, 1 cross I. 1 cross I, because it is a path homotopy, you goes to this point. This whole thing goes to this point. Similarly, this 0 cross I goes to X naught, which means when you lift it, this entire thing, the image must be contained inside these points. Similarly, image contained inside this point. We have chosen the image to be at this point. Okay. Once this point is there, because this is connected space and this is a discrete space, so the entire thing must be this point here. There is no problem. Same way, omega bar we have not chosen where it goes to, second point. It goes to some point and that point is in F inverse of this X1, P inverse of X1. But this entire path here under H bar has to go to the same point wherever this has gone, because this is connected and this is a discrete space. Okay. So that is the property that we use. So this point is the same thing as the end point of this one. Okay. So that is the proof of part 1. All right. And part 2 I have already shown how part 2 comes from part 1. Full detail is written down here with notations. Actually, in this detail there may be some typos, but what I have told you there is crystal clear. Okay. So omega prime is path homotopic to omega. Two different representatives of the same class. First of all, starting points are the same. End points are the same. That is what you have to notice. Now H is a path homotopic from omega to omega. H bar is a lift of H such that 1.00 goes to X0 bar. The rest of them are automatically defined because of what? I cross I is connected. Right. And the uniqueness of the lifting. Okay. By the uniqueness of the lifts, whatever happens we know. It follows that the first horizontal interval goes to omega bar. The second one at one level goes to omega prime bar. Okay. And H0 of 0, s is inside this fiber. 1, s is inside the other fiber. These two are discrete. Therefore, and this is a connect, it is a, it is just I, right. This is just connected space. So the image must be a single point. Let us go to the next result now. This is the most fundamental guess, the discreteness of fibers are used here. Now suppose you have two paths in X with initial point X and end point Y. Instead of X1 and X2 or X0 and X1, I will change the notation. You will be able to do that without notation change. Now there are some problems here with type Y. Just change it, but still it hasn't come. Sorry. So suppose you take omega first and then come back by omega 2, right, omega 2 inverse. That is a loop. Suppose this loop lifts to a loop at X bar. Okay. Little X bar is one where P X bar is the first point X. Little X is not the full capital X. Let omega 1 bar and omega 2 bar be the lifts of omega number respectively at the point X bar, little X bar. Then the end points of this one are the same. You see, the first lemma said if omega 1 and omega 2 are homotopic paths, then the end points are the same. That was the conclusion. Here what I have, I just have omega 1 composite omega 2 inverse lift to a loop. There is no homotopic or anything. So how to get this one? Okay. Let us see. So this is not all that difficult, but it looks somewhat charming. Suppose gamma be a loop at X bar such that P composite gamma equal to omega 1 composite omega 2 inverse. So this is the hypothesis that omega 1 star omega 2 inverse lifts to a loop. That loop, I am calling it as gamma. Okay. So this is a loop at little X bar. By the uniqueness of the lifts, it follows that the first part omega 1 bar, you have lifted omega 1, right, omega 1. What is the definition of omega 1 star omega 2? Omega 1 bar at T equal to gamma of T by 2. That is the definition of the composite here. P of this one equal to first part must be gamma of T by 2, must be this one. Because if I apply P to this one, P to this one what I get is the first part of gamma here, omega 1. For every T inside I, because T is half between this one. It follows that gamma of 1 minus T by 2 when you come back, must be omega 2 bar, right. So P of, if you apply P to this one, then what you get is this part. And that part is defined 1 minus gamma T by 2 for every T. So in particular, omega 2 bar of 1 will be equal to gamma of 1 by 2. But gamma of 1 by 2 is also equal to omega bar 1 of omega 1, okay. So this gamma wherever, if it cannot be loop, if two different paths have been lifted, two loops have been lifted at one point, and the end points are different, you start from one point to go, you cannot come back. You will have to come back by the same path, not from the other path. When you are coming from the other path means the end points of the other two must be the same. So that is geometrically, it is simple, okay. But we have verified now. So this is, this looks like nothing to do with the covering space here. This is simply geometry here, okay. If you combine these two, you get some wonderful result here. And this is our fundamental result now, okay. So start with the covering projection between path connected spaces. Fix base points x and x bar as usual, put f equal to P inverse of x. These notations, I told you I have fixing again and again I am repeating it. The first part here says a loop omega at x can be lifted to a loop at x bar, where x bar is in the fiber, okay. Any x bar is already in the fiber at here. You find only if the element omega as an element of what? As an element of pi 1 of pi 1 of x, x. It must be in the image of P check of pi 1 of x bar, x bar, okay. So this was the first part. And when this loop, this is what it is. So I am just summing it up. This is nothing more than that because I am combining the two lemmas here. Now second part says there is a loop at x bar, which is a lift of omega. If and only if some conjugate of omega belongs to a subgroup, okay. So in the second part, I am not stating the base point for x bar, you see. A loop in x bar, which is a lift of omega. Where is that loop taken? I am not telling. If and only if some conjugate of omega belongs to the subgroup pi 1 of x bar, x bar. But this x bar is fixed. It is sitting over x. The loop may be at some other base point, but the other base point will be also in ZF. You cannot have more freedom because P of that base point should come to x because omega is a loop at x, okay. So you might have taken a different base point. So this part is the one which you have to work out here. The first part is only part of the first lemma, okay. Let us work out what is the proof of B. To see the part B, suppose omega can be lifted to a loop omega bar, some other point x 1 bar, not x bar. Choose a path lambda from x bar to x 1 bar inside capital X bar because capital X bar is also path connected. This is the first time we are using that x bar is path connected, okay. Now put tau equal to peak composite lambda. Look at this lambda is not a loop, but peak composite lambda will be a loop at x in capital X because both the endpoints are in P inverse of x. So this is a loop. So look at this class in the fundamental group. Call that class tau. Then all that you have to check is P check off lambda composite omega bar composite lambda inverse. This makes sense because omega bar is a loop at x 1. This lambda is from x to x 1 bar then this path then this loop. Again you come back by lambda inverse. So this path now this loop becomes a loop at x bar. The omega bar was a loop at x 1 bar. So this is converted into loop at x bar. So P check off that is P check off lambda which is tau, P check off omega bar which is omega and then then I see the inverse. Over, okay. The converse if and only if is there, right. This part is if and only if, right. Let us work out the converse. Suppose there is an element tau in the fundamental group such that when you conjugate by that element, the started with a loop represented by, started with an element represented by a loop omega, you conjugate it and suppose that is in this inside this subgroup. You start with some element. It is none of this conjugate maybe in this subgroup, okay. That is group theory. Suppose this happens, okay. Let theta be a loop in x bar at x bar such that P composite theta equal to this one that is the image. This belongs to image means there is a there is a theta such that P, P check off the class of theta is same thing as the class of P composite theta, okay. That is equal to tau omega tau inverse. Now lift, let lambda be the lift off at x bar, lift off this loop, okay. Loop gamma representing the tau. Tau is an element in in in pi 1 of x bar. So it is a loop pi 1 of x which is a loop at x, okay. Lift it. This lift of course is some part at x bar, okay. So it follows that once you have lifted it theta which is a lift of this one by the uniqueness cannot be anything other than this this lambda is a lift of tau. Omega bar is a lift of omega and lambda 1 is the inverse path for that one, okay. So where omega bar is a lift of this one at x bar. Lambda 1 is what? Lambda 1 is the lift of omega bar at the end point of this lift, right. Omega bar of 1 is some other x 2, x 2 bar, okay. But now look at lambda 1 composite lambda star. Lambda 1 is from x 2 and ends up with x bar. So I can take lambda here, okay. This is the lift off gamma inverse composite gamma, right. Because lambda was a lift off. Gamma was a lift off lambda. Lambda is a lift off what is this? Lambda is a lift off tau, okay. Representing tau. So but this is whatever it is. This is gamma inverse gamma. This is both of them are loops. This is null homotopic. It is a trivial element. Therefore this lambda 1 composite lambda is a loop because the lift of a trivial map is a trivial path. It is a trivial path. Trivial path has both end points same. So all its lifts must be in the homotopic class must be loops. So this is a loop. That means this x 1 bar is equal to x 2 bar. Once they are these two points are same, it just means that omega bar is a loop because omega I have lifted and I did not know either omega bar of 1 is x bar itself, friction bar itself, okay. So these two are same means omega bar is a loop. So what we wanted to show is if some conjugate belongs to the subgroup, okay. Then you have a loop as a lift. That lift is already at x bar. Then you do not have to conjugate because then it is in the subgroup already, okay. If it is not in the subgroup, you may have to conjugate it something, okay. Then the lift will not be at that point but at some other point in the fiber. So this is the theme. All these lifts happening. What is happening? Completely converted into what happens to the fundamental group level and what happens to that inside the subgroup P check of pi 1 of x bar. So this subgroup becomes important for us, okay. So the relation between the fundamental group and the covering space starts revealing itself. The subgroup P check has a special role to play in the lifting property of covering projections. So obviously you would like to know how this subgroup is related to the fundamental group of pi 1 of x bar itself more closely. It is image of that one under P check. So how is this homomorphism P check behaves? That is our next topic, okay. So let us stop here and take up the study later on.