 Okay, so let's start. Last time I wrote an exercise and I wrote, so this exercise, you have a covering, and you have a mapping from X, continuous mapping. And if you prescribe one point, if you prescribe one point for this mapping, okay? You will see with F tilde of X0 equal E0, okay? And then we want to show that this is unique. And it seems that I wrote B connected, okay? That's, but we need X connected, yeah? So, F tilde is unique. So it's not B, it's X. If X is connected, okay? And probably I wrote B, that's X, we need X is connected, okay? So the idea is the usual one for connected spaces. You assume you have two liftings, okay? Two different, well, no, two liftings. And then you take all points in X where they agree, okay? These two liftings. And there's one point where they agree because we describe in some point here, we say what we want. So this set is non-empty. And then what one proves is that this set is where they agree, the two mappings, is connected, is open. And also the complement, or it's closed. Also the complement is open, and it's non-empty. So if X is connected, it's a whole X, okay? So they agree everywhere. That's the structure of the proof. And so we need X connected for this, okay? I missed, I wrote B connected, but we need X is connected, okay? But the idea is this, okay? Take all points where they agree. This is non-empty, open. The complement is also open. X connected implies it's the whole space. So they agree everywhere. So they are equal, the lift-to-liftings. So this was just the correction to this exercise. Now I prove a fundamental group of some surfaces were too many folds. But a little bit more general. This is surface. We start more general, so here's a theorem. We want the two sphere. We have S1, okay? The circle, fundamental group are the integers. And we gave two applications. One is the Brouwer fixed point theorem. And the other is the fundamental theorem of algebra, okay? Now you want the other spheres, S2, SN. So this is a special case of Fonkampum's theorem. I will not prove the general case, but Fonkampum's. It's very known under the name of Fonkampum's theorem. So if X is the union, A union B, AB open, the intersection pass-connected. That's important. So we have X is the union of two open sets. And the intersection is pass-connected. And we choose the base point in the intersection, X0. So then this special case says if pi one of A, X0, this is our base point. And also pi one of B, this is base point, are trivial, trivial groups. Then also pi one of X, X0 is trivial. This is a special case of Fonkampum's theorem. So the general Fonkampum's theorem describes the whole fundamental group, also in non-proven cases, okay? So proof, so what we have to prove is that every path, so let W be an element of the fundamental group of X with base point X0. So W is a path, no? W from zero one to X is a path with a closed path with initial point and final point. It's a base point X0. We have to prove that W is homotopic to the constant path, okay? We have to prove W is homotopic, path homotopic to the constant path. So this is constant path in X0. That means the fundamental group has only one element, which is the trivial element, class of the constant path, okay? So it's a trivial group. Okay, so it's again the lemma, by the way, if W, the just a remark here, if W from I to S to, no, sorry. That comes later, that remark. I was already in an application of this theorem, okay? We will apply them to SN, to the M sphere. It's homotopic to constant path. So the Lebesgue number lemma. Applied to the open covering. So we have an open covering of X, which is AB, so we take pre-image, W minus one of A, W minus one of B. This is the open covering of the interval I, zero one. This is a compact metric space. So there's a Lebesgue number. Associated to this open covering, that means every subset of diameter smaller than the Lebesgue number is contained either here or here. That means the image is contained either in A or in B, okay? That we applied already for the homotopy lifting lemma, for the path lifting lemma, then the homotopy lifting lemma. Okay, this implies, so we subdivide the interval. Into small intervals, and such as the diameter of small interval is smaller than the Lebesgue number lemma. So this is subdivision. So this is T zero equal to zero, smaller than T one. Smaller than, smaller than T L minus one. Strange. Smaller than T L, the last one, which is one. There's a subdivision of the interval I, such that W, of each of these small subintervals, let's call them T I, T I plus one, sorry, interval, closed interval, T I, T I plus one. It's contained in A or in B. It's contained in A or B. Either it's here or it's here, okay? For I, for all I. I equal from, what is the first interval? Where are all the, from zero to L minus one, no? Zero, alternately in A and B. Alternatingly, if one is in A, the next one is in B. Otherwise we can forget the point in the middle, okay? Alternatingly and B, so alternating E, alternating E means if one is in A, then the next one is in B, then the next one is in A. If two are in A, you just forget the middle point, okay? Then we have a larger interval and we are in A. So this is okay. So we can say alternating in A and B. That means that the end points are all, and the initial points of this are all in the intersection, okay? If one is in A, the next one is in B. So this implies that VI of zero and VI of WI, oh, sorry, it's not zero one, that's a problem here, okay? Because that means that W of TI and W of TI plus one are in A intersection B. The initial and final points are all in the intersection. X zero is in the intersection, it starts in the intersection and then it goes on to the intersection and then it starts, okay? Maybe I'll make this picture look like this. So our picture looks like this. This is A, this is B. Here's the base point in the intersection which is path connected, by the way, that's important. So we start, maybe in A. Then the next is in B, okay? So the end point of the first one is in the intersection. Then comes the next one. The end point is again in the intersection. Then comes the next one in the next one, okay? And finally, we come back to X zero. Here are three pieces, okay? Divide it in three. So we have a problem with the product a little bit, but so I have to, the product of these is not defined in our notation, okay? Because our products are defined on the, it's zero one, okay, of passes. This is not zero one, okay? So we reparameterize linearly this W on this to zero one, okay, each of these small. So let W, I, W restricted to TI, TI plus one, okay? So this is our original, this is this small little piece here, but we want zero one, reparameterized to zero one, okay? Reparameterized to zero one in a linear way, but it doesn't matter, okay? So zero goes to this, one goes to this, and then in a linear way, you're interval. So that means WI now goes from the interval zero one, okay? And then, so this is W one, no? This is W two, and this is W three. Here are just three of these, okay? Now, then, now the product is well defined, okay? W is homotopic, not equal, of course. It depends on the parametrization. So W, now we have the product, it's well defined of all these because we have to find the product only for this interval, not for our particular interval. We could define it also for others, but it's more or less the same. So W one, no W, where does it start? The first one is, sorry, maybe I missed something. WI goes from TI, so we start from T zero, unfortunately. So this is not W one, but W zero, no? This is W one, and this is W two, okay? This is our notation. I should, because we start with T zero and not with T one, okay? So now the product is defined. W one times, times W, how many do we have? L minus one, no? L minus one, yes, L minus one. W is not equal to this. This depends on the parametrization also, okay? On the product, in which order, which is associative. Because this is just a reparametrization of W, okay? This is a reparametrization of W. It's the same path, the image is the same, but we go with different velocities, okay? This is a reparametrization of W, so they are homotopic, okay? This is, if you reparametize, the paths are all homotopic. So they are homotopic, the paths are homotopic. This was a homotopic. This was an exercise, okay? If you reparametrizize the paths, then the homotopic class does not change. Okay, so what? Now we have all these small paths. This is in A, this is in B, this, or, let's start in A here, okay? Doesn't matter, A, B. A, this is in B, this is in A, and so on, no? B, A, B. The fundamental group of A and B are trivial, okay? However, these are not close paths for the moment, okay? That's the problem. So we make them close, okay? That we can do since A intersection B is path-connected. That's the important hypothesis here. It's path-connected, okay? That's, otherwise it's not true. So what do we do? We go with W, zero, okay? Then we are not in the initial point, unfortunately. So we choose a path which goes from X zero to this end point and we go back. From here we go back to X zero. Then for the next one, we go back this path, then we take W one. Then we go back to the X zero, then we go back to this again, and then we take the last one in this example, okay? So let whatever is called, doesn't matter. Let K I from I to A intersection B be a path from X zero to, now we want the end point, okay? So W I the end point of one. So this is W zero of one, no? This is W one of one, the end point of these. And here we are close anyway, okay? So we have to do this from I from zero to L minus two is sufficient. Because L minus one, the last one, finishes anywhere in the right point, okay? So L minus two is sufficient. So I will not make these passes here because otherwise we cannot see anything. Okay, and then of course I may write W is still on the topic to what? So it's W zero. So what you said is, W zero, we close it, okay? So we close it, how is it called? It's K zero minus one, okay? K zero minus one, we go back. And then we take K zero, okay? And then we take W, the next one is W one. Then we go back with K one to the minus one. Then we go back, and so, so where does it finish? The last pass is W L minus one. This goes back to the right point anyway. And this before, what is it? K L minus two, all right. And why is this homotopic here? Because this here is homotopic to constant, okay? And the constant, so we use our lemma, okay? So this is homotopic to a constant pass. This is homotopic to a constant pass, and so on. So this doesn't matter, I can forget this, okay? This is like, it's not a close pass. But if you take any pass with a constant pass, then it's homotopic to this pass, okay? So we can cancel this, and then we are left with this, okay? This is the lemma, we prove the lemma. Two lemma, lemmas, no name, associativity first, no. We use lemma one, two, and three, okay? When we introduce fundamental group, one, two, three. This is associativity, this is neutral element, okay? And this is inverse. So you look at this lemma, and now, of course, associativity, we may do it in a different way, okay? So I should use colored chalk, which I don't have, but it doesn't matter, so now I make this, of course, this kind of, this, and the last one is, okay? And now all these passes, here, here, here, and so on, these are closed passes, okay? So these are all closed passes. In A or in B, okay? First one in A, next one in B, and so on and so on. They are all closed passes. But the fundamental group of A and B is trivial, okay? With base point x zero. So they are all homotopic to a constant pass. They are all homotopic to C x zero. They all go from x zero and go back to x zero. So all homotopic to x zero. So I may write C x zero, C x zero, C x zero, C x zero, okay? So w is homotopic to C x zero, which means that w is a trivial element, okay? Of the fundamental group. So w is equal to C, that's the only element of the group. So this means that the fundamental group is trivial, also of x. This is a proof. Again, the Lebesgue number lemma is the main point. Application, that's the end of the proof. Application, so theorem. For n bigger than one, the fundamental group of SN with some base point x zero is trivial. So recall what is SN? SN is the n-sphere, unit n-sphere. So these are all points x in Rn plus one, such that the norm of x is equal to one. This is the n-sphere. So this is the subspace of Rn plus one. That's the topology, of course. And this is the unit n-sphere, okay? So this was, okay, so proof. So we divide SN into, let SN plus be all points in xn. No, let it just be SN minus the south pole. What is the south pole? Zero, zero minus one. This is south pole. And SN minus be equal SN minus the north pole, zero, one. Okay? Then SN plus is homeomorphic. So this is homeomorphism to what? SN minus the south pole is homeomorphic to what? N, yes, Rn. And SN minus is homeomorphic also to. What is the mapping? Famous mapping? Yes, stereographic projection. From the south pole or from the north pole? And this of course implies that a fundamental group is trivial, okay? So this implies that pi one of SN plus with some base point x zero and pi one of SN minus base point x zero are trivial. These are trivial groups, okay? SN plus intersection SN minus is pass-connected. This is homeomorph to what? So this is equal to SN minus zero both. Plus minus one, no? So this is homeomorphic to Rn minus one point, zero. Which corresponds to Rn minus zero. So it's Rn minus zero, which is pass-connected. Why? Since N is bigger than one. Of course R1, the real minus zero is not pass-connected, okay? It's not connected even, okay? But R2 minus zero, that's pass-connected. So since N is bigger than one. Not for an equal one. This is, but otherwise we prove that the fundamental group of S1 is trivial, okay? That's not the case, that's the integer. So here you see that's the hypothesis that intersection is pass-connected. It's important in this theorem, otherwise not true, okay? Otherwise we prove something also for N equal one. That's the point here, okay? It's pass-connected. So the theorem then implies, Fankampian theorem then implies pi one of the union with x zero is trivial. Okay, that's all. So that's the application of the calculus theorem. Now I can, if I want, we have this remark, okay? For remark, S2, pi one of S2. Let's take the two sphere, which we see, okay, in three space. With some base point x zero is trivial. That's a surface, that's a two-manifold. That's why I write, that's a surface, S2, okay? That's a two-manifold. This is trivial. So I wanted to make this remark, take any pass here, okay? So W is again from I to S2, okay? If surface initial point and endpoint or base point, wherever it is, okay? If W is not subjective, then it's easy that W is homotopic to a constant map. If W is not subjective, that seems strange, no? W from the interval to S2. If W is not subjective, then it is easy to see that W is homotopic to a constant map. Why? So here you have, this is a two-sphere, no? S2, here's some base point x zero. So it's not subjective, okay, W. By homeomorphism, there's some point not in the image, okay? And then we can assume that it's the antipodal point, doesn't matter too much, okay? You just take a homeomorphism and you can assume that it's not so important. I mean, some point is not in the image, okay? And we assume that this point is the antipodal point of x zero, that minus x zero is not element in W or phi. It's not in the image, okay? The antipodal point, which is this one. And now we have a path, so they avoid this path, W, okay? So we have a path here, which is very complicated, but it never comes to this, okay, to this point. And now what we do? We don't have straight line homotopy here, okay? But now we project W along great circles back to the constant path, okay? That's like straight line homotopy, but on the two-sphere. These are not straight lines, but these are great circles, okay? Now project W back to x zero to the point. Contract, contract, contract W to x zero along great circles, okay? Along great circles as two. So here's a point, maybe here's a point, W here, some point, so here's a point of W, so I'm projecting along this, okay? I have to parameterize from zero to one. At zero, everything remains at one half, I go one half at one, I go all the way back to x zero. This is like straight line homotopy, but on S2, which are not really, what is this straight line on S2? This is like straight line homotopy, but now along great circles. We don't have segments, okay? Like we have in RN or something, okay? Along straight line, but along great circles. But it's very similar, no? I hope you see that, okay? In RN, you go along the segment back. Here you go along the circle back, okay? So this implies that W is homotopic to the constant path. This is homotopic, you see the homotopy, okay? You see it, you go, this is missing, and so you shift everything to x zero, to the point x zero, the whole two sphere, okay? Between terms zero and one. So this is a nice example, okay? So you say, but now it's not surjective, okay? Then we can do this, okay? Unfortunately, passes are very complicated. So there exists surjective passes. This would be, otherwise, a very easy argument, okay? Unfortunately, there are surjective passes from I to S2. But, so this proof, I mean we proved it, okay? But Fankampian theory, which is not so trivial as this, but there are surjective passes. That's something, seems strange, or no? W from I to S2. Because this seems one-dimensional, no? And this seems two-dimensional, so how do you can get everything? But there are these strange phenomena in mathematics always, okay? So the first famous name as well, they are surjective passes from I to the square in R2, okay? There's a name associated with these kinds of passes. Somebody already saw this name. What? Piano curves, yeah. Type piano curves. He constructed such strange surjective mappings from the interval to the square. And then you can go to S2, that's no problem, okay? So this is a remark, okay? In some cases, it's clear if it's not surjective. But what is bad are these bad paths, okay? So then you have to work in some way, okay? And that's, you make it non-surjective, maybe, but then you have, it's technical, okay? Anyway, we proved this in general, using the back number number, okay? So other examples of surfaces. So we should, we take Rn minus zero now, example, other example. Rn minus zero is homeomorphic to Rn minus zero. It's homeomorphic to Sn minus one, which is subspace of Rn minus zero, times the real numbers which are bigger than zero, okay? The positive real numbers. What is homeomorphism? So you take a point X here, and then, what is this point? You divide just by the norm, and then you take the norm, okay? So these are polar coordinates, kind of stuff. So X norm, okay? That's just what you do. This is the subspace of Rn, okay? Sn minus one. So this is homeomorphism. Rn minus zero is nothing else than Sn minus one times the positive reals, okay? And now there's a general proposition. If you have a product, the fundamental group is easy. If you know the fundamental group, so what? For proposition, the fundamental group, pi one of a product X times Y, con base points X zero times Y zero, it's isomorphic to the pi one of X, X zero, times pi one of Y, Y zero. So this is nice to say, the fundamental group of a product is a product of the fundamental groups, okay? The fundamental group of a product. It's almost nice to write informally. The fundamental group of a product is a product. Well, finite product here, okay? And if it's infinite, you have to think a little bit. Then you have a product, but it's not the product of the fundamental groups, but probably the, well, whatever, two. Of a product of two spaces is the product of the fundamental groups, or finitely many. Otherwise, there's a difference between, if you have a building group, for example, with product and sum, okay? The product is large, the sum is not large, okay? So one has to be careful with this kind of thing. But that's not our, you have two spaces, there's no problem. So I will not, the proof is, this proof is boring, because you just define the natural maps, okay? So we define, there's no, nothing, the quantum is interesting, okay? The back number, lemma you use, and so on, this is, so you define a mapping from here to the product, pi one x, x zero times pi one y, y zero, and you do it in the most natural way. So you have w here, okay? And then you have the two projections, okay? Pi one, pi one is not so good, because we have too many pi ones. P one and P two, P one from x cross y to x, and P two from x cross y to y, the projections, okay? First factor, second factor. And then what you do, we just project our paths, we have our paths, we have w, and then we project it to x. And we take the path, the class of this, this is here. We just project it, and we project w also to y, so P two. So we have this element, okay? So, what we have to prove, well, everything is well defined, maybe, okay? It's well defined, and we're not proof. I mean, if w is homotopic to w prime, then w, P one, w is homotopic to P one, w prime. And you take the homotopy, P, homotopy, okay? You compose everything with P, just so. It's well defined, homomorphism of groups. Well, if you have a product here, w one, w prime, then what you may first take the product, and then project, or first project, and take product, then the result is exactly the same, okay? The paths are the same, okay? Even the paths is not the homotopic, the paths are the same, okay? First the product, then projecting, or first projecting, then taking the product. The result is exactly the same path. This means this is a homomorphism of groups. It's injective. So what is the inverse? Then we have the inverse, of course. Inverse homomorphism, that's maybe the easiest. Inverse, isomorphism. Phi, phi, inverse isomorphism of phi is psi from pi one x, x zero, times pi one y, y zero, two, we're finishing the blackboard, x times y, x zero times y zero, okay, so what you do is you take w one here, and you have w two here, and what you take? You take just the product of the passes now, okay? W one times, in this sense a product, okay? Not the product of passes, in this sense a product. So w one times w two of t of a point is of course w one of t times w two of t, okay? So if you go by this and then you project, then you get back w one and w two, okay? And also the other way around. First this and phi is the identity. First phi, then this is the identity, okay? That's the whole proof. So there's no idea, it's just formal, okay? However, this means then some corollaries. Pi one of r two minus zero. So it's isomorphic to pi one of s one times pi one of r positive. I forgot the base points. I have a problem with base points now, because when I take x zero and s one anyway here, okay? It takes the same x zero, and here I don't know. What group is this? What is this? So this is, yeah, and this? The positive is trivial, no? Straight line homotopy, trivial group. Well, I don't write zero. I don't know what zero is, okay? Trivial group, whatever is, okay? This is again straight line homotopy. So this is pi one of s one x zero, okay? This remains, so this is that. And I mean you project onto s one, right? So the isomorphic is again given by rotation number, okay? You have, first you are in r two minus zero, and then you project to s one, and on s one you have the rotation number, okay? Which doesn't change. So isomorphism is given again by the rotation number. You define it in this way, okay? That's it, and then you call it rotation number. Around zero here, okay? Rotation, you are, how many times you go around zero, okay? And then you see projecting to s one, and then how many times you go around s one, okay? And that you get by lifting to the reals, and the path from zero to some integer, and this integer is the rotation number, okay? That's the isomorph. Okay, so this, this is also surface, of course, but non-compact. The other fundamental groups are trivial, no? I don't write. If you take s n minus zero, okay? R n minus zero, then it's trivial, okay? The fundamental group. The normal corollary of another surface is s one cross s one. This has a name, the surface. Yeah, the torus. Except it's not so clear where is this living, this torus, okay? If you have it in this way, we have s one cross s one is contained in c times c, or r two times r two in this version, no? c times c, which is, if you want r four, okay? It's a subspace of this, okay? This is one version. But of course, you can do it in r three, okay? But not, this is a very canonical way, okay? And then you have this in r three, and you make this picture. This is one s one, the first. And the second one is this. This is another s one. And if you take the product of this s one, with the second s one, you get exactly the torus, right? So this is not equal maybe, but it's homeomorphic to s one cross s one, okay? But this is living in r three, okay? Formally, it's living in r four. Did you do differential geometry? Not yet. Second semester, yeah. Because this is interesting torus. There are many torus, okay? And then you have all types of curvatures, okay? Flying around. All types of curvatures that we will do. Toe is the main example, okay? In differential geometry. S two is the first one, and the second one is the torus, okay? At the surface. And then you have all kinds of curvatures, okay? Positive, negative, where you have this kind of points and zero, okay? How are they called? Yeah, curvatures, okay. Whatever it is. This torus in r four is a very special torus, because this, what is curvature here? Well, whatever it is, this is a s curvature zero, okay? It's a flat torus. This is a flat torus. This is not flat. Here's curvature, okay? So, seen from point of view of differential geometry, this torus in r three, and this torus in r four are very different, okay? From differential geometry. This torus in r four has curvature zero, okay? It's a flat torus. Flat means curvature zero. This torus in r three has all, it's not possible with curvature zero, okay? You always have some curvatures, okay? But this is differential geometry. Anyway, that you will see. The most interesting torus is maybe this one. It's a flat torus, okay? It has no curvature, this one. Whatever two curvatures. It's locally like r two in r four, okay? It looks locally exactly like r two. Where? What? Well, you want to make this identification. Well, anyway, pi one of s one cross s one, whatever, one times one, I can write now it's also morphic to pi one of s one, one times pi one of s one. One, so this is isomorphic to z times z. So this is written as z squared. That's the fundamental group of the torus. And you see the isomorphism. Because you project a path first on, if you have a path here, first you project onto this s one, and see how many times do I run around this. And then you project onto this s one, and then you see how many times do I run around this. And this is, you have to get two integer, two rotation numbers now, one around this, and one around this with the two projections, okay? From s one to cross s one to the first s one, and then to the second s one, okay? Two rotation numbers, and this is z two. That's the isomorphism. That's okay? More or less. Well, the projection is defined, no? From s one to the first, and then you have a path, and then you project this path. And then you are running just around this s one, no? And you count how many times do I run, okay? And then you do the same here. So you get two integers. And these are the two rotation numbers, okay? And this is the isomorphism. So corollary, so this is the corollary. It's a fundamental look at this, okay? And so you have another corollary, that they are not homeomorphic, no? And this is completely clear to everybody, but s two and the torus, s one cross s one. So as a surface, of course, we want to be in r three. That's our space where we see what is a torus, so what? So we make the picture here. In r four, you have to imagine how it looks in r four. Anatomyomorphic. Why not? The fundamental group of s two is trivial. And the fundamental group of s one is z two. So they have different fundamental groups, okay? Of course they are path connected, so it doesn't depend even on the base point. So this is a proof that they are not homeomorphic. S two and the torus, okay? If you want to do that in a direct way, it gets somewhat not so easy, okay? The fundamental group is okay. So you call s two is the surface of genus, is the orientable surface of genus zero. That's what they call, no? And s, the torus, what is this? The orientable surface of genus one. And then maybe you see what is the orientable surface of genus two. Wait, wait, this is a picture. One picture is this, now you have two of these, okay? Oriental surface of genus two, and so on. So you see the orientable surface, okay? Oriental surface of genus two, and so on. They're also non-orientable surfaces, okay? And now I will introduce a non-orientable surface. So what is a real projective spaces? I want to define real projective spaces. So you did geometry, okay? Projective geometry. So if you have a vector space, you have a projective space, okay? So each vector space, that's projective geometry. R3, for example, okay? This is a vector space. So real RP2, RPN, that's my notation, RPN. RP2. In geometry, what do you take as a set? The projective space associated to R3, this is, okay? So you take all lines, so you take all lines in R3 through zero, okay? All lines in R3, zero. These are the points, okay? And in general, for any vector space, you take all lines in the vector space, all subspaces of dimension one, that's the same, okay? All subspaces of R3 of dimension one. That's what it is. In geometry, in projective geometry. That's the definition in projective geometry. Now we notice that, I want a topological space now. All subspaces of dimension one. Each line through zero intersects the two three and two points, okay? So this is a remark. Each line through zero intersects S2, intersects S2, which is, okay? In two antipodal points, clear? In two antipodal. Then as a set, I take equivalence relation on S2, okay? Equivalence relation on S2. So I give the equivalence classes. That's an equivalence relation. I just give the equivalence classes, okay? So equivalence classes, the equivalence classes are X minus X, X and S2. So this is a class of X. Equivalence class of X. Contains two points, which are antipodal points. We want to say this is one element only, okay? And then we have, as usual, we have a canonical projection from S2 to this. So we have a canonical projection P from S2 to the equivalence class and that's RP2. RP2, our definition. What is this? This is equivalence class of X, X and S2. And this is just X minus X, two points. Each equivalence class has two points. So the mapping is P of X is the equivalence class of X, obviously, okay? Each point goes to its equivalence class. So this is, if you want our definition, no? What is corresponds to the division of geometry, no? Where you have lines. But each line intersects S2 in these two points. So assets, they are bijective in a natural way, okay? The lines now on S2 correspond to these pairs, right? So it's the same set. And now we have a topology. So what S2, it's a subspace of R3. So we have a topology, subspace, no? We have a topology here and we want a topology here. So what to define? Quotient, the only reasonable way in general is the quotient topology. So let the quotient topology. So V in RP2 is open. What is the definition of quotient topology? If and only if, yes, P minus one of V is open in S2. So this is what, which topology is this? Quotient topology. This is the finest topology on RP2 such that P is continuous, okay? The finest topology. So this exercise on quotient topology. The finest topology on RP2 such that P is continuous. Okay, that's the quotient topology, okay? And then this is, you can do the same for SN, okay? RPN, exactly the same. I did it for S2, okay? From S2 to RP2. It's a covering. So here's S2. That's easy to make a picture, okay? Now I have to identify antipodal points. That becomes one point, okay? So P projection, okay? Well, I can represent everything on the lower hemisphere, right? A representative. But I still have to identify something here, okay? Because still here we have antipodal points, which we have to identify now, right? And then I cannot make a good picture. This, of course, is homeomorphic to the disk. No, that's the projection, okay? And then I have to identify antipodal points on the disk, okay? So I have to identify still to identify antipodal. This is maybe, so here you have more, still more equivalence relation. You still have to identify to the stock of. This is maybe the standard presentation of projective space, okay? A disk where on the boundary you have to identify antipodal points. You cannot make a picture, a good picture now, three. You can try, but it looks. So what means covering? Covering means just given any point here, I take a small neighborhood, U, okay? And how does the preimage of this U look like? Well, you get one copy here, no? Okay, so, and you get one copy on the other side, no? So here, on the other side, right? This, so this is P minus one of U. Of what? So this is evenly covered, okay? And this is a small disk, these are two small disks, okay? That's all, open disk, open sets. So this is a covering, two fold covering. Each fiber has two elements, okay? That's a covering, two fold covering. Each fiber has two elements. And these are antipodal points, okay? Antipodal points, the two elements. You recall, if this is connected, then each fiber is the same number of elements, okay? That's what I proved, okay? This is connected. This is connected, this is subjective, this is past connected, sorry, past connected, okay? This is past connected. Well, it's connected, so I proved for connected, no? If the base is connected, then each fiber has the same number of elements, okay? Here's two. So theorem, that's a surface. This RP2 is a surface, okay? There's a too many fold, because each point has a neighborhood, which is like an homomorphic to an open disk in R2, okay? You may be saying, these points are strange, this point, no? But it doesn't matter, I can make this where I want, okay? I can make it this. I can avoid my points, okay? That's not the problem. Okay, so theorem, pi1 of RP2, X0. So it takes some class here, no already, okay? It's isomorphic to Z2. So this is, by definition, Z mod 2Z. So a group is two elements, okay? This is a group with two elements. The fundamental group of RP2 is a group with two elements. And in the same way, you prove RPN, I didn't define RPN, but it's exactly the same, RPN. You do everything the same, and this is also, ah, this, well, I have to be careful. Well, I will prove this, but the same for RPN, but we have to be careful with RPN a little bit. So I may give this remark. So we define, in a similar way, define RPN. RPN is all classes X in SN, no, no? So these are, again, antipodal points, X minus X. And then we have also here projection from SN to RPN, the same kind of projection, PX equal class of X. And this is also two-fold coverage. So you define RP2, I did RP2, no? RPN, you define in a similar way. There's one interesting special case. This is RP1. Example, RP1 is equal, so the question is what is RP1? That should be one manifold. One manifolds are very easy, no? So this is X or classes, but X now is in S1. So just, again, X and minus X. So what you have to do is, you have to identify, this is S1, and you have to identify antipodal points, right? This becomes one point, okay? So if you do this, this you can see, what do you get? So what I can forget half of these, then I have each equivalent class still, but I still have to identify these two points, okay? These are the remaining points which I have to identify, to one class, this, this, and this becomes one class. And now if you identify these, what do you see? So you can do it, okay, in R2, and what do you see? S1, okay, S2, S2, S2, S2, S2, S2, S2, S2, S2, S1, you see the S1 sphere, the one sphere, okay? And that's true, RP1 is just the one sphere, S1. RP1 is S1. So remark, this implies RP1 is homeomorphic to S1. That's just the one sphere, that's a special case, okay? And now you say maybe RP2 is a two sphere, no, that's not the case, but how to prove that. Well, I wrote the theorem, which are unfortunately okay. The two sphere is a trivial fundamental group, and the RP2, okay? So a corollary, you have another surface, and this means S2, RP2 is not homeomorphic, RPN. Well, this is true for N also. RP2 is not homeomorphic to S2, why not? Because S2 has trivial fundamental group, and RP2 has Z2, this is not the same group, okay? So RP1 is S1, but RP2 and S2 are very different, okay? In fact, RP2 is a non-oriental surface of genus, let's forget, of genus one, this is one. This is the first of the non-oriental surface. What is the second one, the next one? That has still a name, what, I didn't understand. I still don't understand the name. No, no, maybe spent is with boundary, okay? Here we have without boundary, there's no boundary, okay? Maybe spent is closely related to RP2, okay? But this is with boundary. Surface with boundary, without boundary, no? We have without boundary. Other surface, of course, is a disk, two disk, no? But this has a boundary, okay? So I didn't mention the two disks. S2 are closed surfaces, without boundary, okay? The torus has no boundary, okay? Whatever it is. So not RP2, there's one more famous name of a surface. So you have the two sphere, you have the torus, the oriental ones, you have the projective plane, projective plane, this one, no, RP2. And then the next one is the Klein bottle. That's what it's called, okay? Klein bottle, whatever it is. But maybe you heard this name, no? If your surface is so Klein bottle, I don't define, but this is the next one. Is non-orientable surface of genus two, okay? That's the Klein bottle. There's another name, then I have no name. Then the next one is non-oriental surface of genus three, and so on, and so on, and so on. But without special names. The first one is the projective plane. The second one is the Klein bottle, okay? The non-oriental ones, and the oriental ones, the two sphere and the torus, the next one. Without boundary, okay? Closed, without boundary, only. Well, anyway. So proof. What do we have to prove? The theorem, right? Of pi one of, well, I write even RPN here now. It doesn't matter, X0 is isomorphic to Z2. But here I have to be careful, that's one special case, okay? If N is equal to one, this is S1, so it's Z, okay? So for N, bigger than one, then it's always Z2, for the higher ones, okay? The only exception is, so pi one of RP1 with X0 is isomorphic to pi one of S1, they are homeomorphic, with some base point, whatever it is, X0. And this is Z, that's the exception, okay? Otherwise it's, okay. So now let's prove it. What would be interesting is to do the classification of surfaces, okay? All closed surfaces. However, we don't have enough time for that, okay? That's the interesting point, okay? To classify surfaces. That's a classical argument. Some combinatorial argument, of course. You need some combinatorics here, okay? You have to present the surface in a good way, by a triangulation. So we take the covering, P. I'm doing the case RP2, okay? But the case RPN, for N bigger than one is the same. You will have to see where N equal to one is not good, okay? Maybe. So RP2 to S2, no. That's the wrong direction. From S2 to RP2. So P of X is the class of X. And the class of X consists of two points, which are antipodal points. So this is a two-fold covering. What do we have to prove? We have to prove this one. So I define a map from pi one of RP2 with base point X0. So X0 is a point in S2. I want an isomorphism with Z2, okay? And Z2 I write multiplicatively, okay? So I write, what is this? Multiplicative with one and minus one. And with a product. So this is isomorphic to Z2. The same group, no? With two elements also. So I want to define this mapping. So I take a class here, W. It's the same proof for S1, no, okay? You will see. Not exactly the same. The structure is exactly the same, but we need something else. It's a little bit more difficult, maybe, because we have to use the theorem before. So what is W? W is a path. WI from RP2 is a path with initial point and end point. The class of the point X0, and X0 recall for a point in S2. It projects to its class. So, what did we do? We use a path lifting lemma to get a lifting. Let W tilde from I to S2 be a lifting. So this is a path lifting lemma. Of W with initial point, W tilde of zero. So I have to take a point which projects to the class of X0, so I'm taking zero, obviously, okay? X0, which is a point in S2. So I have to take a point which projects to the class of X0, so I'm taking zero, obviously, okay? Which is a point in S2. Fixed. Let W be a lifting of W with this. Then, what is W? Then, W tilde of one, the end point. What is the end point? If you project, it becomes a close path, okay? W is a close path, right? We lift it, this is X0. So what are the two possibilities for the end point of X0? Two possibilities. X0 or minus X0, yes. So this is epsilon X0, where epsilon is equal to one or epsilon is equal to minus one, okay? X0, the same, or minus X0. These are the two points which project to our base point, okay? And now I can define my mapping. I didn't give a name. Phi. So I introduce a name here, Phi from this, okay? Now I define Phi. So let Phi of W be, this is a definition. So what do you take? Epsilon. If it goes to X0 as a lifting, then we take one. If it gets to minus X0, we take minus one. So it's epsilon in one minus one. We just take this epsilon. We look where it doesn't fit, okay? So that's the definition of this mapping Phi. And now we just only have to prove that Phi. Well, Phi is, now it's the same as before. So then what we have to prove? The first is, Phi is well defined. And this is the homotopy lifting lemma, okay? By the homotopy lifting lemma. Or it's corollary. Phi is very fine by the homotopy lifting. That's the next important step, okay? Then Phi is the homomorphism of groups. How much time do I have? Not too much time. So I don't prove it. It's the same as for the reals. In the reals, one path, you have to translate, okay? Now, if the first path ends in minus X0, you have to multiply the second path with minus one. So it starts in minus X0, okay? That's, it's the same otherwise, okay? So the time is more or less finished, so I don't write it now. So Phi is, this is now interesting. Phi is injective. So this I will do. Well, it's also the same as before, except we have to see the difference between n equal one and n bigger than one. I'm doing the case two, but it's, otherwise it works for every. So Phi is injective. So let W be in the kernel of Phi, okay? And we have to prove it's trivial. The kernel is trivial, no? So what does it mean? This implies that W tilde of one, the endpoint of the lifting, which point is it? It goes to one. So X0 also. It's one times X0. So this is X0, okay? Epsilon is equal. Epsilon of W is equal to one. So this is, of course, is also the endpoint, no? By definition, this is by definition. So this means W is closed. This means W tilde is closed. It's a closed path. Where? In S2. So W tilde is an element where in the fundamental group of S2 with base point X0. And now this is more difficult than in the case of S1. This is a trivial group, okay? In the case of S1, we have the reals here, okay? And this is trivial by straight line homotopy. Here, this is a trivial group. This is a trivial group. That means that W is equal to the trivial element. W tilde, sorry. W tilde, there's only one element, which is the constant path in the point X0, no? That's the only element of the trivial group. So this implies that W tilde is homotopic to CX0 by a homotopy H. They're homotopic. Another project. So this implies that W, which is a projection of W tilde, is homotopic to the projection of CX0 projected. What is this? This is the constant path now in the class of X0, RP2. And the homotopy, of course, is H and then projected. So this is trivial, this step, okay? What is not trivial here is this fact here, okay? This is an addition to that. So this means that W is the class, many parentheses here. W is also the trivial class, okay? And this means the kernel is trivial. So this implies the kernel of phi is trivial. And this, of course, means that phi is injective. And then the last thing is subjective, okay? So objective is trivial again. So the last thing is phi is surjective. So it's the same as before. Maybe I'll do that to two lines. We have to get minus one, okay? Phi goes to one minus one. One we get, no? That's the unit element. We get from the unit element in any case, no? So we have to get minus one. So let U be any path, be a path in S2 from X0 to minus X0. Well, there's no problem, no? It's path connected. We go on the great circle, okay? In one or in the other direction. There are two ways not to go on the great circle. Let U be a path in S2 from X2 to minus one. And W be the projection of U. So this means W is a closed path. It's a closed path. So it represents an element in the fundamental group of pi1 of Rp2 with base point X0, the class of X0. It's a closed path. And now what is phi of W tilde? So W tilde of one. What is W tilde? This is just U because we project it from U, okay? And now we lift back, so we get back U, okay? W is a projection of U, no? W is a projection of U. So we lift back, we get back U, okay? So W tilde of one is equal to U of one is equal to minus one. By our choice, okay? And this implies phi of W is minus one. No, minus, this is minus X0, yes, thank you. And phi of W is minus one, yes, exactly, okay? And this, we get minus one in the image, and this means phi is subjective. And that's the end of the proof, okay? The main difference between this and the case of S1 is that the fundamental group of R is trivial, and that's easy. The fundamental group of S2, we have to prove that it's trivial, okay? And that's not completely trivial, okay? That's Fankampen, we use Fankampen theory in a special case, okay? So this, now we have many surfaces. Now we have RP2, that's a projective plane that's called also, okay? Fundamental group on two elements. We have S2, fundamental group trivial. So this is in some sense to, so fundamental groups, our surfaces, fundamental groups. So we have the surfaces, S2 is trivial. We have torus, this is Z times Z. We have RP2, so this is the surface, and this is fundamental group. So we have what else, RP2. This is called the projective plane, no? Projective plane also, okay? That's the projective plane, real projective plane, if you want. Fundamental group is what? Z2, not this Z2, but this Z2, okay? That's it for the moment, I cannot, we don't have others, okay? So this is, so they are not homeomorphic, okay? They are not homeomorphic. These three are not homeomorphic. Okay, that's application, okay? Of fundamental group. This is difficult to prove without fundamental group, okay? It's not so easy to prove that these are different here, okay, otherwise, okay? The fundamental group is nice for this, okay? And also for the other surfaces, by the way. Fundamental group is the most important invariant, okay? Algebraic invariant. That's the fundamental group. And so Thursday, we have last lecture, no? Thursday. Thursday from, not tomorrow, no? Thursday, from 11 to 13. And then I will finish with homotopy equivalence, probably. Maybe I talk also about mobius band, okay? That's also an interesting example, but it's a surface of this boundary, and we will see also the mobius band, okay? What? Is it? This? Well, more or less distributed, so you have, yeah, it's in the book more or less, but it's in different chapters. I'm not going in a linear order through the book here, okay? Because there's not enough time to do everything in the book, okay? It should be in the book, yes. Projective spaces, real projective plane, that is in the book, yes. Fankamp theorem, in some version, it's later, in the later chapters, okay? Yeah, in some way it's everything in the book, but distributed, you don't, okay? I went a little bit diagonal through this, because we had only five hours of this part, and if you follow one chapter, then, okay? But the important things are these, no? The fundamental group of S1, okay? The fundamental group of S2, SN, okay? Fundamental group of the torus as a product, that's easy. Fundamental group of RPN, okay? Projects, these are a nice example. What we should see maybe also next time is, these are all Abellion groups, right? All Abellion groups, okay? So you may ask them, why do we see only Abellion groups, okay? So we should give one example of a space which is non-Abellion fundamental group. So that I will do first also. At least one example, okay? And typically, fundamental groups are not Abellion, okay? Abellion groups are easy. But fundamental group can be everything, okay? So in general, it's not Abellion. Here it happens that for these first easy cases, okay? It's always Abellion, trivial Abellion groups, okay? But if you go on with other surfaces, it becomes immediately non-Abellion and get more difficult, okay? So I will give one example of non-Abellion fundamental group also next time. Okay, that's it.