 Welcome to the lecture number 8, module 8. Last time we started the computation of fundamental group of the circle. We have introduced exponential function from R to S1, theta going to e power 2 pi i theta and verified a few basic properties of this map, which is going to be used very heavily in this computation. We also saw that any two lifts of functions from a connected space to S1 through the exponential function, they will differ by an additive constant. In other words, if you fix one point, value of one point, then the lift is unique, there cannot be two lifts with the same value at a single point, this much we have seen. This will be used in the proof of the existence theorem. Now we want to show that given any function continuous map from i to S1 such that, let us assume this is just a easy way of carrying on this one, F0 is 1. If it is not 1, we can also do that, that is not an essential part. Then there exists a unique map G from the interval to R such that G at 0 is 0 and exponential composite G is F. Namely, every map can be lifted uniquely after specifying the starting point. The unique part we have already seen, we also saw that, I mean we also introduced what is our plan of proof. Look at the set Z, set of all points inside the closed interval such that G is defined in the closed interval 0 to that point T. So, this is a subset and it is non-empty because I can take T equal to 0, then of course we know that G0 is 0 can be given. Then exponential of G0 which is exponential of 0 is 1 which is F, F0. So, that verifies. So, Z is non-empty. Our idea was to prove that Z is both open and closed. I being connected, a non-empty open and closed set must be the whole space. So, that is the plan. Using the order in I, we can even simplify this idea even further by the following idea. Namely, let us look at the supremum, the least upper bound of the set Z. The set Z is bounded. Therefore, it has a least upper bound. That upper bound may not be inside Z. It will be inside 0, 1. We will claim that this upper bound T0 is actually inside Z. That corresponds to almost proving that Z is closed here. It is a closed interval. So, the old idea is slightly changed into proving that the supremum belongs to Z and the supremum is nothing but 1. So, that will prove that Z itself is the whole interval 0, 1 which is the same thing as saying G is defined on the whole of interval 0, 1. So, we want to prove that the supremum of Z is inside Z and that supremum cannot be smaller than 1. It has to be 1. Of course, it is bigger than 1 because already 0 is there, right? Okay? Now, consider the open set V which is S1 minus F of T0. T0, T0 is some point in the interval which is the supremum of the set. F of T0 is defined. Look at the minus of that and throw away that point. Then you get a big arc. On that arc, we have a log function. Log function is from the arc to arc, back to arc, okay? So, there are many log functions, right? Which branch you would like to choose? So, let us look at this one. For 0 less than epsilon less than 1, let us have this notation. Namely, I epsilon is T0 minus epsilon to T0 plus epsilon. I would just like to have this much but it may happen that my choice of epsilon is somewhat big and it may go out of the interval I. So, I will intersect with I, okay? If epsilon is sufficiently small, this will be completely inside I. There is no need to intersect with I. That is all. Given epsilon, look at this. This is in the neighborhood of the point T0. Therefore, by continuity, okay, there exists some epsilon positive such that the entire of F epsilon will be contained inside this open set V. So, this is all not continuity. This is actually, yeah, okay? Some open set is there. I am taking this M to A. T0 goes into that. So, some F of epsilon is contained inside V. That is continuity. V is open. T0 belongs to, F of T0 belongs to this side. Okay? I have thrown a minus of F of T0. F of T0 belongs to V. So, F of I epsilon will be contained inside V. For some epsilon. Now, if we use lemma 2.3, okay? Look at the log functions. They are several of them inverse, inverse of exponential function. There will be one copy from V to U, okay? Which is the inverse of X where U is the interval containing G of T0 minus epsilon. G of T0 minus epsilon makes sense because T0 is the supremum of the set Z. So, everything smaller than T0 will be inside Z. Therefore, G is defined there. So, G of T0 minus epsilon makes sense. This will be in one of those various inverse images. So, I am choosing U to be one of those intervals, okay? And there, the logarithm function now from V to U will be a inverse of exponential. So, it is the one-one mapping, okay? So, G of T0 minus epsilon is such that this interval U is contained instead of exponential of V. Exponential inverse of V as all these disjoint intervals. All that I do is now put H equal to logarithm of N namely logarithm of F, namely ln composite F. Come F from F is defined on this subset F of I epsilon will go inside V, right? We will take F inside I epsilon over here. Then we have G of T0 minus epsilon by 2 will be H of T0 minus epsilon by 2 and exponential of G equal to exponential of H on this interval. Therefore, by uniqueness, at one point if they agree, they must be agreeing everywhere. Hence, by uniqueness again we have G equal to H on this interval. There are two functions now. At one point they agree. So, they must be agreeing on same interval. Therefore, we can extend G into Z. On Z it is already defined. Now, union I epsilon, this interval, okay? So, this is the trick. Pick up a point. If that is the end point of the definition, you can extend it slightly. That is the whole idea. Now, the end point I do not know even it is not defined at the end point. I do not know because it is a premium. So, premium may not be inside. But any small value, if you move to the left, that will be also there inside that. So, from there I can extend it. So, now Z union I epsilon, this will have T0 that that sorry the point T0 in the interior because I epsilon goes beyond that one, right? Okay? So, this first implies that T0 is not T0 is already inside Z. But if epsilon is in the 1 plus epsilon, if this intersection, if T0 is in the interior, then T0 will not be the optimum. Therefore, T0 must be the end point. The end point is, end point of this interval is has to be 1. So, the moment it is smaller, that epsilon will be larger. So, it will go beyond. Therefore, T0 is less than 1 will give you a contradiction because there will be a larger number inside Z than T0. So, that is the episode. So, T0 must be 1. So, in one single go, we are getting both essentially showing that the set Z is both open and closed. So, this is easier than that part. So, we are just fine. Okay? So, essentially we are using the local behavior of the logarithmic function. So, if it is partly if the function is defined there, we can extend it to the full thing. Okay? Inside that, inside that intervals, once your chosen logarithmic function, it is a homomorphism. It is a unique way of extending. There is no, there is no ambiguity about that one. So, these things are all used very rigorously here. Okay? So, that proves the existence of lifts up to additive constant. They are unique also. All right? All paths can be lifted up to additive constants. So, now we make a divination. Having prepared this match, namely passage from loops inside S1, now we can go to some maps into paths inside R. Okay? So, what happens to them? Start with a function map from I to S1 such that F0 is 1. We take the unique map G from I to R such that G composite F, sorry, exponential composite as G is equal to F and G of 0 is 0. Further, if F is a loop, suppose F, you started with a loop means F of 1 is equal to 1. Then G of 1 will be also an integer because e raised to pi i of G1 must be equal to 1. Okay? We call this integer the degree of F. Now, this is a string thing. We took some G but that G is unique. Okay? And that is why we can define the value of G1 as the degree of F. In particular, suppose you start with a map from S1 to S1. That is a loop. We remember that one. We have a map from I to S1 with both 0 and 1 going to the same point can be thought of as a map from S1 to S1. We can view it as a loop via the parameter t going to be power 2 pi i t. Okay? So, that is t going to F of 2 pi i t take the corresponding map G from I to R and call it for G1 as the degree of F. So, here we would like to convert a map from S1 to S1 into a map from I to S1 and then lift it. Okay? From S1 to S1, we do not want to lift it. Map from S1 to S1 may not be, you cannot lift it because the end points may go to different point. That is the whole idea. Often the end points will go to different points when you lift it. Okay? So, to allow that, you have to think of this as a path from I, a map function from I to S1 rather than S1 to S1. Okay? Okay? The whole is reversed here. Justification for this terminology, why the degree? That needs some explanation, right? So, here is an example. Let us take Fz equal to Z power n from S1 to S1. Then the map G is nothing but Gt is equal to n times t and hence G1 will be equal to n. E power 2 pi i, say some theta raised to n. What is it? That is all I am looking at. E power 2 pi i n times theta. n times t, that theta, instead of t, that is all. Okay? So, the lift of the map Z1 to Z power n from S1 to S1 converted into a map from I to S1 is nothing but n times t. So, that n is the degree, the degree of this polynomial Z power n. Okay? So, this is, right now, this much of justification is good enough. Slowly you will understand that this degree is really a good name. Okay? Later step by step, this concept will generalize to maps from S1 to S1 also and then to maps from manifolds to manifolds. So, this degree concept is very, very important. Okay? The first thing is we have taken a function from S1 to S1, map from S1 to S1 and then assigned a degree to it. This degree, this association, the number is a homotopy invariant, path homotopy invariant. So, this is what we want to do. Okay? So, this is the purpose in 2.2. If F1 is path homotopic to F2, then the degree of F1 is degree of F2. Okay? Let us go ahead, we will take care of this edge later on. Let us start with a homotopy edge from i cross i to S1 to a map such that h of 0 s equal to F1, h of s1 comma s to F2 Fs, h of t0 equal to h of t1 equal to 1 for every t and s belong to, that is the definition of a homotopy, path homotopy from F1 to F2. Okay? Now, let G from i cross i to r be a function such that exponential of G equal to h. Okay? And for all t inside i, the function s going to GTS is continuous and G of t0 equal to 0. I am taking a function G from i cross i to r, I am not telling that it is continuous. But if you fix t, then as a function of s, this is continuous. Okay? And G of t0 is 0. For each t0, I am defining a lift of the function h of ts where s goes to h of ts. So, that is a path. So, that path can be lifted. Each fixed path for when first coordinate is fixed, you can lift it by the previous proposition. So, such a G is there, right? G of t0 starting point is 0. That is what I do. Okay? By proposition 2.1, namely what we have done, such a function G exists. Okay? So, in order to prove the proposition, we have no other choice but to prove that G itself is continuous. If I show that G itself is continuous, then what I get is a homotopy of the lifts. Okay? So, this is what we wanted to prove. Okay? So, that capital G itself is continuous exponential of G composite. So, capital G will be a lift of h. Okay? So, we have no other choice but to prove that G itself is continuous jointly in both variables. That is the point. Only when you have fixed s, fixed t, s goes to G t of s is continuous. That much we know. Okay? So, this is what we are going to prove now. Put u plus minus after throwing away minus plus. Namely, u plus is s1 minus minus 1 and u minus is s1 minus plus 1. Okay? Minus is throwing away the point. So, these two are open arcs u plus and u minus. They will cover the whole of s, whole of s1. Okay? The inverse image of these things under capital H, started with a homotopy h. Right? That will be an open cover of i cross i. Right? Right? So, for compact sets with an open cover, you have the Lebesgue number. Okay? There exist number delta positive is called Lebesgue number for this cover such that if s is any subset, sub square of i cross i of side length less than delta, then the entire of hs is contained in either u plus or u minus. So, how do I manage this one? All that I have to do is take the Lebesgue number and take square root of delta divided by square root of 2, delta to be less than the Lebesgue number divided by square root of 2. Then the sub square of side length i cross i with less than delta, if that is the case, then its diameter will be root 2 times delta which will be less than the Lebesgue number. Therefore, this open subset will be contained inside h inverse of u plus or h inverse of u minus, all these sub squares. It is the same thing I say h of the sub square is contained inside either u plus or u minus. So, let me show you the diagram first, then explain this one, let us more clearly. So, what I have done is I have cut down i cross i into smaller squares, each one small square here under h will go either inside this whole u plus here or the u minus, remove this point and take all this part. Either it avoids this minus 1 or it avoids plus 1. So, it will not be a subset like this starting from here to here that is the whole idea. It avoids either u the minus 1 or avoids the plus 1. The image of each square here, is that understood? So, this is what the Lebesgue number does in all several cases in analysis. So, this part of this analysis here. So, once you have such a thing, what I do is I will cut down the squares, cut down i cross i into squares like 0 less than 1 by n less than n minus 1 by n. I have to choose n to be sufficiently large so that 0 to n is side length, this 1 by n is less than this delta that is all. Then each square will be having side length less than delta. So, this hypothesis will be true. To show that now g is continuous, g is defined as a function which is defined. But to show that g is continuous, it is enough to prove that restricted to every sub square which is continuous. How does a sub square look like? Something k by n to k plus 1 by n cross L by n to L plus 1 by n, where L and k are between 0 and n. So, for each of each square if it is continuous, then you are done. There are finitely many squares, they are all closed squares. So, I have to show that h is continuous on each such square. So, how do we start? We start from the bottom here. We first show that the function h is continuous on each of these bottom squares. Using that we will put each of it is in second stage, third stage, fourth stage and finally. So, proving the continuity of h will be also done inductively on this value of S here. How do you keep moving along across that? So, what we have done? L equal to 0, consider k S of k 0. S of k 0 may be this square here or this square, this square, whatever, one of them. S of k 0, this whole thing, remember all these are initial points, they are here, they are mapped to this point. Therefore, this entire thing has to avoid this point which means they are in containing the U plus, S1 minus 1. They have to avoid S minus 1 because they cannot avoid plus 1 because all these points are mapped to this point here. Therefore, G of S k 0, these things is contained in a disjoint union of the inverse image of U plus. Inverse image of U plus is nothing but N minus half to N plus half because half is the point which goes to minus 1 under exponential function. So, that is avoided. So, all these intervals of length 1 disjoint union of that, one of them must contain G of k 0 because G of k 0 on that square, all these lines, see this is a trick here. G may not be continuous on S k 0 on the whole of it, but on each vertical line it is continuous. It is continuous, the vertical lines are connected. So, if it is contained in the union like this, it must be contained in one of them, only one of them, which one, wherever the starting point is. The starting point, we know where they are. They are all been chosen inside 0. Starting point, little G has been chosen to be starting point of that. So, that is what. So, G of t 0 is 0 for all t. We know that G of t cross 0 1 by N is contained minus half to half, but continuity of G restricted t cross i is what is happening. So, G of s k 0 must be inside minus half to half. N must be 0. Once you have one single thing, the exponential function is a homeomorphism with its inverse being log here. Therefore, G itself will look like log of h, but log function is continuous here. Therefore, G is continuous on the whole of s 1. So, on each s n 0, all of them it is continuous, where log is the inverse of exponential s 1 minus s 1. So, in particular G is continuous on each s k 0 G, the first blocks here along the x axis here, t axis. Now, what happens in particular, if you look at this line, this is 1 by N, s equal to 1 by N. So, on this one it is continuous. Now, use that fact, prove it the same now. Look at one of the squares here, same argument, will tell you that it will be continuous on each of these squares. In particular, it will be continuous like this on this line. Once you have that, it will give you for the next squares and so on. So, induction starts. So, it will be continuous on each of these squares. This argument will be used in a much larger context later on as to argument this all we need. But details will be slightly more complicated, technicalities. We will do a big generalization of this one later on when we study covering spaces. So, what we have got is G itself is the continuous function from I cross I to R exponential of G is the original H. Now, look at degree of F 1 is G 0 1, but that will be also equal to G 1 1, but that is degree of F. Difference is a constant, that constant must be the same for both of them because they are all in one single G continuous function. Difference is a constant integer, that is what we knew. So, that integer must be the same. Okay, we just took at the last coordinate G that will give you the various integers. Degree of starting, when you put the starting value here, 0 that will be F, the F 1, that end will be F 2. So, it is what we are doing here. The 0 1 will be F 1, G 1 1 will be F 2. Then whole thing is continuous. So, yes, because we have a very defined function from the set of which actually set of homotopy path homotopy classes of maps from S 1 to S 1 to integers, namely the class of F goes to the degree of F. Irrespective of what class you choose because they will be homotopy, the degree will be the same. So, this is our first attempt in the computation of this group. We have got a function out of that, synthetic function. The next step is that we want to show that it is a group homomorphism and it is surjective and injective. So, it is an isomorphism. Okay, that we will do in the next module. Thank you.