 Okay, so let's start. So we have, start with u over zon, matrization theory. So X normal, if X is normal with a countable basis, the second countable, right? Countable basis implies X matrizable. So it's stronger with regular, okay? Regular is okay, but we prove this. A regular space which is with a countable basis is normal, but that I will not prove. So proof, lemma. There is a countable collection of continuous functions, fn from X to zero one to the interval zero one. So n and n, countable. Such that, given any point X in X and neighborhood u of X, given any point and every neighborhood of that point, there is an index n such that fn, so it's X zero, well. Given any point X zero in X and any neighborhood u of X zero, there is an index n such that fn of X zero is bigger than zero and fn of the complement of the neighborhood is zero. This is an important lemma, okay? It's on the rest is easy. Okay, proof. This is the orison lemma, proof of the lemma. So we have a countable basis, okay? So let b be a countable basis for X, second countable. For each pair of indices, a, b, so sorry, for b equal bn, n and n, so this is also countable. Be a countable basis for X, bn, for each pair of indices, let's say nm such that, so we have this bn, this is on our basis. Such that the closure of bn is contained in bm by the orison lemma, so we apply the orison lemma, choose a continuous function, f and m, so it depends on these two indices, f and m. This is not the same n, okay? Then this is countable also in two indices, so no problem. f and m choose a continuous function such that, so orison you can separate closed disjoint sets, okay? One is bn closure, such that gnm of the closure of b is equal to one, so it's one on this, and the other closed set, it's a complement of bm, the complement of bm, and gnm of the complement of X minus bm is zero, so this is a lemma of orison, okay? You can separate disjoint closed subsets, okay? Yeah, all right, now it's g, okay? So I call it g now, so, but these are all fs of course, but in another order. So this is a countable collection of continuous functions, okay, which depends on two indices, g and m, this is countable, okay? And now we need this property, there is an index, for any neighborhood, there is one of these functions with this property, okay? So this is the property. So let's see, let X, so what is the notation? X zero in U, any neighborhood U of X zero, so let X zero, so consider now X zero and the neighborhood U as in the lemma. So there's a basis element which contains X zero as contained in U obviously, so there is choose a basis element, bm such that, such that X zero is in bm contained in U. So this is the basis, no problem, yes, thank you. So now we have this situation and we choose obviously a basis element, bm such that bm X is contained in U, but X is regular, okay, normal, X is normal, regular sufficient here, what does it say? Well, maybe it's regular, it's better here because normal is for close ups, that's regular as for point, a point, okay? So regular, anyway, it's normal, but maybe better to try and figure it out. So given a point and the neighborhood, so this is a lemma, no, which you prove, not, I made by picture, okay? If you have a neighborhood of a point and there's a smaller neighborhood, such that the closure is still contained, and given a neighborhood of a point, there's a smaller neighborhood of the point, such that the closure is still contained in the larger neighborhood. So we have the smaller neighborhood, the smaller neighborhood obviously contains, again, a basis element, still smaller, okay? Because it's a basis, okay? So there is, X regular, there is a BN, a basis element, BN in our basis such that X zero is in BN closure still, and this is contained in, and then of course G and M has what we want. So this implies that G and M, where's our G and M? It's bigger than zero, G and M of BN bar is one, and G and M of, what were X minus BM is zero. So X minus U is zero, okay? So this proves the lemma, right? So now we have G and M in two indices, but it's countable, and now we call it FN in one index, okay? This is our, this is the lemma. And now the proof of the theorem, so the idea is, I didn't say that, but the idea is the following, so we define a function F, continuous function F from X, X, we have to prove X is matrizable, okay? And so we embed it into our omega, and this has a product topology for the moment, product topology. Well, also it works with the uniform, but let's take the product topology. So how do you find, F of X is equal to, so we have this countable collection of functions, FN, okay, so we take these functions, F1 of X, F2 of X, and so on, so it's N of X, N and N. This is a definition F, okay? And now we want to prove that this function is an embedding, okay? The claim is, F is an embedding. What is embedding? Well, we will see in the proof. It's a homeomorphism onto its image, okay? Homeomorphism onto its image, embedding, homeomorphism onto its image, that's an embedding. And then of course, that's nice, no? Because if it's embedding, homeomorphism, what does it imply? This implies that X is homeomorphic, homeomorphism, to what? To its image, F of X. And what is this? It's a subspace of X. No, subspace of what? Of our omega. Our omega is matrizable, okay? This is matrizable. I gave the metric, okay? The theorem and what is the right metric for our omega is the product topology. So our omega is matrizable. So this implies any subspaces matrizable. The subspace of matrizable to space is matrizable, okay? So subspace of a matrizable to space is matrizable. It's the same metric. And this implies X is matrizable, okay? F of X. That's the idea here. This is the idea, okay? This is embedding. F of X is matrizable. So we embed it into our omega. So this is a proof. So I have to prove that F is an embedding, right? Also the uniform topology would be okay, right? Because this is a metric space even, okay? Defined by a metric, okay? So it works also with some modification. But here we take the product topology. So I have to prove this claim. F is an embedding, okay? Then we are done. Okay? This is the idea here. We construct enough continuous functions to the real numbers. And then we take all this countable number, okay? Uncountable, our omega, our uncountable is no longer with the product topology, you know? The uniform topology is in all, okay? So, well, anyway. So this is, now I have to prove only this claim, okay? Well, what means embedding? Homomorphism, so proof of the claim. So embedding, homomorphism, so it's injective, okay? F is injective. So that's the first, F is injective. That means if you have two different points, then you should find one of these functions which has different values in different points. So let X, Y be different points. How are they called? Whatever. X and Y, yes? Let X, Y be in X, X different from Y. So X is T1, T2, okay? X is T1. As before, X is T1. What does it mean? There's an L value of X, which does not contain Y, the other one, okay? Now we do the same game as before. We find two indices, L and M, such that as before, what I did in the lemma, okay? As before, as before what? There are BN, BM in our basis, such that X, what is our point? X zero before, now X is contained in, no, I forget, BN closer, contained in BN or what did I write? BN closure, yes, BN closure contained in BN, okay? That's as before with X zero, exactly the same. They are BN, BN. So there is an FN and then we have our GNM, okay? So GNM, so what was FN, this is some FL, some, I don't know which one. FN, M of X and BN closure, what was it? It's one, I think now, and GNM of Y. But Y is the complement of BN, okay? So it's zero. So this function is one here and zero on the complement of BN, that's the original lemma, right? These are the two different closed sets. This is one here and the other one is a complement of BM and that's zero. So GNM, X is zero, so for some, FN this happens, okay? And this means that F of X is different from F of Y because these are all these GNMs. So it's injective. The second is, it's continuous. Well, this is easy, no? F is continuous, since Y, why it's continuous? What? All component functions, this is product topology, okay? Since all component functions, coordinate functions, whatever, FN are continuous. And here in the, if it takes the uniform topology, we have to prove something still, okay? Because this is a larger topology, you know? So continuity is stronger, okay? But for the product topology, no problem. And then the last thing is that F is, so what is the notation? F is a homeomorphism onto its image, homeomorphism onto its image. So we have to prove that F is a homeomorphism onto its image. So what remains to prove here? Remains to prove onto its image, which I call Z. That's the image of F of X. So this is a subspace of our omega, the image, okay? We prove that for each open Z, which is called U in X, F of U, open, okay, open. It's open, so for each open Z, U in X, F of U is open. The image of open Z is open, where in Z. Not maybe in X, that I don't know, in our omega, okay? In the image, we are restricting to the image. It's open in Z, which is the image, that's important. So that we have to prove, right? Okay, that's the last point. So we have to prove that for each open Z, F U is open in, so take any point here in F of U, okay? Let, which is called Z, whatever it is called. Z zero in F of U. F of U is open here, so we have a point here now. We have to find an open Z in Z, which contains this point as it's contained in Z, okay? The usual thing, open, okay? Open means, for every point, we find an open Z, which contains a point, is contained in our Z, which we want to prove open. So let zero be in F of U. So let X zero be in U, so it's the image, F of something, such that F of X zero is Z zero. It's in F of U anyway, okay? So we take a pre-image in U. Now, we have this point X zero, okay? And we have our collection of functions. And we have a neighborhood of U. So what was the lemma? For each point, each neighborhood, we find an index N such that F N is bigger than zero in the point and zero in the complement of this function. That's the main point, no? So a chosen index N such that F N, G N M, G something, F N of X zero is bigger than zero, one. It was one, but bigger than zero, okay? And in the complement of U, F N of X minus U, it is zero. Again, the lemma, okay? The lemma, so we apply again the lemma. So let V, by definition, V is a pre-image of, so pi N, what is pi N? That's a projection on the Nth coordinate of r omega, okay? Projection onto the Nth coordinate as usual of r omega. So one here, projection of the Nth coordinate, pi N minus one of the interval, zero infinity. So we avoid zero, okay? We take the, so this is open, open, where's it open? In X for the moment, okay? Pre-image of an open sector. But that means, of course, that W, which I'm not interested in X, I mean open in X here. I'm not interesting, so this is V, intersection Z, there are too many notions here, V, Z. So V is this pre-image and Z is the image, okay? Sorry, I wrote something wrong, open in where? R omega, okay? I was, something was not okay. But I didn't, so this is a projection from r omega onto its Nth coordinate, okay? So this is open in r omega. But we are not interested in r omega, okay? We are interested in the image. And so I intersect now with the image, okay? So this is open in Z now, clearly. This is open in Z, this is the image of F of X. So this is in r omega, we are now, okay? And now this is the set which we, so the claim now is again, well, one claim after the other. That's okay, this is our set. So claim again, what is our point? Our point is called, where is it? That's zero, okay? This is the point we choose in the image. So the claim is X zero is contained in W. This is open, no? This is open in Z, open in Z, in F of X. And this is contained in, where it should be contained. What is it called? F of U. And then we are done. So this implies F of U is open, Z, which is F of X, no? Then we are done, okay? So Z zero is in W since, so this is immediate. Since pi N of X zero of Z zero, this projection, what is this? This is pi N of F of X zero, yeah, this is FN of X zero. So this is bigger than zero, yes. Pi N of Z zero is bigger than zero. So we are, so this is okay. And second, so what remains is W is contained in F of U. So take a point in W, let Z be in W. Then Z is equal to F of X for some X in W. And X, we are in F of X, okay? But F of X is not sufficient, we need F of U for some X in X. And pi N of Z is where in zero infinity. Z is in W. That many notions, too many notions, now W. You always have a new name for everything, no? But that's from the book. So this is the image, okay? We are in the image anyway of X, and now we are also in V, positive coordinate. Now we have these two things here. Okay, since now, pi N of Z is equal to what? Is pi N of F of X, no, F of X is Z, anyway. And this is F N of X. And F N vanishes is zero, F N vanishes outside. Where does it vanish? Outside U, this was our neighbor, what? This means that X has to be in where? In U, F of N vanishes outside U, no? But F N of X, so this is bigger than zero, sorry, okay? I might better add that. So pi N of Z is this, is F F of X bigger than zero, okay? This is bigger than zero, because we are in V here, in the pre-image of volume zero. And where N is also, so this means X is in U. It has to be in U, because F N vanishes outside U. So we have to be in U. And this means that F of Z, Z, what is that? Z, which is F of X is in F of U. So we are proving this here, no? Any point here is also here. And that's the end of the proof. Okay, well, we have to concentrate. There are many, this is not taken from the book. There are many names here, no? So one has to look, what was 1W? What is V? What is that, no? Maybe too many names, but one has to concentrate. What is our set, okay? Anyway, the proof is not difficult. So the main part is the orison lemma, okay? To have all these functions, F1, F2. So this is the proof of the orison matrization theorem, okay? Well, and all the last theorem is much easier. It's a consequence of this one anyway. I finish with the following theorem, which is also very nice. Also orison matrization theorem. So this is a complete solution. If x is compact, suppose x is a compact space, and x is matrizable, if and only if x is matrizable, then, well, anyway. If and only if x has a countable basis. For a compact space, this is a complete solution, okay? Of the matrization problem. If x is compact, then matrizability and second countability are equivalent completely, okay? For compact spaces, this is the solution. Here the proof is now easy. That's nice, okay? We don't prove the general matrization theorem, no? We know second countable in general is not sufficient, okay? It's not necessary, okay? First count, it's between first and second countable, the general matrization. Well, the proof is we have two directions. The interesting is this one. So x has a countable basis. This is a horizontal matrization. So compact implies normal, okay? X compact, that you have to prove, implies x is normal, anyway. And if x is normal and second countable, then x is matrizable. And this is a horizontal matrization theorem, okay? Which we proved. Normal, second countable, x is matrizable. So this is corollary, x matrizable. This is horizon matrization theorem, okay? So the other direction is also interesting because this is one of our exercises, which I like. So what now, x is matrizable, compact, okay? X compact matrizable, so exercise. X matrizable, compact implies, what you have to prove, x is second countable, okay? X has a countable basis. This is the easy direction. You remember this exercise, okay? So there are two exercises, one we did already. This is if x is matrizable and it has a countable dense subset, then it is a countable basis. That's what we did. And this is the second of this. If it's now it's matrizable and compact, then it has also a countable base. What? Say again, I didn't. So if x is compact, compact space is normal, right? A compact house of, of course, okay? Is that what you mean? Compact house of, oh, sorry, sorry. There's something wrong, right? Compact house of, yes. Very often compact includes house of, but not with us, okay? So I forgot that. Of course, house of we need, okay? Okay. Compact house of, we need a compact house of space, okay? If it's not house of, it's not matrizable, right? So that's certainly the first necessary condition for being matrizable, okay? So we need a house of space, yes. Thanks. Compact house of means normal, right? And then normal second countable implies matrizable, yeah, okay. And this is our exercise, right? I hope you remember that, okay? Yes. So what is the idea, so proof? Well, let's do it as an exercise, proof. This was the exercise, it's the exercise in the book. So matrizable compact. So let BN, a finite covering, of what is X by open balls of radius N, okay? One over N, one over N. By open balls of radius one over N. This exists now because it's compact. You take all open balls of radius one over N, this is an open covering, so you have a finite sub-cover, okay? So no problem. And let B be the union of all of these BN, N and N. This was a hint in the exercise, right? This was a hint. Let B be, so B is countable, okay? This is clearly countable. Even a countable covering would be okay, still for countable, right? Well, anyway, so B is the basis, okay? Now we have to prove that B is the basis. B is the basis for the topology of X. So what means basis? So now it's okay, okay? This is the, ooh, the matrization, the second. Compact house dot space. So what means B is the basis? This means that if you have an open set U open, if you have a point X in U, you shouldn't be able to find the basis element which contains X and is contained in U. Then the lemma says this is the basis, okay? For the topology. So we have this situation again. However, X is matrix space, matrizable, okay? So what you find is, I make my picture. You find the ball, okay? Which contains, we center X and contain in U. So this is the ball, D, X, for some N, no? One over N, epsilon. And then I take some one over N immediately, okay? One over N. Now we want to find a basis element. So what we do, what I do is, I consider half the radius again, okay? Half the radius. One over two N. No, no, sorry, one second. I apply for two Ns, okay? So I have an open covering of balls of radius one over two N now, okay? So this is our N, okay? And now I pass to two N. So let X, it's one over two N, they cover. Finally many, but they cover, okay? Let X be B, D. So now I find another point which is called, whatever it is called, Y, radius one over two N. And this in B, so this is in B, how do I call this? Two N, no? B two N, one of these, so contained in B. So I consider these balls, okay? N is this one. So here's some, so it's in this, obviously, but it's somewhere, this is one over two N, so the point Z, Y, what is it, Y, it's here, okay? Half the radius, somewhere here. And now we have this ball here, okay? This is this ball, Y one over two N, which contains X, okay? This contains X. So what we need, it's contained in this one. Which is clear from the picture, and then it's contained in U, and then we are done, okay? So the point is that Y one over two N however is contained in this one, B, D, X one over N. Why Z? Well, you take any point here, okay? Z, so take any point here, where in the, this one, Z, okay? Then the distance of Z and X, okay? You have triangular inequality. So this implies that the distance of X and Z, since distance of X is smaller or equal to what? This is one over two N, and this is also one over two N, okay? One over two N plus one over two N. So one over N. We need smaller, okay? And that's it, so we have this. So this is the lemma, okay? This is the exercise, okay? This is the proof of the exercise. Okay, that finishes the general topology part. And now we start immediately with, so the nice theorems obviously are, lemma of orison, orison matrization, and this second orison, okay? By the way, I looked, so orison was from 1898, 1924. And the general matrization theorem, it proved in 24, all this stuff, okay? It's, there are three names or two, three names by, so one is snagata, Japanese, independently, okay? Smirnov and Bing also has some version. They all have some version, okay? General, so there are three, two, three groups of people who independently prove the general matrization, which I didn't tell you, okay? We have to find the condition between first and second counter, okay? Normal, okay, normal we need anyway, okay? But we need something between first and second counter. And this was quite subtle. So this was more or less from 1950, okay? Proof, 1950. So that means it took 25 years to find this condition, okay? After orison, it took another 25 years, no? To find the right condition, which is subtle, okay? You can have a look in the book, what is the condition? Some existence of special kind of basis between first and second counter, okay? So this is also historically, it's quite interesting, okay? Okay, maybe, he died with 26 years now. Maybe if he would have gone on doing mathematics, maybe this would be, I don't know, when he obviously would have also thought about that, okay? After this, after his first paper, when then maybe we have 1935 or something about this, of course, history. Okay, so now, it's good. So this, we can forget not everything, but almost everything. No, no, something we need, but homotopy and fundamental. The first part will be very formal, okay? And maybe a little bit boring, but the second part, then we have applications and that is very interesting again. But the first part is formal, okay? So I have many definitions in there, so. So definition, what is homotopy? To continuous maps, f and f prime from x to y are homotopic, and we write that f is homotopic to f prime. So this is the notion of homotopic. Not isomorphism, okay? Homotopy is one, just one law. If there exists a continuous function, a continuous map, map and function, let's just say most. Capital f from x cross i, i is the interval zero, one, so this is interval zero, one, as you know, zero, one. Close interval to y, such that f at time, so this is a time interval. f at time zero is f of x, and at time one, f of x one is f prime of x. So f is called a homotopy between f and f prime. And sometimes you write also f homotopic to f prime, and we buy a homotopy f, okay? f is a homotopy. What is a homotopy? Homotopy is a continuous deformation from f to f prime, okay? And so a continuous deformation from f to f prime. That's intuitively what a homotopy is, okay? Can the form continuously this map f to a map f prime? That's the definition of homotopy. Example, first of all, this is a equivalence relation. So homotopy, so observation, remark. Homotopy is an equivalence relation on the set of continuous maps from x to y. What is equivalence relation? So f, the first is f is homotopic to f, no? By constant homotopy, okay? By the constant homotopy. So constant means f of x, t is constant, is f of x. Constant does not depend on the time, okay? So this is trivial. The second, what is the second? So f homotopic to g, f prime. By a homotopy, f implies that g is homotopic to f by a homotopy g. And how do you define g? g of x, t is equal to f of x. You just reverse the time. From zero one, you go from one to zero, okay? So this is the second one. And the third one is if f is homotopic to g by a homotopy f. And g is homotopic to f, g, what's the next, h, no? By a homotopy g, then we have to define a homotopy from f to h, h, okay? So then you can think of how to define h. So how do you find h? So what you have is we have f from, we have f, which go, so this is my kind of picture which I will use, okay? Here we have f. Here we have g. And here's the homotopy between f and g, which is called f, right? This is the kind of notation, f, g, homotopy, f. So this is a picture of x cross i, okay? This is a time interval here on this side. And then we have a homotopy from g to h. And this is called g. So it's defined here, okay? This again is a time interval, okay? So we have to put it together in a good way. So h of xt is equal to between zero and one half, we take f, f of x, two t, no? And between x of, between one half and one, we take the other one, which is g of x and now two t minus one, yes. h is continuous, y by the pasting lemma. We have on two parts, it's continuous and then, okay? So you have this lemma, pasting lemma, okay? So it's an equivalent, right? Examples. First, every map from f, x, every map, a continuous map, of course, f from x to rn, let's take rn here. Standard, rn, analysis, rn. Every continuous map, f from x to n is homotopic to a constant map. So let's see this. This means it's not very interesting. Seen from point of view of homotopy, maps to rn are not interesting, okay? They're all homotopic to constant maps, which are trillion, you know, trillion maps. So let's see this. So we have to find a homotopy from f to a constant map. So we have, so we have f of x, which constant? Well, I don't care, x zero, okay? Here we have an rn, this is an rn, right? f of x and some constant, zero. But it doesn't depend on the constant, so I call this z zero, okay? And now we just move f of x for each x continuously to this point. So what is the homotopy? So f is homotopic to the constant map. I write this way in z zero, okay? This is a constant map. In the point z zero, so we are in rn here. The constant map, okay? And so this homotopy. So what is the homotopy? H of x, t is equal to, so this one, okay? So at time, whatever. At time zero, we want f of x. f of x, at time zero, so one minus t. That's good, no? At t equals zero, we have f of x, okay? And at time one, we want z zero. So plus t, that's it, okay? So now at time one, we have constant, okay? At time one is constant. So what is this? This is a line, no? This is a segment. So we just go on the same. This works only in rn, okay? In outer space, we don't know how to do something like this. So this is called straight line homotopy, okay? Because it follows straight line, okay? Straight line homotopy. So we will use this. This is a straight line homotopy. In rn, only in rn, in outer space, if you don't know what is a straight line, right? So we have a vector space. We have, okay? This sum in rn is a vector space. So this is straight line homotopy. Second example. So take a path, f from zero one to x. Path is continuous map, okay? Take any path. So this means then f is homotopic to constant path. Again, not very interesting. Why is this? Well, it's almost the same as this, no? So we have homotopy, h, x, t. So what do you take? f of, x is bad now, okay? Because this seems we are here, sorry, okay? x is bad, x is the image now, okay? So I take s, t, okay? This is better, s, t. So what do you take? Well, depends where you want the constant, okay? You can take any point, but the first point. Let's take the first point. So then I have to write s times, let me try, one minus t. So if t is equal to zero, we have f, okay? If t is equal to zero, we have f. If t is equal to one, we have f constant in the end point, okay? That I don't want, I want the initial point. So what is s? t times s, well, sorry. One, I wrote one, let me try, one minus t times s. So if t is zero, we have f. If t is one, we have f of zero, the initial point, okay, yeah? So the constant map in the initial point, okay? This is the constant map. In the initial point, f of zero. In some sense, you see the homotopy, okay? It's clear what is this homotopy. It looks like this. So here you have, the picture is this, no? You have this and you have some f, okay? And now you have this homotopy. What happens in the homotopy? Well, you multiply with, intuitively, this homotopy with this, at time one half, you just go to half of this, okay? Between zero and one, the reparameter was. And then, at time three over four, you just go to this point, okay? And then, finally, at time zero, you stay all the time at this point, okay? So you see, so in time zero, you go all the way, okay? In time one half, you go half the way. Half velocity, okay? Or maybe you stay here. You gave us the same velocity and then you remain here, okay? At time three over four, you go just to this point, one over four, and stay there. And at time one, finally, you stay all the time at this point, okay? So one homotopy is this one, okay? This means, this is not interesting for pathos also, okay? So we have to, but we want, so what we do now, this we don't want, okay, for pathos. The general definition of homotopy is what I gave, okay? For pathos, I have to modify now, okay? Why? Because this is not interesting, otherwise, okay? Every path otherwise would be homotopic to a constant path and constant paths are trivial pathos. No, not interesting pathos, okay? So now, modification. Modification of the notion of homotopy for pathos. Only for pathos, okay? Only for pathos. What we do, we fix this point and we fix the end point, okay? We are not allowed to move these two points. So in some sense, we fix it here, we fix it here by some nail, and then we can move in between what we want by continuous deformation, okay? But these two points, we are not allowed to move, okay? So we cannot go back to the initial point, okay? With everything, so a definition. And this is the definition of path homotopy. That means, of course, that the two pathos, if you fix these points, the two pathos have, we need the same initial and the same end point, otherwise, we cannot compare, okay? Right? So let F and G from the interval to XP pathos with the same initial point, so what is this? F of zero equal G of zero, and the same end point, F of one equal G of one. So we have this picture there, no? We have F and we have G going. Then F and G are the same, okay? Well, path homotopic, but I will not use, for pathos, the homotopy is this definition, okay? So it's path, but we say just homotopic. F and G are homotopic as pathos. If there exists, there is a continuous map, capital F from, now we have I cross I, the time interval to X, such that. So as before, capital F of, now I have to write S, no? Because X is not good. At time zero, we have F, at time one, we have the other one, G. So these two means homotopy in the general sense, okay? That is homotopy. Homotopy in the general sense. But not for pathos, this is not sufficient, so now we add F of zero, this is the initial point, T. We always have the point F of zero, which is also G of zero, okay? And at the end point, F of one, T, we always have F of one. These points, we are not allowed to move. G of one. And so this, all together, this is path homotopy, okay? So the picture looks like this. And I will use these pictures, which is very, so down we have F, here we have G to pathos, and here we have the constant, okay? In the point F of zero, which is also the point G of zero. And on this, we have the constant in the point F of one, which has to be also the point G of one. And here we have our homotopy. And we write, again, F is homotopic to G. And in the book, I forgot, he uses some other sign, which I've not used, I forgot which sign even, okay? He uses something like this or P, path homotopy, okay? I will not use this, okay? So I just write this, okay? Also for path homotopy. If F and G are pathos, this is always path homotopy, okay? So this is the definition of path homotopy. So again, the same remark as before, this is an equivalence relation, but where? So a remark, path homotopy is an equivalence relation on the set of pathos from a fixed initial point X zero to a fixed endpoint X one, okay? These two here. From a fixed initial point, from the set of pathos in X, of course, or space, from a fixed initial point X zero in X to a fixed endpoint X one in X. And the proof is the same as before, okay? You just check also that everything goes well. I mean, if it's constant here on both, then also the map which you defined before is constant on this, okay? No problem. So this is a path homotopy, okay? Homotopy, important notion in general, okay? Homotopy, path homotopy for pathos. Because it's not interesting otherwise for pathos. Okay, for pathos, we have a product. We want to define a group, okay? Using pathos, a group, okay? Fundamental group. And we need a product. So product of pathos, it's another definition. Let F be a path, X is fixed in X from X zero, from X zero to X one, and G a path. So of course, if you want a product, the second one has to start, where the first one finishes, no? Then we have, and G a path from X one to X zero, X one, X two. So we have this picture. This is X zero, X one, this is F. And then we have G, which goes from X one to X two. This is G, okay? Then F, first F and then G. That's a product, the product we write in this way. First F, then G. Then FG from Y to X. So how is it defined? F times G of, well, no, S or T, I don't know. No, the T is the, let me take S, okay, the parameter. So it's equal to what? S before, between, for S between zero and one half, we take the first path, F of two T. So this is twice velocity, okay? And for S between one half and one, we take G, but also with twice velocity, with double velocity, twice the velocity, okay? No, it's okay, yes. So this is a path, which is called the product. It's a path. Why is it a path? Again by the, pacing them, you know? That's why it has a name. So it's used everywhere. It's a path, which is called the product of F and G. So again, clearly defined only if, defined only if, let's emphasize again. Defined only if F of one is equal to G of zero. Then the product is defined, otherwise there's no product. So X zero, X one, so this is X one, no in this situation, okay? This is the X one here, yes. So we have a product now, which is not always defined, but anyway. What do we need for, we want to define a group, right? What do we need? We need associativity, no? And then we need identity element and we need inverse element, these three things, okay? For this product, we have no chance, okay? Because a path is a path, okay? And if you fix a path, you can go back, okay? But this is not, so we have to pass to a homotopy. If you want a group, we have to pass, pass to a homotopy, okay? So definition, for a path, f from X, from I to X, let, so this is not, the equivalence class, you always denote like this, no, in general. F be the path equivalence class of F. The equivalence class of F is an equivalence relation. Be the equivalence class of F with respect to path homotopy. This is a familial notation, right? The equivalence class you very often denote by these parenthesis, okay? Let F be the equivalence, and then definition, let, again, let F and G be pathes with, so the condition F of one equal G of zero. Otherwise there's no product, okay? Now the product is defined. Then we have the definition, define F, but now I want to, not pathes, but equivalence classes of pathes, no? The usual thing, you, so this is the definition, and you take the equivalence class of F times G, okay? So this is a product of equivalence classes, no? Not of pathes, product of equivalence classes of pathes. What do you have to prove? That is well defined, okay? Which is well defined, the usual thing. Which is well defined. Well defined means does not depend on the choice of representatives. So what do you have to prove? You have to prove if F is homotopic to F prime, so if F is equal to F prime, no? That means F is homotopic to F prime by a homotopy F, right? And G, we have another representative. So this means G is homotopic to G prime by a homotopy G. Then we have to prove that F times G is homotopic to F prime times G prime sorry, homotopic to F prime times G prime, okay? And so this implies then that F times G is whatever it implies, F times G is equal F prime times G prime, okay? Yes. So we have to define this homotopy here, H. And now we get a picture, okay? So what do we have? We have F, where is F? Homotopy F between F and F prime. And then we have G between G and G prime. And here it's constant, okay? Here's the same point, F of one is G of zero. So this looks good. Now we just have to go back to our interval. So we define H of S, T equal two. And now the interval, which I have as a parameter which I have to change is S, not T. S, here's the interval, that's okay. That looks anyway good, okay? But here I have to change. So between S between zero and one half. So what I take here, here I take F of two S, T. And S between one half and one, I take the other one G, two S minus one, T. Sorry? Now you say this is a product between F and G. We don't define a product here. Yeah, for any fix T, it is a product of this path and this path. Yeah. But I don't have names for this. So one second, what I want to say now is that then this is this homotopy, okay? At time zero, we get really the product of F and G, okay? So H of what? Of S, at time zero, we have this product, so I'm getting tired. So at time zero, we have the product of F and G. That means for T equals zero, H of S, T, at time zero, this is F times G of S. And at time one, this is F prime times, so then we're here, times G prime of S, by the definition of this product, by definition of H and of this product, okay? We're doing here exactly the same as we did for F and G, okay? By definition, by the definitions of H and the product of two paths. So that's more or less what you, and at the time, in the intermediate time, we have the product of this path and this path, no? Yeah. But I didn't give names, so I don't know. Okay. So again, F is, H is continuous by the pasting, okay? H is continuous by the pasting lemma, okay? It's always the pasting lemma. F is, H is continuous by the pasting lemma. Well, so the time is finished, so we go on tomorrow. This is, omotope is important concept, of course. Probably you know, I don't know. Omotope, path, omotope. Then it's natural product of paths, no? And we want to agree, so tomorrow we define a group. This is very formal, this last second part today and tomorrow also, okay? The definition of fundamental, it's very formal. And then you say, well, many definitions, no? But then the fundamental group is important concept, okay? In mathematics, so, and then we have very nice, so this is the first part here is very formal, but then we have some very nice applications, which are very concrete applications, okay? Or fundamental group. So tomorrow we go on with this.