 So the last thing we proved was the path lifting lemma and the homotopy lifting lemma. So here's the corollary. So what does this do this is? So what is the situation? We have a covering, P, the ERB, a covering. And we fix some points, P of E0 equal B0 as before. So let F and G be pathless in B from B0 to B1, which are path homotopic. So they have the same end, which are path homotopic. Let F and G be pathless in B from B0 to B1, same initial point, same endpoint, which are path homotopic, and what? Sorry, I don't understand. Yes. Let F and G be pathless in B from B0 to B1, which are path homotopic. And let F tilde and G tilde be their liftings to E with initial point E0. So this we can lift, we know, OK? We fix the initial point and then we can lift it's even unique. Then also, F tilde and G tilde are path homotopic. And in particular, it's the same endpoint. Otherwise, we don't, and in particular, have the same endpoint. The same initial point is E0, have the same endpoint. In other words, if you lift homotopic passes with the same initial point, then they have also the same endpoint, the liftings, OK? They are also path homotopic, so the proof. So we have a path homotopic, OK? Between F and G, so there's a homotopic, H, homotopic. And the path lifting lemma, the homotopic lifting lemma, H lifts by the homotopic lifting lemma. H lifts to a path homotopic. Also, the lifting of a path homotopic is a path homotopic. To path homotopic. H tilde. So the situation is this. We have H, and it's a homotopic between F and G. Here's constant, which point, B0. And here it's constant, B1. So this is H, OK? A homotopic from F to G, a path homotopic from F to G. This means constant. So H tilde. Now we have lift H tilde. So the constant path, this is also a path homotopic, OK? The initial point here is E0, so this is E0. This is constant also here. It's constant also here, OK? It's also a path homotopic. I don't know the point, some point E1, OK? But these two paths, this is a path and this is a path. They are liftings of F. This is a lifting of F, and this is a lifting of G, OK? And they have the initial point, E0. Lost path homotopic H tilde. So by uniqueness, OK? So I say that by uniqueness of the lifting of F with this initial point, with the right initial point, this lifting can only be F tilde. There's only F tilde. So this here is F tilde, and this here is G tilde. So by the uniqueness of path lifting with a given initial point, in this point, E0 in this case, no? E0, we have that F tilde is homotopic to G tilde. H tilde is a path homotopic between F tilde and G tilde. This can be only F tilde, and this can be only V tilde. That's the uniqueness, OK? Because it's a lifting of F and G. OK, that's all. So this is a path homotopy, and so they have also the same endpoint, E1 in this case, OK? So this is the corollary of path lifting and homotopy lifting, this corollary. OK, for theorem, pi1 of the circle is base point 1, well complex number 1, OK? The circle we are in C is isomorphic to Z, the integers. So proof. So we define a mapping phi from pi1 of S1, pi1 of S1 to the integers. We want to define a mapping, OK? So what do we do? We take an element here, so that's the path. The homotopy class of a path, OK? So W is from I to S1 is a path with initial point and end point 1, this one here, with W tilde of 0 equal W tilde of 1 equal 1 in S1. So we do want to define the integer. This is a rotation number. How to get this rotation number? This can be interpreted as a rotation number. So what is a rotation number? How can we define this formally here? So we just take, let W tilde be the lifting of W. So we consider the P from R to S1, the covering, our covering. The P of t is equal to E2 pi I t, which we discussed yesterday. We take this covering. Let W tilde from I to E be the lifting of W with initial point, W tilde of 0 equal. So we have to choose one point which projects to 1 in S1. And that 0, the best choice is 0, is 0 in the reals. The lifting of W, this is not W1, W with initial point W tilde of 0, which is 0 in the reals. Then, of course, I mean W is a closed path, OK? So then W tilde of 1, the end point of 1, OK? What? Yes, that's E is R in this case, OK? OK, yes. This initial point, then W1, of course, is in the fiber pi minus 1 of 1, OK? But under this mapping, which goes to 1, the integers, exactly, OK, which is z. So that means W tilde of 1 is an integer, OK? This is a rotation number, OK? The rotation number can be, you can call this the rotation number of W. How to get this, the rotation number? You take the lifting with initial point 0. The end point is an integer, and this is the rotation number, OK? If you go downstairs, n times around, OK, in one direction, in S1, n times, OK? And you lift this path, and you go from 0 to n, OK? And you get the rotation number in this way, at least in this special case. It's just a name, OK? You call it rotation number, and it doesn't matter. So we define phi of W equal to, well, at this point, W tilde of 1, which is an integer. So this is the mapping phi. Of course, there's a big problem here, you know? That's well-defined, OK? First, we have to lift. That's a path lifting lemma. And now, it has to be well-defined, OK? So if you have homotopic paths, if, so phi is well-defined, this is, if W is W prime, or if you have another representative, that means W is homotopic to W prime, OK? H, homotopy, some path homotopy. Then, by the homotopy lifting lemma. So let W tilde and W tilde prime be the liftings of W and W prime with initial point 0, OK? Our initial point is 0 in our, in the way. This initial point 0, in the way. By the corollary, by the path lifting lemma, by the corollary, W tilde of 1 is equal to, they have also the same endpoint, OK? W prime tilde, they have also, because downstairs they are homotopic, we lift the homotopy and upstairs it's also a path homotopy, so they have also the same endpoint. And this means, of course, that phi is well-defined, OK? So it doesn't depend on the representative of this cloud. So phi is well-defined. OK, this is the main part of the proof, OK? The rest is true, almost. Now we have to prove something still, but the rest is very easy. Here, this is an important part, OK? We use the path lifting lemma and we use the homotopy lifting lemma or the corollary, OK? So this is the main part of the proof. Now we have to prove something. What we have to prove, so after the definition, phi is a homomorphism of groups, OK? Phi is a group homomorphism. That's the first one we have to prove. So what does it mean? It means phi of w times v should be equal to phi of w plus in z now, plus phi of v, OK? That would be a homomorphism. So what is phi of w? This notation as before, this is w tilde of 1. So w tilde is a lifting with initial point 0 and then we look at the endpoint with the rotation number. So this is w tilde of 1. What is this? It takes the same for v. We lift it, v tilde with initial point 0 and takes the endpoint, so v tilde of 1. v tilde is the lift of v, obviously, in the same way, OK? Now we have to look at this. So this is phi of w times v. So this is defined as we have to lift now the product and look at the initial point 0, or this, and then look at the endpoint. So this is the lift w times v tilde, the lifting of this product and the endpoint. That's the definition, three times the definition of phi, three times the definition. Now we have to look, who is this pass? So we want to understand this lifting. Well, this means first we lift w, this initial point 0, that we have already, now that's w tilde. This lifts w with the right initial point, OK? And then we have some endpoint, which we don't know. And then we have to lift v. But not with 0 initial point, because maybe the endpoint here is not 0, it's the rotation number. But the initial point is the endpoint of this, of course, OK? The endpoint of this is w tilde of 1. So I cannot write v tilde here. This is the wrong initial point, OK? That's not defined in general, that's not defined, OK? But what I have to do is I have to translate v tilde, which starts from 0, so that it starts at the endpoint of this one, right? It must start in the endpoint of this for the product, OK? So what has been, I don't take w tilde, but I translate v tilde, OK? I translate it with which number? With the w tilde of 1, with the integer, OK? So w tilde 1, so this is this integer, which is some number in that, which I don't know, OK? So this means a translated path v tilde, OK? A translation, OK? A translation of v tilde. And now this is defined, OK? Now this starts, this w, v tilde of 0 is 0, OK? So it starts here, and it ends somewhere. So w times v tilde of 1 is, of course, this is first, this ends this. It's the endpoint of this, OK? The endpoint of this is v tilde plus w tilde of 1. This is just a constant, OK? This is just a constant, we add, we add the reals. Add 1, right? This is w tilde of 1. And then we have this translation plus w tilde, v tilde of 1 plus w tilde of 1, OK? So this is the endpoint. And this is, of course, exactly what we want, OK? This is w tilde of 1 plus v tilde of 1, OK? So this is what we need. So they are equal here. And that's the homomorphism. So this is just looking at the situation. There's nothing difficult here, OK? Just definitions. So this is a homomorphism of groups. And then we want an isomorphism, right? So second, injective. So we prove that the kernel is true here, OK? So we take something in the kernel, w in the kernel of 5. What does it mean? It's in the kernel of 5. This implies that w tilde of 0 is 0 anyway, no? What is w tilde of 1? So in the kernel, that's everything which finishes in 0, right? So it's 0 also. That means exactly this, OK? If and only if, OK? So this means w tilde is a closed path. It starts in 0, and it finishes in 0 because it's in the kernel. So it's a closed path. So it's an element where w tilde is an element in, it's a closed path in the reals, no? It's an element in the fundamental group of the reals with base 0, OK? That's our initial point, and also end point here. But what is this group? The fundamental group of Rn is trivial. So this is a trivial group. That's a straight line homotopy. This is a trivial group. So what means? That means this is a trivial element. So this implies that w tilde is equal to the trivial element. What is a trivial element? That's a constant path in 0. Or if you want, w tilde is homotopic to the constant path by some homotopy f, maybe h. And then we want to prove that w, what is w? w is w tilde and then the projection, no? It's homotopic to what? To c0 and then the projection. What is this? c0 and then the projection. This is c0 to 1 in s1, c1, constant path. This is constant path in s1. This is in R. This is projected. And what is the homotopy here? I mean, look at this. It's clear what the homotopy is. You project h. You project everything. h and the projection, OK? This is very trivial, this direction, OK? What is difficult is lifting in the other direction, OK? Here we just project everything, and it's OK. So this means that w is equal to the constant path, OK? And this means that the kernel is trivial, OK? This implies the kernel of phi is trivial. And of course, this means that phi is injective, because we know already that it is a homomorphism of groups, OK? So this is injectivity. And the last thing is surjectivity, which is even easier, 3 phi is surjective. So let n be in n. Let n be in zeta. No, not r. Sorry, that makes no sense. We take any integer, OK? And then, of course, now we want to find a path, OK? Which gives this n. So we take let w let u be any path in the reals from 0 to n. Let u from i to the reals be any path. u of 0 is 0, the initial point. And u of the n point, n. This is the rotation number. Now we project, OK? Now we project. Let w be the projection of u, u and then the projection. Then w is a closed path, because the integers go all to 1, OK? So let w, this is a closed path, because integer goes to 0. It goes to 1 in S1. And that means that this w is an element of pi 1 of S1, 1. It's a closed path. And now what is phi of this w? Then phi of w. So how do we find this? We take the lifting of w with initial point 0, OK? And then we take the n point, so w tilde of 1. But who is w tilde? I mean w is a projection of u, OK? So this is u of 1, because w tilde is u, OK? The lifting is because we project it. And u of 1 is what is this? n. So in other words, it's easy to find path with rotation number n in S1. You just go n times around, OK? And if you lift this, you have the linear path from 0 to n. That's trivial, OK? And this means phi is projective. And that's the end of the proof. That's an isomorphism of groups, OK? And this is a rotation number, no? You have any path in S1 which goes very complicated, no? I mean there are easy paths which just go n times with constant velocity around n times, no? Then the lift of this goes from 0 to n. That's clear, OK? That is trivial. But if you have a very complicated path, no, which there's a lot of, then it's not so clear. Now first we have to lift, and then we look at the end point, and that gives us a rotation number, OK? And that's all. And then it has to be well defined. These are the two main points, the lifting of any path, not just of a trivial path, which goes n times around, OK? But the path which may be very complicated, we have to lift this, and then it doesn't depend on the choice of representation. And that's the homotopy lifting lemma, OK? So these are the two main points of the proof. The path lifting lemma and the homotopy lifting lemma, or it's color. OK, so this is a proof. That's the fundamental group of the circle is the integers, and the isomorphism in a sense is given by the rotation number of a path, which we can define in this way, OK? We just define it in this way. That's we call, we give the name, oops, rotation number. OK, so I give two applications now. The first one is a fixed point theorem of Brouwer. Let me define, so let A be a subspace of X be a subspace. This is an important definition anyway. Let A be a subspace, A of X be a subspace, a mapping, a continuous mapping, a continuous map. R from X to A to the subspace is a retraction. That's a definition of retraction. That's a continuous map from X to the subspace, such that the restriction of R to the subspace A, OK, is the identity on A. On A, it doesn't move points. So it retracts X to A, OK, and A doesn't move. So this is the identity of A. That's the definition of, that's the definition of retraction. A continuous map to a subspace, which is the identity on A, which doesn't move points in the subspace, OK? This is equivalent to the fact that this can be written in a different way, that first the inclusion and then the retraction is the identity of A, where I from A to X is the inclusion. This is exactly the same as this. This or this is the same, OK? Here you have the inclusion and here the restriction, but it's exactly the same. So that's the definition of retraction. For example, I define a retraction from R minus 0, Rn minus 0 to Sn minus 1. What is Sn minus 1? These are all points in X in Rn such that the norm is 1, OK? This is a unit sphere. The norm of X is 1. This is a unit sphere. Unit sphere in Rn, that's Sn minus 1, OK? S1 is the unit sphere in R2, OK, on C, if you want. So what is a good definition of such a, what is a retraction, a retraction of this? For example, if you take any point X here, no? And what should be the image? Yes, exactly. X by the value of the norm of X. This is a retraction. That's an example. So small lemma, if R is a retraction, by the way, then if there exists a retraction, A is called a retract of X, OK? If R, this R, not V, the R, from X to A is a retraction, then the induced map R star from pi 1 of X, with some base point, which I choose here, this is a subspace, no? A0 to pi 1 of A, A0, is subjective. And the inclusion induced map, I star, which this is inclusion as before, from pi 1 of R, A, A0, to pi 1 of X, X0, A0, is injective. So here A0 is a point of A, which is a subset of X, a subspace of X. So our star of a retraction is subjective, and I star. This is very trivial because, to the proof, what means a retraction? That means that first the inclusion, and then the retraction, is the identity map of A, no? That's our definition, the second version of our definition. That by the factorial properties, OK? That implies, of course, that I and then R, the induced map, is the identity of A, induced map. And these are equal, the induced maps are equal, of course. And now we have the factorial properties, OK? This is I star and then R star. And this is the identity of the fundamental loop, A, A0. So they are equal. And this is by the factorial properties. There are two, no? By the two factorial properties. So we have this, and what does it mean? This means that I star is injective, and R star is subjective, no? As usual. So this implies I star, R star is subjective, and the inclusion star is injective. Because this is subjective and injective, OK? If this is subjective, injective, then this has to be injective and this is subjective, no? Because this map is both. OK, so for a retraction, we have the induced map is subjective. You mean this might be an isomorphism? This is an isomorphism. But that doesn't mean that these are isomorphisms, no? This composition is the identity. That's an isomorphism, of course. Yeah, but that doesn't mean you need more identity. OK, it's clear, right? The proof, yes. So now we give you an implication. So these are general stuff. And all the first application, proposition, oh, this is easy. There is no retraction. No, well, there is no retraction R from the unit ball. So what is the best name in the book, I forget? B or D? Disk, B. So these are all points. Z in x in R2, such that x norm is smaller equal to 1 now, OK? That's a unit disk. What? Well, yes, imagine 2 is disk, OK? And then it's unit ball in all dimensions. So R, B to S1. S1, by the way, which is the boundary, OK? So this is the boundary of B2, S1 B2, OK? So B2, this is the boundary of B2. There's no retraction from here to here, OK? There's no retraction. Proof is easy. I mean, it follows from what we proved. So proof, if R would be a retraction, would be either a retraction, by contradiction, if you want. If you have a retraction, suppose you have a retraction. If it's a retraction, then R star is subjective. Then R star from pi 1 of B2, base point, I take 1 also. And the same base point as here, no? So this then is Z in C, Z smaller equal to 1, the complex version, no? So pi 1 of S1, 1, the same base point, OK? 1 as a complex number. Then it's subjective. But what is this group? Now we have this group is isomorphic to the integers, OK? What is the group of B2, the fundamental group of B? What? Trivial, yes. So this is the trivial, not trivial, trivial, trivial, trivial, trivial group, right? This is by straight line homotopy, like Rn, OK? Here we have straight line homotopy. The same proof as for Rn. So this is a trivial group, OK? And of course, there's no surjection from the trivial group to the integers, right? So this is a contradiction, so contradiction. That's not possible. The important fact here is that this is not trivial, OK? That's the only, this is not trivial. That's the important fact. That this is trivial is trivial, almost, OK? This is easy. The important fact is that this is not trivial. So there's no retraction. By the way, there is a retraction, of course, remark. There is a retraction from R from, that's almost what we did before. If you take b2 minus 0, then we have a retraction 2s1, OK? And that's x goes to x over the norm of x. That's a retraction. How does it look like this retraction? So we have b2, and here's 0, right? Which is not there. So you take any point, and then you project it, OK? So if you have this x, then this is R of x, OK? That you see this retraction, this is this retraction, OK? But if you have 0, there's no retraction. You don't see any retraction. You don't know how to move 0, no? 0 cannot go to everything, OK? But you cannot move. So this is the fact that there's no retraction if we, OK, this is easy. And this uses the fact that it's a fundamental rule. So theorem, and this has a name. It's a famous theorem. It's a Brouwer fixed point theorem in dimension 2 law, OK? It's in all dimensions, but we have dimension 2. Brouwer fixed point theorem in dimension 2. So what does it say? Every continuous map, continuous map, f from b2 to b2 has a fixed point. Very easy, no? Formulation. Every continuous map from b2 has a fixed point. That's a famous Brouwer fixed point theorem for 2. This holds for bn, OK, for each n. Let's prove this first. So prove. Suppose f has no fixed point, OK? And you want the contradiction. So I make a picture now. This is b2. The unit disk, and we have f. So any each point, x, goes to another point. f of x, which is different from x. It has no fixed points, OK? So what you take is you take the segment from f of, well, to x, from f of x. So these two points define a segment, a line. And then you go in this direction on this line, from f of x to x, OK? And this meets the boundary, which is s1, in two points. And you choose this one where you go in this direction. And you call this r of x, OK? That's the definition of r. So this defines r from b2 to s1. So if you take a point x here on the boundary, f of x is different from this point, no? I don't know where it is. It's here, f of x, somewhere, but not the same point. Then you get the same point back. So then rx is equal to x, OK? So this r restricted to s1 is the identity of s1, OK? And r is continuous. Why is it continuous? Well, I don't want the formula, because since f is continuous, OK, since f is continuous. Well, very, I mean, you can compute, if you want, you have x coordinates, two coordinates, no? You have f of x, which depends in a continuous way of these two coordinates, OK? f is continuous. And then x and f on x, you can compute the line and then you compute this point. And it depends on the continuous way of the coordinates, OK, of the coordinates of x. Because f of x depends on the continuous way, and then also there's some square root of some expression, OK? In other words, if you vary x very little, OK, then f of x varies very little, because f is continuous. And then also this varies very little, OK? But you should see the formula, OK? You have x two coordinates in r2, x1, x2, xy, no, xy is not good, x1, x2, OK? And then you have f of x, which depends in a continuous way, f1 of x of these coordinates. And then you can compute this point, OK? And it depends in a continuous way on these coordinates, OK? That's clear. So this means that r is a retraction. But there is no retraction, OK? So this is a contradiction. There is no retraction from b2 to r, to s1. No retraction. There's an exercise which I gave or not, no, I don't remember even. The same for b1. What is b1? That's minus 1, 1, OK? This is in r, r1. That's a ball in r1, OK, the unit in some sense, no? So every continuous map, f from b1 to b1 has a fixed point. That's the problem of fixed point in dimension one. This is much easier, OK? Was it an exercise? No? No, it's an exercise. I forgot. Maybe you forgot also. 0, 1, 0, 1, it's the same. f from 0, 1 to 0, 1. Now I have minus 1, 1, OK? Yeah, it was 0, 1, 0, 1, maybe in the book, OK? So how do you prove it? I give the same proof as this, OK? What? No, no, no, that was a homework. Every continuous map from 0, 1 to 0, 1 has a fixed point. Yeah, and now it's minus 1, 1. Long time ago. Yes, that's the same exercise. Except that now I take minus 1, 1, b1, OK? But no difference. That's homeomorphic to 0, 1. All closed intervals are homeomorphic, OK? So this was an exercise. Yeah, I know I'm doing the exercise in this line, OK, along these lines. So what? So I define the retraction here in the same way, OK? I define r. Suppose f has no fixed point. So prove. It's the same proof. Suppose f has no fixed point. And then I define a retraction, OK? Now I have b1. This is b1, minus 1, 1. This is b1, no? And I define it exactly in the same way. What was the way? I take a point x. I look at f of x, which is different from x. And then I take the retraction, which goes from f of x to x, and goes to this point, OK? So you go from f of x to x, as here. You go from f of x to x, and then you meet the boundary. So db1, what's that called? That's s0, no? There's two points. s0 is minus 1, and minus 1, OK? And so this is r of x, in this case. So now we can write a formula. Here it's easier to write a formula, OK? No? What is r of x? Let me try. Don't write that immediately. f of x minus x, which is not 0. So I can divide by the absolute value, no? f of x minus x, absolute value. It's not the, wait a second, I have to check, OK? If this is this mapping, no? So where goes 1? I'm interested in 1. No, in minus 1 first, OK? In my, well, doesn't matter. f of minus 1 is on this side, it's not minus 1, OK? So r of minus 1 should be, what I want is r of minus 1 is minus 1, OK? This is the definition, of course. And r of 1 is 1, right? That's what we need. So is it true here? So I have to write minus, yeah? Let's see, if it's minus 1, if it's 1, x is 1. This is smaller, so this is negative. So it becomes minus 1, yeah? I need the other one. x minus f of x, x minus f of x. So that's OK now, OK? This implies r of 1 is equal 1, r of minus 1 is equal minus 1. Or if you want, r, restricted to s0, is the identity of s0. So this is a retraction. r from b2, from b1 to s0 is a retraction. As before, in particular, surjective, no? Retraction is surjective. Where's the contradiction? We need the contradiction, as in the first case. In the first case, we use fundamental rule. No, where's the contradiction? So r goes from minus 1, 1 to 1, minus 1, no? Yeah, exactly. This is connected. The image of a connected is connected, OK? This is not connected. Not connected. Obviously, it's not connected. No, you get a separation then. What is a separation? r minus 1 of 1. So you write minus 1, 1. You have a separation, which is r minus 1 of 1, union r minus 1 of minus 1. If they're both non-empty, there's a separation, no? But there is no separation. So we have a contradiction again. So it's the same proof, right? Formally trivial, straight line homotopy. Pi 1 of r n is trivial, pi 1 of b n is trivial. They are all trivial groups. You always have this straight line homotopy. Like b2, OK? So in general, yes. Now, fundamental group P is not very useful, no? Because here we have just two points, OK? Pi 1 of r n, 0 is trivial. Pi 1 of b n, 0 is trivial. Pi 1 of whatever you want, OK? Of any convex subset of r n is trivial, OK? Pi 1 of a convex subset of r n with some base point is trivial. Because in a convex subset, you don't need convex subset. What you need is star-shaped, OK? But let's say convex, that's OK, OK? That means with any two points, you have a straight line, OK? So you have a straight line. It's always straight line homotopy. Pi 1 of any convex subset of r n is a trivial group by straight line homotopy. And this is a special case of this, of course. And this is also a special case of this. Now, a fundamental group here, this is trivial, fundamental group, OK? That was your question, no? This is trivial group. And this is trivial also, OK? So everything, fundamental group doesn't help here. But it's easier. It's connectivity, no? So now I'll prove another application. I'll give some easy lemma first, one easy lemma set. If a continuous map, f from s1 to x extends to a continuous map, well, whatever, f, capital f, from b2, that's a unique disk, to x. So this means f restricted to s1 is f, no? That means this extends this, OK? Extends a continuous map. Then f star, the induced map, from pi 1 of s1, some base point, 1, to pi 1 of x. Well, I don't know, f of 1 I have to write. You have to preserve the base points, OK? f of 1, I don't know which one. Then this is a trivial homomorphism, OK? This is an exercise, or a lemma, whatever. Typical lemma, typical exercise also. So proof, very easy. So what you have, we have that with the inclusion that I, as before, from s1 to b2, b is inclusion, as before. Look at the inclusion. And then this means that f is nothing else than the inclusion and then the extended. That means exactly this again, OK? First inclusion, then f is f. And now we have functorial properties, OK? So this implies f star is f inclusion star, which is f star inclusion. And now I look at this, OK? Where does this inclusion go from? This goes from pi 1 of s1, 1 to inclusion, OK? Goes from this to pi 1 of b2, the same point, 1. And here we have f star, which returns to pi 1 of where f goes from there to there. No, x, f goes from b2 to x, yes. x and f of 1, whatever that is, OK? And this, what must it be, this mapping, this is f star, the composition. First i star, then f star. This is f star, right? That's exactly what it says here, no? This is a functorial property, OK? But what is this group here? This is z. What is this group? Trivial, no? Straight line homotopy. Trivial. And this we don't know, of course, no? That's x is any space. We don't know anything about it. However, so this is this, and this is this, no? And this is f star. So it goes from this to the trivial group, and then it goes on. So this means it says trivial, no? f star is trivial, because it factors over the trivial group, OK? So it can only be trivial, f star is trivial, OK? So this is easy. The second exercise, or lemma, small. So this is the first one, and now the second one, lemma. If f is homotopic to g by a homotopy h, and f and g are continuous maps from s1 to x, OK? We have two maps from s1 to x, which are homotopic. Then if f extends to b2 as before, if we can extend this f from s1 to b2 to x, OK? If f extends to b2, then also g extends to b2, OK? If one of these extends, then the other one also. And then by the lemma, f star extends as trivial, no? And then also g star is trivial, OK? So f star is trivial, and this has also g star extends. So also g star is trivial, and also g star is trivial. That's what we used the lemma before. So proof. So I make a picture. So what means h? h is a homotopy from s1 cross 0, 1 to x, OK? And now one time, instead of 0, 1, I take 1, 2. You will see why. It doesn't matter, no? Very much. If this is 0, 1, or 1, 2, no, 2x, b homotopy between what? f and g. So s is written here. So now I make a picture. So this is not a unit. This is radius 2, OK? Ball of radius 2, not 1. But here's also b, no? This is 1. So this is b2, radius 1, OK? But this is twice the radius, no? There's a bigger ball. So now I want to define this mapping. If x extends to, let me call it f from, sorry, f from b2 to x, right? Then also g extends to b2, OK? This is a bigger b2 here, but it doesn't matter. So I want to define a mapping from this to x, the extension. So here I have, on this, I have f on this, OK? And this extends to b2. So what I do here, f, OK? So on this part, it extends, OK? This is f on this circle, s1. And this extends to f from b2 to x, OK? What I want is g here outside. Extend to a continuous map, no? This part I have. I have to define it here. But this part is homomorphic. It's equal to what? What is this part? This part here. That's an annulus. It's homomorphic to what? Or it is equal to homomorphic to s1 cross 1, 2, OK? That's s1 times an interval, not clear? This is b2. And this part between these two, that's s1 times the interval, right? Well, you can write homomorphism, no? Of course. I mean, you see here the intervals, and this is s1, OK? So this is s1 times the interval. But I have a map from s1 times interval 2x, which is h, our homotopy, OK? So here you take h on this part. h goes from f to g, the homotopy. So by the pasting lemma, this mapping g is continuous, OK? And extends g. By the pasting lemma, g is continuous and extends, well, it extends to what? Not to b2 exactly, no? Because this is a larger b2, OK? But it doesn't matter. It's the same, it's the aromimorphic, OK? This is radius 2 now, but we don't care. It's a close ball. They are all aromimorphic, independent of the radius. So this is the picture. If f extends, so we have this f, OK, to b2, then g extends. And in this part, I take the homotopy between f and g, OK? So this is all. Now this next application, the second application, well, theorem, even more famous, which will prove in complex analysis, fundamental theorem of algebra. What does it say? I forgot the notation. Maybe not the fundamental theorem of algebra. I need some notation. So what does it say? Every polynomial, complex polynomial, non-constant complex polynomial has a root, has a 0, OK? Has a root. So every non-constant complex polynomial, so x to the n plus a n minus 1, x to the n minus 1 plus a 1x plus a 0, has a root. How do you say root or 0? What would you say? Root in C. This means the coefficients in C, a n minus 1, a 0, it's a complex polynomial. And n bigger than 1, OK, of course. This means non-constant, OK? n bigger equal to 1. The degree has to be at least n1. For n equal 1, it's trivial. It was just 0x minus a 0 divided by a 1. So this is trivial. For a 2, it's trivial, OK? You have the formula. For 3, there's some formula. For 4, there's some formula. For 5, there's no formula. What? Sorry? I don't understand the English. Then for 5, it's not, yes, right? Maybe in algebra. It's not resolvable, no? Because the alternating group A5 is non-solvable. Yeah, exactly. That's 5, starts with 5, OK? Exactly. So proof. And you understand why this is a case, OK? Hopefully. So there are two cases. The first case is an interesting case. The other case is easily reduced to the first case. So suppose we will see why this is strange. But in the proof, n minus 1 plus a 0 is smaller than 1. The absolute values of the coefficients, the sum is. So they have small coefficients, OK? Small coefficients. But the other case is you just reparameterize. I mean, you multiply with some. You divide by something or something. Then you go to this case anyway. Small coefficients. Suppose, so these are called f of x, sorry, f of x, or p of x. Whatever. p of x, that's a polynomial, OK? p of x, OK? That's my polynomial. p of x, complex polynomial. Suppose p x has no root. And we need a contradiction. So I define a mapping G from B2. So this is in C, the unit disk in C, in the complex numbers, to R2 minus 0. So this is our polynomial, but on B2. So it's just a polynomial. And it has no root, so no 0. So G of x is just p of x. That's our polynomial, OK? But we consider it on B2 only, not on the whole complex, OK? And even we restrict it to S1, OK? And so this is G and f from S1 also, if you want. This is not so nice, no R2 here. It should write C, of course, OK? Sorry. What? No, what I say is I should write C here, not R2, OK? Because we're in the complex setting. So it's R2, but it's better C minus 0, OK? So f from S1 to C minus 0, and it's again the polynomial. f of x is p of x. So f is what? f is the restriction of G to S1, no? Yes. It's obviously a polynomial, OK? One time considered on B2. First on C, then on B2, and then on the boundary, OK? S1 of B2. It's always a polynomial. So this means, so what is this? f, this is our mapping, f, extends to B2. That's G, OK? So f extends to B2, no? G. It's always a polynomial. If you want, the lemma then says that f star is trivial, no? f star is trivial. The use method is trivial. So let's see. f star goes from, here we see why it's true, OK, already. f star goes from pi1 of S1 with some base point 1 always to C pi1 of C minus 0 with some base point. I don't know which one. f of 1. So I didn't say what is this fundamental group, OK? This is z. Both are z. We don't need that necessary. But this is also the integers. And we have rotation number also here, OK? Around 0, here along S1, here around 0, OK? Also rotation numbers. But now you look at your polynomial. This is polynomial of degree n. This is not the proof. I mean, this is the idea, OK? You have a polynomial of degree n, OK? And now you take this S1, this S1. This is a unit, OK? But you take some S1, which is very large, OK, in C. Some other, which is very, very large, OK? Then what does it mean? Then the only thing which is important, if this S1 is very large, then this is completely dominated by this power. This get, OK, are no longer important. If the larger we get, OK, then these are not important. This x to the n is a big thing, OK? The other things you can neglect then, OK? Then we can suppose that our mapping is just this, OK? And forget this for a very large, OK? And then you say this would be trivial, OK? In any case, this would be trivial, f star, no? Also for a very large, no difference, OK? For a very large S1. This would be trivial. But x goes to x to the n. How does that look like? Now we have just x goes to x to the n, OK? You say this is not important for a very large x. Small, so forget. So x goes to x to the n, like x goes to x to the n. But what is the rotation number of x goes to x to the n for large? If you go one time around 0 in C and you have x goes to x to the n, you go n times around 0, right? n times around 0. And if these are rotation numbers, that means you go one time here, you go n times around 0 here. So 1 goes to n. That's not the trivial homomorphism. That cannot be the trivial homomorphism, no? You said it's trivial, but that's not possible. If you go to a very large S1, it shouldn't be trivial. It should be 1 goes to n, OK? That's all in some sense. That's your main idea, OK? If it extends, F star is trivial, OK? But this map should not be trivial. Because if you go one time around S1 here, then x goes to x to the n for getting the small stuff, OK? It goes n time around 0 here. And this is not the trivial homomorphism, as you will see. But that's the idea. That's the whole idea here, OK? That's the idea of the proof. That's not the proof. I will prove it now. That's a good question. I said we have to take this S1, very large, OK? This circle, then if x is not so large, then these are important, OK? If x is very large, then this is not so important. And the fact that you take S1, very, very large, corresponds. If you take S1, radius 1, then you take the coefficients very, very small. This is the correspondence, OK? In some sense. That's not the proof, OK? So very small coefficients, then this S1, the standard S1, seems large, OK? If the coefficients are arbitrary, then you have to go to a very large, OK? That's the connection uses. More or less, it's not the proof. Intuitively, it's so, no? If your polynomial is very small coefficients, then either you have a polynomial with very small coefficients, or you go to some S1, which is very, very large, no? This is sort of that's only the idea. But you can understand the idea, OK? Why the fundamental theme of algebra should be true, OK? All here, that's the idea, OK? If there's no 0, then you have this mapping, and it is trivial, because it extends, OK, to B2. And then it's a lemma, that's easy, OK? But this should not be trivial for this polynomial, because you have a rotation number here one time. You have here n times around 0, OK? And that's not a trivial. 1 goes to n, not to 0. That's the whole idea. No, I give the proof anyway, OK? I mean, we are in the proof. Where are we? So f extends to B2, OK? So f star is trivial. f star goes from here to here, anyway, OK? I don't know if this is that. We can prove that, but it's not important here for the rest. Not too important. So I go on with the proof, no. There's a formal proof. So we want a contradiction. So the contradiction, now, we follow this informal proof, OK? We say these coefficients are not very important. The other ones, OK? So we must put them to 0. So what we do is we define k. Let k from s1 to c minus 0 be the mapping x goes to, well, I take just x to the n now, OK? That's the dominating first term of our polynomial. That's I call k, complex numbers. Then, and this is important, f is homotopic to k. And these are maps from s1 to c minus 0, OK? Not to c. To c doesn't help, because the fundamental group of c is trivial. So these are maps from to c minus 0. They have to go to c minus 0. Any two maps to r2 are homotopic. Any two continuous maps to c, to r2 are homotopic. Important is c of c minus 0. That's an important fact, OK? For c, it's a trivial fact. So I have to define a homotopy here, OK? And this uses this condition. I will use now this condition. So the homotopy is the following. The homotopy is the most trivial you can imagine. What is the most trivial you can imagine? So you define h of, I call it h, no, h is OK. H of xt, h of xt. So h should be a map from where to where? From s1 cross i to c minus 0, to c minus 0, OK? So what I say, I just take the most trivial, which I can think of, OK? So I put here some small, this t, and let the rest disappear. a n minus 1, xn minus 1, plus a 0. So for t equal 1, it's all polynomial, no? For t equal 0, it's just k, OK? So let this disappear. Then there's one point to check now, because this has to go to c minus 0, OK? In c, everything is trivial, but we need c minus 0. And this I will check now. This is our condition, OK? So h of xt, absolute value. So this is this computation, which is this is bigger or equal to x to the n. And now I subtract as much as I can. All the rest, t times the rest, a n minus 1, xn minus 1, plus a 0. Is that OK? Yes, I subtract as much as I can, OK? So this is smaller equal to, this is 1, anyway. We are on s1, right? We are on the unit circle. So it's 1 minus t. And now I make it also this bigger, as big as I can. So I take a n minus 1, xn minus 1, value, absolute plus. I make this even bigger, plus a 0, OK? What I subtract, I make this larger and larger, OK? So this is equal to 1 minus t. And now x has value 1, OK? So it remains this value, a n minus 1, plus a 0, right? And the hypothesis was that this is smaller than 1. That's the hypothesis. That's exactly what I wrote. This is smaller than 1, means that this is bigger than 0. This is 1 minus t is between 0 and 1, OK? So this is bigger than 0. And this implies that h of xt is not 0. It's not 0. So we avoid 0. So this is the condition. Here we have the condition of the first case. That we use here, only at this point. So we have almost finished with the first case. So f extends to b2, no? f extends to a map. Well, we have to always be careful. f from, no, what was it called? g, was it, right? From b2 to g minus 0. We have to take always g minus 0, otherwise everything becomes trivial. f extends to a map g, right? f is homotopic to k. Then that's why I proved the lemma before. This lemma, which we said, that k extends to b2, OK? That's the lemma. k extends to a map from b2 to c minus 0. This was x in the lemma. f extends to a map. f is homotopic to k. Then also k extends. This means k star is trivial, OK? This implies another lemma implies k star. So let me write from where to where. Pi 1 of b2, sorry, s1. No, no, no, b2. k goes from s1, no? k goes from s1 to c minus 0. Extends to a map from b2 to c minus 0, OK? So k star goes from pi 1 s1 with some base point 1 to pi 1 of c minus 0 with some base point, which I don't know. No, no, k, now I can take the same. k is just x goes to x to the n. So 1 to the n is 1, OK? The lemma implies k star is trivial. It's trivial. It's a trivial homomorphism, OK? But k is this very simple map here. And now I'll show you immediately that this is not a trivial homomorphism, OK? So that's the last step. So I'm interested in k star. So I take pi 1. I make a diagram. Pi 1 s1, 1, this is k star. So this goes to pi 1 c minus 0. I can take the same base point here because 1 goes to 1. And as I said, I don't know this. This is also z, the fundamental, but I don't do this here. So what I take is I have this retraction on 2s1. So z is r star. What was this map? C minus 0, 2s1. So I go to pi 1 of s1, also 1, OK? So r of x is equal to what? x over the norm of x, again, OK? That's the retraction. So I come back to this. What is this map? Here, I mean this k. k goes x goes to x to the n. This retraction, x to the n, goes to the same x to the n. Right? Doesn't change because it has norm 1 anyway. So this mapping here is induced by the mapping x goes to x to the n. Right? This is this mapping. Continuous, x goes to x to the n. That's k. x goes to x to the n. This doesn't change. Just we go from c minus 0 to s1. But we are anywhere in s1, OK? So this is also induced by the mapping x in s1 goes to x to the n. So this is z. This is z. Now, the generator here, you go one time around, right? Rotation number 1, the generator, no? One time around. So 1, now you go x, go to x to the n. You go n times around. So 1 goes to n. That's a rotation number. So 1 goes to n. 1 here goes to n. So this is multiplication with n. But n is not trivial, no? n is bigger than 0, OK? And bigger than 0, that's not a constant polynomial, OK? So n is bigger than 0 implies that this cannot be. This is not trivial, OK? Then this cannot be trivial. Otherwise, also this would be trivial, OK? Means k star cannot be trivial. k star is not trivial. Homomorphism. If this would be trivial, this is trivial, no? Because it's a composition, right? But this is not trivial. This is 1 goes to n. This is very easy, this k, OK? This is x goes to x to the n, and you see what happens. So it's not trivial, but it has to be trivial because it extended. So there's a contradiction. Where are we? Contradiction. So this is the end of the proof of the first case, OK? And the second case, well, I finished the second case. Two minutes, OK? The second case is not interesting, OK? But here, the idea is nice, OK? If you avoid 0, you extend to be 2. So it's in use map, it's trivial, OK, on fundamental groups. But this is rotation numbers, no? If you take this S1 very large, then it's like x goes to x to the n. If you go one time around, the image goes n time around, OK? And that's not the trivial homomorphism. That's all. So in two minutes, I give you the second case. It's really nothing, the second case. This is a general case, OK? So we have the equation xn plus an minus 1 xn minus 1 plus a0 is equal to 0. We want a solution of this equation, no? For some x. This is, and now what I take, I let, I call x equals c1. x equals cy. So c is bigger than 0, some real constant, big positive real constant, OK? I make a change of indeterminate. So then this becomes equivalent. This equation is equivalent. So I just put c to the y. cny to the n plus an minus 1 cn minus 1 yn minus 1 plus a0 is equal to 0, OK? Now in y. So I divide by c to the n, OK? y to the n plus an minus 1 c to the n. This is not 0. This is bigger than 0, OK? y to the n minus 1 plus a0 divided by c to the n equals 0. If you find a solution here, we have a solution here, we have a solution here. A root. But if you choose c large, clearly we are in the first case, OK? Because the sum of the coefficients c we can see as large as we want, OK? If you choose c large, we are in the first case. The sum of the absolute values of these coefficients becomes arbitrary small with large c, no? Clearly. If you choose c large, we are in the first case. We are in case i, in the first case. Whatever that was, i was in the first case, OK? So the last equation has a solution, has a root. So the last equation has a root but then also the first one. So this implies the last equation, so this one, this one. The last equation has a root, has a solution but then also the first one because it's just giving other names to some variables and so on. So also the first one has a solution. But this is a general case, OK? This is a general case. So this was a good question, no? You asked, why is this condition, no? This condition is, this polynomial, if you take the standard as 1, then the polynomial has to be small, the coefficients, OK? If you take an arbitrary polynomial, then what you could do is take a very large s1, OK? So this is exactly the condition. And if the coefficients are small, then if, or we take a very large one, then the coefficients of xn minus 1, xn minus 1 are not very important. It's completely dominated by x to the n, the nth power. That's important, OK? The rest. This is our homotopy, which makes this appear the rest, OK? Just by multiplying with t, the most stupid homotopy, OK? Only you have to be sure that you avoid 0, OK? One minute. I give one exercise, last exercise. You have exercises, right? I give the last exercise now. I have to write it because I didn't find it in the book. Just one exercise and then it's finished. I looked in the book today. I didn't find it. So p from e to b is the covering. p of e0 equal b0. You have to write, OK? p of e0 is equal to b0. And b connected. That's important. B connected. If let f, so we have a continuous map, f from x to b continues with f of x0, so we fix all of our points here, b0, OK? We have a continuous map from x to b. Then a lift, if it exists, then what you have to prove, a lift f tilde from x to e, sorry, to e, OK? Then a lift f tilde from x with, again, the base point that we have to preserve, f tilde of x0 is e0 now, OK? It's unique. The important thing is that b is connected. Then the lifting is unique. Then the lifting, OK? That's all. So we have this diagram, the usual diagram, which is this one. We don't say existence here, OK? It's only uniqueness. If it exists, it's unique, OK? So existence is not true always. So we have this diagram. We have e to b, the projection. We have x, the usual diagram, f, and here is the lifting, OK? And we preserve base points always, OK? So we have f tilde of x0 has to be e0, where e0 goes to b0 and f, OK? So f of x0 is b0, and obviously p of e0 is b0. So this is the situation schematically, OK? If it exists, it's unique, but it doesn't always exist. It exists if x is a path, if x is an interval, then it exists. That's a path lifting lemma. It exists if x is i cross i. That's a homotopy lifting lemma. But in general, it will not exist, OK? There's some condition, which we will not see. Because last week is the last week, OK? Next week is the last week. So this is the last exercise, OK? In the book, I didn't find so many. I mean, I made some exercises here, no? Some kind. Next week, I will do some exercises. But in the book, I don't know. There are many chapters, OK? If you look in the book, we are going diagonal a little bit, and there are not so many useful exercises. So next week, we will have some. OK, sorry. So that's it. Yeah, you can collect. There were three exercises, which I wrote also, right? Yes.