 So, last lecture, and we introduce, so what we did in the second part was homotopy, fundamental group, coverings, then we computed fundamental groups of the circle, of the two-sphere, of real projective spaces, these were the main examples, okay, nevertheless. This is the last important concept from algebraic topology, which are homotopy equivalents. That's also a very basic notion. So what is the definition? So let f from x to y be continuous, always continuous of course, f is a homotopy, f is a homotopy equivalence, if there exists a continuous map, if there is a continuous map, in the other direction, g, from y to x, from y to x, such that we have first f and then g is, this is homotopic, homotopy, or sign for homotopy, to the identity of x and first g and then f is homotopic to the identity of y, so these are the conditions for homotopy equivalence. If the example is homomorphism of course, okay, example, homomorphism, that's, well, f, that's an example, then we have equal here, right, then we have equal, but now we have weaker and we write not equal but homotopic, okay, so we will understand, try to understand a little bit of this notion, that's a very basic notion of algebraic topology, homotopy equivalence, one of the most important. Okay, so then one says that, if there exists, one says, if, that's a way to talk, if there is a homotopic equivalence. So x and y are homotopy equivalent, so spaces, x and y are homotopy equivalent, and sometimes one just says homotopic, maybe, but this is homotopic spaces, okay, not maps now, okay, there are homotopic equivalency spaces, there's just homotopic spaces, so we have, sometimes there exists a homotopic equivalence, okay, if, if there exists, there is a homotopic equivalence, f from x to y, then they are called equivalent with respect to homotopy, homotopy equivalent or just homotopic, okay. This is by the way an equivalence relation, this homotopic, this is an equivalence relation for topological spaces in general, very general, for topology, for the, for topological spaces. Well, I will not prove this now, okay, so this you can say, if you want an exercise. Because I don't want to, examples, so we have to, to get a feeling for this, we have to do examples, right, examples, so we have a feeling, what, what, what does it mean, also homomorphism is an equivalence relation, right, but if two spaces have homomorphics they are equal in some sense, okay, here they are not equal, it's weaker, much weaker, so the classes are much longer, so definition, then there is another definition, then starting examples, a topological space is contractible, x is contractible, if the identity map of x is homotopic to a constant map, so it's homotopic to a constant map, so this means identity x is homotopic to constant maps in some point, x zero and x, that's contractible, okay, so what has it to do with homotopic equivalences, so, so there is a little remark, another exercise, this I will do that, but it's very easy, so f, x is contractible, if and only if, x is contractible, if and only if, x is homotopic equivalent to a one point space, okay, these are the one point spaces, x is homotopic equivalent to a one point space, so here also we write, sorry, x and y are homotopic equivalent, if there is a homotopic equivalent, and one writes also here, which is convenient, x is homotopic to y, okay, this means homotopic, so somebody writes for homeomorphism, okay, in general for isomorphism, homeomorphism, we write x this for homeomorphism, okay, this is homeomorphism, but this is just a notation of course, okay, this is homeomorphism, this is homotopic equivalent, okay, this is isomorphism in algebra, isomorphism, we write very often in this way, somebody writes this way, but interpolation is not good, because this is homotopic, okay, well, so this means to one point space, this means x is homotopic to just one point x zero, wherever that is, in x or not in x, doesn't matter, just a name, so exercise, well, let's do then this exercise to get used to this, this is very trivial because there are no maps to one point spaces, okay, almost no maps, okay, so just to get used to the definition, so proof of this, so we have two directions, x is contractible, so let's suppose x is contractible, x is homotopic to one point, but there are no maps, okay, so this means we have said the identity map is, identity map, x is homotopic to constant map, x zero, x zero in x, that's the definition, no, that is the definition of where is it, this one, okay, this is, for some x zero in x, no, maybe, it's also clear if it depends on x zero, which one, okay, so that we have now, okay, and now we have to define mappings, f from x to x zero, but we have no choice here, no, this is a constant mapping of course, okay, constant, no choice at all, and then g from the, in the other direction, x zero in x, and x zero is a point of x already, okay, so I take an inclusion, because we only have a reasonable choice, and then what we have of course is if you take first g and then f, then it's equal to the identity, f first g and then f, then it's equal to the identity, it's equal to the identity of this, okay, and if you take first f and then g, so what is this mapping, so first f, everything goes to x zero, and then we are in x zero, and we go back to x inclusion, so that's the constant map x zero, okay, so this is the constant map x zero, constant mapping x zero, okay, from x to x, and this is from a topic to the identity of x now, so we have these two conditions again, no, so this means that f and g are homotopic equivalents, okay, f is homotopic equivalents, and g is inverse, it's also homotopic, it's symmetric, okay, there's a one direction, easy, and the other one is s easy, easy, so this direction we have to prove, so f is homotopic to one point space, so what does it mean, we have a mapping from f to a one point space, now it's y zero, I don't know which point it would be, so I give another name, and then we have a mapping g from y zero, that's a point here, to x such that first f and then g, what is this, y zero, y zero goes to x, so this is even equal, no, sorry, first f then, which mapping is this, that's first x goes to a point, then the point goes back to x, so this is a constant map in which point in g of y zero, g of y zero, g of y zero, everything goes to g of y zero, okay, and of course this, by definition, is homotopic to the identity, okay, so this is homotopic, because these are homotopic instances, so this is homotopic to the identity of x, so this, okay, so that's all, okay, so this is an easy exercise, the contractible spaces are exactly the spaces which are homotopic to one point spaces, so these are the trivial spaces in the sense of homotopy, okay, example, standard example of course is, Rn is contractible, so seen from the view of point of homotopy, okay, Rn is trivial, okay, from homotopy, it's just as a point, nothing else, this is our usual straight line, homotopy, okay, so we have to find mapping contractible, so what we have to prove, we have to prove that the identity of Rn is homotopic to what, to a constant map, okay, to some constant map, so constant, well, let's take zero in Rn, okay, constant map, so we have to define such a homotopy, h, and so h of xt, one minus t times x, or t times x, but it's a metric, okay, let's try this, one minus t times x, so at time t equals zero, this is the identity mapping, okay, at time t equals zero, this is just x, so this is the identity mapping, and at time one, we have zero, everything goes to zero, so this is this mapping, okay, so this is just this homotopy, again straight line, okay, straight line, homotopy, we go from each point x back to zero, okay, so this is a straight line, homotopy, going from x to zero on the straight line during the time interval, from zero to one, okay, so Rn is correct, but another example, many small examples, almost trivial examples, so for example, x is always homotopic to x cross i, if I take a product with an interval, the homotopy type, the reason is that this is correct, okay, so again it's a trivial proof almost, so we have to find f, well now I write R instead of f, so x cross i to x in one direction, okay, I need a map, and that takes the most natural map, x times t is equal to x, okay, that's, well let me call it f, sorry, and g in the other direction, from x to x cross i, so x goes to, any point goes to g of x equal to, well we can choose x zero, doesn't matter here, right, we can x times zero, okay, for example, and then of course we already have one thing which is that, what do we have, first f, first g, and then f, first g, and then f, this is equal in this case to the identity of, first f, first g, so the identity of, this is equal to the identity of x, okay, and then we have to take f and then g, the other direction, so x cross i projects to x and then everything, x t goes to x zero, that's the mapping now, okay, so x times t goes to x times zero, okay, that's this mapping here, okay, the composition, so we have to prove that this is a homotopic to the identity of x times the interval, right, it's not equal, obviously now, because x t, it's like a projection, and so we have to define our homotopy again, our homotopy of, so I wrote t now, okay, I have two parameters in the interval because it's x cross i already, times another i for the homotopy, okay, so I should have got this s probably and then use t for the homotopy, sorry about that, so it doesn't, it's not big difference, okay, just write s here and then we have s, x cross s goes to x cross zero, right, and now I have x, it's x cross i, okay, that's the point here and then t, the time parameter, okay, times t is equal to x, so we don't leave x times, so try s times t, something like that, or minus one, okay, s times t, both in the interval, right, so at t equals zero, this, at t equals zero, this, at t equals zero, this is zero, so this is this mapping, okay, at t equals one, it's the identity of x cross i, x cross s goes to x cross s, okay, so it's an homotopy between these two, so exercise, if y is contractible then x is homotopic to x cross y, if I multiply with the contractible space, it doesn't change the homotopy type, okay, this is like a point from, okay, so that's the very reason, okay, so this I will not do, if y is contractible then x, okay, that's, what else, another example, so I'm doing only examples, examples, so rn minus zero, we know already, that is homomorphic to what, sn minus one times real, all right, bigger than zero, real, positive reals, so this is contractible, okay, so we do the same proof here, only we have to, this is positive reals now, but this proof works, we cannot take zero, but we have to take one or some point in the positive reals, doesn't matter too much, okay, however, so this is homotopic, would be homotopic to sn minus one, but I can give a direct proof without using this, so let's, this is also very trivial, but let's do it, so we define a map, these are the standard maps, x run minus zero goes to f, goes to sn minus one, so what mapping do we have, of course, we have x goes to x over x, that's in one direction, and the other one is g, which goes from sn minus one to rn minus zero, and what to take, well, this is a subspace anyway, so we take just inclusion, okay, so this is inclusion, so now I write different letters, okay, r, this is a retraction, so this is a retraction, that means, so r, and this is inclusion, so that means first the inclusion, then the retraction, this is equal to the identity, no, of sn minus one, so this implies r is a retraction, okay, and the other one we have to prove, so first r and then i, so this should, it's not equal to the identity, but should be homotopic to the identity of rn minus zero, this is the other space, okay, so we have to define a homotopy here, and we again have straight line homotopy, it's always the same because we are in rn, okay, so these are the two conditions, one time we have equally even, okay, that's the retraction, and the other time we have homotopic, okay, this has also a special name, which I write, first let me define the homotopy, so the homotopy h of, which is another space, rn minus zero x, t is equal to, so this should go from the identity, so from the identity of t equals zero, one minus tx, what is this mapping, r, first r, the retraction, and then the inclusion, that goes to sn minus one or everything, okay, with the retraction it goes to sn minus one, and then it remains sn minus one, so this is, the other point is x over x, that's t plus t, so this is again straight line homotopy, always the same, so we have to be sure that this is different from zero, however, okay, straight line homotopy, again straight line, because we have rn, only in rn, so the picture is this of course, I mean I can make a picture in r2 or r3, now, so if we point x here, then what we do, we feel the point x over x norm, right, and then we just let the point walk on this, avoiding zero, okay, we go from x to, on this segment, between zero and one, for each point, and the inside we go the other direction, okay, if we have a point here, x, then this is x over x, and we walk in this direction, so this is another definition, all, this is first, maybe the most important class of homotopy conclusions are these, okay, so a definition, still a definition, many definition, a retraction r from x to a subspace a, so what does it mean, that means that first inclusion and then the retraction is identity of a, that means a retraction, okay, that's the definition of a retraction, it retracts onto the subspace, without changing the subspace, identity on the subspace, okay, that's the retraction, and a retraction is a deformation retraction, so this is an important definition also, it's a deformation retraction, if we have, in the other direction, first a retraction, then the inclusion is homotopic to the identity of x norm of the large space, so here's one condition, equal again, and here's the other one, only homotopic, okay, that's a deformation retraction, this is an example of a deformation retraction, right, this is a retraction, and this, okay, that's an example, so a deformation retraction is homotopic, okay, that's the most important example, or deformation retraction, okay, so this implies a deformation retraction, it's a homotopic, of course not all examples of this type, because this requires a to be a subspace of x, right, if you don't have spaces which are not subspaces, one of the other, then we don't have this notion of deformation retraction, only of homotopic results in general, right, so this is the most important, however this is still the most, this is easy to see in general, okay, so somebody, example, another example of deformation retraction, somebody of you mentioned the Möbius band, okay, last time as a surface, so intuitive example, okay, so the Möbius band, how is it defined, I mean there are many Möbius bands, okay, so what you do is, you take, it's a band, and then you close, if you close in this, you take a band, okay, a band is just an interval times i, okay, you take a band, and then you close in this way, what you get is a cylinder, right, and now you close not in this way, but you close in this way, and then you get the Möbius band. So to make a picture, we have one twist here, well everybody knows, because it's a famous, maybe the most famous symbol of mathematics, okay, is the Möbius band, which you see, well there's one other, which is a tree-fold knot, okay, the other one, the tree-fold, that's the Möbius band, okay, the tree-fold is this one, so under, over, this is the tree-fold, okay, this is the other, these are the two most famous symbols of mathematics, which you see everywhere, okay, anyway, we are not interested in this one. So I say, so this Möbius band, M, M, of course there are many Möbius bands, so large, small, but they are all homeomorphic, okay. This is the subspace of R3, of course, okay, subspace. That's a subspace of R3, that's a topology, okay, we are in R3, like the torus, right, that's a subspace of R3 also, or of R4, okay, that we said last time, okay, a torus S1, close S1, which lives in R4, but the picture we make in R3, okay, from point of view of topology, these toroid are the same, homeomorphic, from point of view of differential geometry, they are very different, that you will see. So now I want to say this is homotopic to S1, this is just homotopic to the circle. So from point of view of homotopy theory it's a circle, well it looks like a circle, but you have a rotation number again, right, if you take a base point and then you go here, well you can go in this direction, but that doesn't matter so much, so you just count how many times you go around, okay, and the fundamental group should be the integers again, given by, okay, that's the case. Which S1 do we take? That's an interesting thing, which S1? There's a boundary which does not work well, okay, the boundary is not okay, so we have to take the middle line, the central line, so this is our S1, it's a circle, right, the central line, the boundary line here goes two times around, this goes one time around, okay, you don't want to go two times around, just one, so we take the central line, and now we define our attraction in the most obvious way, we just project everything onto the central line along the intervals, okay. We have these intervals all, so you take a segment times interval, okay, so you have all these intervals, and then you close, still you have all these intervals, okay, vibration by intervals, right. So if you have a point x here, then this point is r of x, we just project along this straight line, and here the same, in the other direction, okay. So this defines the map r from the Mobius band to S1, which we have considered the subspace of m in this way, okay, and this is a retraction, this is a retraction, but it is a differential retraction, so we have the inclusion and we have the retraction, so this is this map what you see, everything gets, okay, you go down to S1, okay, you, how do you say in English, you squash or whatever, the Mobius band and to the center line, okay, that's a very intuitive map of course, right, and now we want to prove that this, in fact, is homotopic to the identity map of the Mobius band, okay. So I will not write a formula now because I make it by picture, so at time zero we have the identity map, at time one everything is in this point, right, and it's clear what we do at time one half, what do we do, we just take half here, okay, so we let this interval get smaller and smaller at the end of the point, that's our homotopy, again some kind of straight line homotopy if we consider this as an interval, right, so this is our homotopy, at time one we have the identity, okay, of this interval, at time zero we just have a point and now we let this between zero and one go down to a point in a linear way, okay, it's an interval, so this implies that this R is the deformation retraction, that's an example of a deformation, so the Mobius band is just seen from point of homotopy, it's S1, okay, okay, so that means that M is homotopic to S1, of course, S1 is a cylinder, we can do the same now, it's a cylinder, it's S1 cross I, that's when we close without a twist, okay, that's a cylinder, and this is called, we proved even X cross I is always homotopic to X, okay, and this is an equivalence relation, so this implies then of course that the Mobius band is homotopic to the cylinder, okay, seen from point of view of some of these, they don't differ, okay, this is not a retraction because these are different spaces now, okay, but the easy space is this one, okay, in this class, and not this one, and not this one, okay, the easiest representative is S1, you will not find something easier in this equivalence class, okay, of S1, but you have the Mobius band, you have the cylinder, you have many things, okay, so this is a homotopic, just an example of homotopic equivalence which is not a retraction, which is not a deformation retraction because these are different spaces now, okay, and now as I said already, so fundamental group, we go to fundamental group now, the Mobius band, what you suspect anyway is a fundamental group should be the integers, okay, and you again have something like rotation, okay, you count how many times you go around, okay, it's like S1, okay, so the question is, if we have homotopic spaces, what happens with fundamental groups? They should be isomorphic, okay, that's the idea, that here, that also here, okay, so that we will prove now, okay, so we have to prove something, and now I prove this, this is an important definition, okay, homotopic equivalence, homotopic spaces, retraction, deformation retraction, okay, contractible space, these are very basic definitions of algebraic topology, these are more algebraic than general topology, okay, so proposition, the problem is that homotopy between maps, okay, does not preserve base points in general, okay, so we have problem with fundamental group, so it moves the base, homotopy can move the base point of the fundamental group, okay, so we have to compare fundamental groups with different base points, so let H be a homotopy, X cross I to Y be a homotopy, homotopy between, between, so we call Ft of X, we may call Ft of X equal to H of Xt, no, a time t for fixed time t, okay, for fixed t in the interval, so what is H is a homotopy between F0 at time zero and F1, so this is a homotopy between these two maps, okay, and now let alpha, it's a little bit technical, this one, but it's important, let alpha from I to X be the path, so we fix the point anyway, so what is it called, X zero, so let fix X zero in X, that's the base point, okay, but this during the homotopy is moved in general, okay, it's not fixed, so let alpha from I to X be the path which X zero moves during the homotopy, so this path, how does it move, so be the path alpha of t equal of Ft of X zero, so we fix our point and look, so this is just H of X zero t, X zero is fixed, no, that's the path, it's a continuous mapping because H is continuous, okay, it's a restriction to one point, then, so this is now the conclusion, then, so we want to compare F zero star and F one star, the induced maps on fundamental loop, okay, I will make a diagram, then it's clear, so they are not equal because F zero of the base point is in general different from F one of the base points, they cannot be equal, okay, this map because they go to different base points, okay, however, we have this mapping alpha hat, so now I make alpha hat is the isomorphism from one base point to the other base point of fundamental loop, we have a path, we define this, and this is exactly this, so I make a diagram which makes it more clear, anyway, this is the point, but the diagram is this one, so we have pi one of X, X zero, then we have F zero star, the induced map which goes to pi one of F, F goes from what to what, to Y, pi one of Y, what do I have to write here? F zero of X zero, okay, I have to preserve base points, F zero of X zero, Y zero, but I don't give you, on the other hand, we have F one star, so this goes also to pi one of Y, and what do I have to write? F one of X zero, okay, F one of X zero, it has to preserve base points, and maybe they are different, these things, so they cannot be equal then even because these are different groups, they maybe are isomorphic, but it's not the same group, okay, they are different, what is the isomorphism? So we say F zero, then we go with alpha head, and then we go to this group, okay, because alpha is a path from, alpha is a path from F zero X zero to F one X zero, yeah. So this is an isomorphism, so that we know, okay, so this is an isomorphism, so this is equivalent to the fact that this diagram is commutative, that's exactly the same, okay, this is short and this is with all spaces, groups. It's commutative, it's a commutative diagram. In algebra, maybe you saw maybe it's a commutative, you have all kinds of commutative diagrams, that's very important in algebra, that means following this way and following this way is the same, okay, and then we have large diagrams sometimes in algebra. What? Thank you. Because applying H will go too far, yes. Okay, so this is the proposition, a little bit technical proposition, I indicate the proof now. So we have to prove, I'm looking at this diagram, we take an element here, okay, let W be an element in the fundamental group of pi one X X zero. We have to prove that, we have to show that, to prove that, going this way, going this way is the same. So that means that, we have to prove that, first alpha minus one, first let me start here, so this is W and then F zero. What is this? This is F zero star of W. That's the definition of the star, okay. You take an element and you just compose a path and you just compose with a map, okay. This is the definition of the induced mode. That's exactly what you said. On the other hand, we have W F one. So this is nothing else than F one star of W by definition of F one star, okay. And so we have to prove, first this one, so we are here, then this one is equal to this one, okay. What is this isomorphism? Now you have to recall what is this. So I take alpha minus one star this one and then I go back with alpha. So alpha minus one comes from here to here, then we have this element and then we go back and then we are here. So this is equal. That we have to prove. That's the definition of alpha hat, this conjugation, okay. That's the definition of alpha hat which we did some time ago. So we have to prove that it's associative, so I don't care about, okay, from left to right, but whatever. So we have to prove that alpha minus one, which is equivalent to the fact that alpha minus one times F zero w times alpha. So this represents this, okay. It's homotopic, it's the same class. It's homotopic H to F one. That is the same, okay. This equation, this equation is the same. So we have to prove, we have to find this homotopy, okay. And this I do schematically as I used to do by a picture. That's the diagram, that's the proposition. So I want to define the homotopy, okay, H. So here we have to define what is H? H should be a homotopy between what? Between alpha minus one. So we have here alpha minus one. So now we have a parametrization, okay, problem. So we take one-fourth, one-fourth, one-half. But we see that reparameterizing doesn't change homotopy class, okay. Parameterize as I want in some sense, okay. So I take one-third, one-third. I mean, if you make parentheses in this way, then it's one-fourth, one-fourth, one-half, okay. But if you make it parentheses in this way, then this is one-half, one-fourth, one-fourth. It's different parametrization, okay. I take here one-third, one-third, one-third. It doesn't matter. So here we have alpha minus one. Here we should have what? F-zero times w, and here we have alpha. That's on the left-hand side, okay. On the other side we have F-one, w. That's it, okay. So here we want to define our homotopy. So this is a time interval, of course, and this is x, okay. It's a homotopy class so far. So what do we do? Well, we make our picture. We try this one. So we want to define it here at a time t, okay. Here we want this, here we want this. Now we have to do what do we want to do at time t, right. But it's, we have F-t also, right. We have at time zero, this mapping. We have F-time one, this map. It's a homotopy, F-t. So it's clear that we should take F-t times w here, okay. Here's zero. This is zero. This is one. So here F-zero, w, F-t, w, F-one, w, okay, between zero and one. And then we have to say this piece here and this piece here. Well, I can make a picture also. We have zero and one. So we have at time one, we have w, F-one. At time zero, we have w, F-zero. We have alpha. In which direction does it go? From identity. So it goes in this direction, I suppose. No. Yes. Okay. And so we have at the time t. What do we do? We go t, okay, on this path. Then at time t, we have here in this w, F-t, of course. Okay. So if we go here, we follow this and then we go back, okay. So this should be alpha with friction to zero t. We don't go all alpha. We just go to alpha t. Then we go around here and then we go back, okay. So this should be, we go back. So we go back. So this is alpha. Well, this is alpha minus one, okay. Sorry. So this is alpha minus one restricted to, this is alpha minus one, okay. So alpha, this is alpha minus one restricted to t one or one minus t one. Which one? Alpha minus one is this one. So this is one minus t. So we want alpha minus one here, okay. That's just, in terms of alpha minus one is. And this one is an, and then we go here and then we go back. So this is alpha restricted to what? t zero. Something like this. Something is wrong here. So most of people, so I'm mixing up something. I'm sorry. Let me try again. Something is wrong here. So I want the homotopy. Let me try again. We go from this path here is alpha minus one. What is alpha? Alpha is this path, right? From this f zero of x zero. This is one, f one of x zero. And so this is alpha. So here I go alpha minus one. Then around this one and then back with alpha, okay. And then it should be this one, homotopic to this one. That's a time, a time zero. I take alpha minus one. Then this one and then I go back with alpha here, okay. At time one, we just have this one, okay, at time one. And this is, so we have to prove that this path alpha minus one around back is homotopic to this path, right? So what we do again, again. At time t, we go with alpha minus one. Alpha minus one, zero, we are here, okay. Then I'm getting confused with the directions, okay. At time zero, we have this, this and going back, okay. At time one, we have this and we have to find the homotopic. So it's a very small time. What do I do? Epsilon. At time zero, I go all the way, okay. At time very small, I go up to this point, okay. And then I follow this and then I go back. So maybe it's more clear. So this is a very small time. So alpha, so alpha, what is this? That's the question here. What is this? Alpha minus one, restricted to what? Alpha minus one, restricted to one, zero, one minus t. And this one, so we go to this point. Then we go around here, okay. And now we go back. And this is alpha. We go back with alpha. But alpha from where? From t to one. From t to one, okay. That's, it's clear what the homotopy is, okay. It's not so clear because alpha one minus one, alpha goes in one direction, the other direction. So that's the concentration. But now it seems to be okay, hopefully. Of course, we have to re-parameterize everything, okay. We have to re-parameterize. So we have to, all these three, we have to re-parameterize, okay. Because this is not the unit interval any longer. This is unit interval. This is not, and this is one-third only, okay. So we have to re-parameterize everything in a good way. But that's no problem. The formulas are in the book, but they are not interesting. I mean, you cannot, never remember. I can't even, here you have to concentrate, okay. But this is, form is a definition, and this is a picture to this definition, okay. So you compare this and this. Well, our theorem, if f from x to y is a homotopy equivalent, then. So let me fix base points with, you choose base points. f of x zero is y zero. Then, if you have a homotopy equivalent, then f star, the inducement from pi one x, x zero to pi one y, y zero. It's an isomorphism, okay. So homotopy equivalence, choosing base points in this way, which one is made to the other base point, then the inducement is an isomorphism. It's like for homomorphism, right? For homomorphism, it's the same, no? But for homomorphism, it's easy, because these are just the funtorial properties, okay. A homomorphism induces an isomorphism of fundamental rules. These are just the two funtorial properties, okay. This is very easy. Here we have to work a little bit, because we have this base point problem. The base point is moving around, the base point of the base points, okay. Good. So what means homotopy equivalence? We know the definition now. f goes from x to y, fg goes from y to x, such that first f, then g, is homotopic to the identity of x, and first g and then f is homotopic to the identity of y. Let's take a picture, no, a diagram. Pi 1x, x0. So here we have to prove that f star is an isomorphism. This goes to pi 1 of y, y0. That should be an isomorphism. Of course, we have g star, okay. Some g star. This goes back. Pi 1x, unfortunately, is not x0, okay. But I have to write, so this is g star. So what do I have to write? I have to write g of y0. I have to preserve base points, okay. I don't know which point this is. g of y0, okay. However, this is, of course, gf star, that's the factorial property. It's f star, g star. No, this is first f star, then g star. This mapping is homotopic to the identity, okay. So the identity induces the identity, no? Identity x, x star is identity of, these are the functorial properties, okay. Identity of pi 1x, x0, functorial properties. So this implies that these two maps are homotopic, okay. So we apply the proposition which we proved. So this implies that f star and then g star is not equal to the identity of pi 1 of x. It cannot be equal because we have different fundamental books, right here. It's a different base point. So it cannot be the identity, okay. But the thing that says that we here have some kind of alpha hat, okay. Where alpha is the path from the model. This is a proposition. So this is just alpha hat because the identity we don't see, okay. So we have here, this is equal to alpha hat. And this is an isomorphism. Alpha is the path defined by this homotopic, okay. As in the proposition. And now, so what does it say? What does this imply? This is an isomorphism. What do you say immediately for f star and g star? If this is an isomorphism, the composition is an isomorphism. First this and this is an isomorphism. The identity doesn't occur. So what does it say about f star and g star? Sorry. f star is? No. A mapping. f star, if you have f star and then g star is an isomorphism. What property has f star? What? Almost. Injective. Yes, injective. The first one is injective. The second one is surjective, right, okay. If you have this, then this is an isomorphism. Then this has to be injective and this has to be surjective, okay. This implies injective and this is surjective. That's half, right. Because now we want also surjective here. So we may go here. And now I take, again, f star. But this is not the same f star because it's from f. But we have another base point here, okay. So it's not equal. Anyway, we go to pi one of, so what do we have to write? Pi one of y and what is the base point now? The base point is f of g of y zero, okay. f of g. I have to preserve base points. f of g of y zero. Well, many. One, two, three. One, two, three, yeah. Okay, I have to preserve base points. This is not exactly the same map as this, right. Because this is with different base points, okay. However, by symmetry, okay, we again have the same stuff. So we have something here which is some beta star. And this is an isomorphism. That's why the second one. It's the same, okay. First f then g, now g then f, okay. Taking care a little bit of base points. And what does that now imply? This is again an isomorphism. So this says then that g star is injective. And f star, f prime star is surging. So what we proved is not that f star is an isomorphism, but that g star is an isomorphism, right. So this implies that g star is an isomorphism. But now you call gf and fg, okay. So by symmetry, it's symmetric completely, okay. And this implies by symmetry. I should have given other names, okay. Also f star is an isomorphism. So what you do, invert the rows of f and g, okay. Okay. That's it. So corollary. Now the fundamental may depend on the base point, okay. If it's not pass connected, it depends on the base point, okay. But first corollary, sorry. If x is contractible, then pi one of x, x zero is the trivial proof. What means contractible? That means that the identity map is homotopic to constant map. That's the definition of what? Of contractible, okay. That's the definition of contractible. The identity map is... Well, I don't want the definition. I'm not distracted. So this means that this is equivalent to fact. I didn't want to write this even. That's the definition, okay. This means that x is homotopic to a one point space, okay. X zero. That I proved. Exercise, okay. Which I did, which I did. X is homotopic to a one point space. And that means that... What does it mean? The fundamental rule of this is trivial, right? So the fundamental rule of this is trivial, okay. So if you have this mapping here that induces... This is homotopic equivalent. So there's only one map from x to x zero anyway. So it's homotopic equivalence. And this implies f star is an isomorphism. However, the fundamental rule of this is trivial, okay. Trial fundamental. X zero. X zero is trivial. Well, maybe I made... This is trivial fundamental. Oops, all of them. This is trivial fundamental. Now you say, but maybe... Well, it's okay. So there's an exercise here. Another exercise, a small exercise. We didn't give so many exercises for this part. But exercise... X contractible implies X is pass-connected. That's a nice exercise. Not nice, not difficult. So the idea of this proof, of this exercise, if it's contractible, it's pass-connected, okay. So what it means pass-connected? Pass-connected, you have to give any two points, okay. And to have to find the pass. Now the identity map, it's contractible. It's homotopic to a constant map, okay. X zero. I don't know if I can choose this point in an arbitrary way, okay. There's some X zero, so that is homotopic. So X zero is here. But here we have a homotopy, okay. And now I find the pass from here to here and from here to here. From here to here and from here to here, okay. So I can define a pass from, well, I don't know, in this direction and also in this direction, okay. And then I take here the inverse pass and go from here to here and then here. So then we go here in the other direction, okay. What are these passes? So let me call these points X one, X two. So what is a pass which goes from X zero to X one? This I call alpha. So alpha of t would be h of X zero t. Let me try. So at time one, this is X zero, okay. At time zero, sorry. At time zero, this is from the identity. So it's the same point, X zero and X one, okay. So at time zero, this is the identity. So it's X one, okay. X one. At time one, this is the constant of X zero. So it's X zero. So it's a pass in both directions, sorry. It goes from here to here, okay. And then we do the same beta of t. It's h of X two t. And now it goes symmetric way, okay. It goes from X two to X zero, okay. And then of course, first alpha minus one. Alpha minus one times, it's not a composition. Times, you know, alpha times beta minus one. Alpha times beta minus one is a pass from X one to X two. Is a pass. Which are arbitrary points. X zero, it's not so clear, okay. Well, now maybe it's clear that you can talk any X zero. But in the proof, we have to distinguish X zero from these two points. X zero is given to us here, okay. And we cannot, or we have to prove that we can take any point, okay. So this is contractible. That's an exercise in the book also, which I wanted to give but forgot. Last time, okay. In the first chapter here, we have a long exercise on contractible spaces, okay. You can look at this. One of these is contractible is pass connected. Okay. At the corollary, if X and Y are pass connected and homotopic and homotopy equivalent, then the fundamental groups are isomorphic. Pi one of X, X zero is isomorphic to pi one of Y. Now it doesn't depend on the point, okay. Because they are pass connected. So for all X zero, Y zero, we can choose whatever we want. For all X zero and X, Y zero and Y. Pass connected homotopic equivalent spaces have isomorphic fundamental groups, okay. That's homomorphism. Same, okay. Same feature as homomorphism. So contractible space is trivial fundamental group, right. And now there's a problem. If you have trivial fundamental group, it's a space contractible, maybe. What do you think? It's a problem. S two is simply connected. Then we prove, no, from comes here and what's all, no. The fundamental group is trivial, okay. It is pass connected also. Then there's a question. Is it contractible, maybe S two? Is S two contractible? What do you think? I mean intuitively. S one is not contractible, no. Because the fundamental group is not trivial. It's Z, okay. A contractible space has trivial fundamental group. S one has non-trivial fundamental group that's not contractible. RP two has non-trivial fundamental group. So it's not contractible. RP n has non-trivial fundamental group. It's non-contractible. S two, Sn has trivial fundamental group. So what? What do you think? I mean contractible means that the identity map, you see the two sphere, you can homotop two points, okay. Is it possible with S two? It doesn't look that way, right. It's difficult. You have to, in a continuous way, right. If you take one out one other point, okay. Then you can contract, okay. But if you have all points, then intuitively you have, it's not in a continuous way, okay. At a certain point there will be some, how do you say, okay. You have to cut somewhere, okay, the situation. So the answer is no. However, how to prove that, and that's not so clear, no. Because the fundamental group doesn't help here, okay. And now this is the starting point of other things in algebraic topology, okay. So if fundamental group doesn't help, you try other things, okay. And the easiest or the next one is maybe homology, okay. So then you have use homology or you have use the higher groups, okay. One can use, it's difficult to prove this directly, okay. I don't know, I never thought about that. One can use homology. The difference is that fundamental group has this one. And homology hn, this is for all n, okay. Or pi n, there's pi n, okay. Which is maybe more difficult, but we have to renovate. Also for higher dimension something, okay. Or pi n. These are the higher homotopy groups, okay. Pi 1 is the fundamental group, it's the first one of these. And this is the starting point of, and then this is, comes out. It's not correct, okay. Because some of these groups are not trivial, okay. Hn of s2 are not trivial. But contractual spaces, everything is trivial, okay. Contractual spaces are like a point. And everything which you can define is trivial for a point. Almost all of us. Okay, and these for the two sphere are not trivial. H2, you need two, okay. So, but that's not all, okay. So, when should I finish, by the way? I started at 11, 12, 30. That's 12, 30, no? Yes. So these are homotopy equivalences. This is starting of algebraic topology, okay. There's fundamental group, homotopy, of course, basic. But this is, then you want other functions, okay. Also, you'll recall that the Brouwer fixed point theory. We proved only in dimension two. If you want to prove it in higher dimensions, we should use something which is defined for something bigger than one, okay. This one. A homotopy group, these are the homology groups, okay. So, there's a need to generate so high a dimension. There are direct proofs of the Brouwer fixed point theory. But this is very nice, but avoid, you don't have to use this. Okay, so the last thing which I wanted to do is to give you an example of a space with non-computing. All groups are very easy, you know. Z, Z2, Z times Z. Easy abelian groups, okay. So now I'll give you an example just to finish. That's not typical for fundamental groups. Fundamental groups are difficult, okay. This is misleading a little bit, okay. These are easy groups, if you saw. But in general, it's very difficult, okay. Because they're non-abelian. That's the point, okay. Non-abelian groups are difficult. For example, I close with an example of a space which has non-abelian fundamental group. And you take the easiest possible, so you take this and this. Well, I do it, but I will need more space upstairs, sorry. So what is this? These are two circles which have one point in common, okay. This is two circles with one point in common. This is called figure eight. That's eight, right? That's eight. Figure eight, which you use for infinity. What is this? This is a picture in R2, of course, okay. So this is a subspace of R2. Two circles. And this is our base point, X0. So the fundamental group of this, it's not here. These are not this. These are circles, okay. So this is one circle. The fundamental group of this is that, okay. But you count this rotation number of this part. Of this part the same. You count the fundamental group of that, no. Now you take a path, this path, and then this path. And then this, and then the other one, okay. And you see no way how they might commute up to homotopy. You cannot transfer something which goes one time around here to something which goes around here. That seems to be impossible, right? So they shouldn't commute. So I take, I call this A, and I call this B. These are also paths, okay. This goes one time around here. This goes one time around here, okay. And I want to prove that A times B is not homotopic to B times A, okay. That's what I said. So that means A times B is different from B times A. And that means it's non-commutative, okay. The fundamental group is not commutative of this figure. But how to prove that, okay. So I prove it in a very nice way because we use coverings. And I define a covering of this, okay. I hope I don't have a picture here, but no. So I do the following. I have to define a map, a projection, a covering. Okay, so these are four circles. It's always in R2, okay. Substance. So I have to define this map P. Well, I have names here, A, B. So the easiest way, I'll say what is happening. So here I go to A. If I go from here to here, I go one time around here, okay. So this and this point go both to this point, right. So this is E0, this is E1, this is E2. So E0, E1 goes to one time around here, okay. Then here I take another time, A. So E1, E0. So going one time around here goes two times around here now, okay. That looks familiar, okay. And then what I do here, now an E1. And here go also this point, right. So the fiber P minus one of X0 are three points. What are they? E0, E1, E2, okay, three points. So here now I go on the other side, B. So this defines my mapping. Here I go to A, here I go to B, okay. And here I do a similar thing. This goes to B, one time around. This goes to B, second time around. And this goes to A. So one time around here, one time around here, okay. Like this, one time around here, one time around here, okay. That defines my mapping P. So P is the three-fold cover. Well, that's clear because for any point, any fiber three elements, okay. These are these three points, okay. And for any other point, for example, and locally it's a homomorphism, because this looks like this locally, right. Neighborhood of X0 looks like this, right. A small neighborhood. And in small neighborhoods of these, of course look exactly the same, right. So these are homomorphisms. And if you take any other point, maybe you take this point. And I look for the pre-image, okay. Of this point. What is the pre-image of this point? It's this, it's this, and it's this, okay. Again, three copies, okay. So this is a three-fold cover. This is a three-fold cover. And each fiber is three elements. Is it clear how the mapping is defined? P should be clear from the picture, no. This means these points going from here to here go to these points one time around, okay. This constant velocity also. That's the mapping also, okay. So this defines the mapping. The picture defines the mapping. And now, we want to prove this now. Now A and B also passes, okay. Now A and B also passes. So it takes the path A times B and I lift with the initial point. Well, my base point is E0. It has end point. What is the end point? Well, you tell me, what is the end point? First A then B, lift. First A, so I lift A. I go one time around here, okay. I lift. So I go here, right. And then B. So I go around here, B, okay. So the end point is E1. I don't even use the lifting lemma now because it's clear that they lift, okay. Here you see everything. It's very explicit, okay. You can lift passes here. This path, no. And then I take B times A with initial point E0. This is also our initial point, okay. That's end point. What is the end point? Well, first B now, okay. So I lift B from this point. Well, the only way is this. And then A. So I go just one time. E2. I finish on the other side. That's end point E2. That's preview, you know. This picture, okay, defines everything. You see everything. It's very geometric. However, so these two passes, this is A times B, the liftings of A times B and B times A with the same initial point have different end points. This implies that A times B is not homotopic to B times A. If they would be homotopic, then they must have the same end point by the homotopy lifting them up. If they are homotopic, their liftings have the same end point. Have also, they have the same initial point by definition, have also the same end point, which is not the case. They have different end points. So they cannot be homotopic by the homotopy lifting them up. And it's corollary. If you look at this, I formulated this as corollary. Okay, so they are not homotopic. And that's exactly what I wrote here. That means A times B is not B times A. So this implies that the fundamental group is not commutative, okay? So this implies, so finish. Pi 1 of whatever it is called. No name. B, basis of the, okay? Figure 8. Pi 1 of B, space point x0, is not commutative. It's not obedient. This is not an obedient group. So you ask, what is this group? Okay, so what group is it? It's not an obedient group, no, okay? It's not that. Well, you suspect two times that, okay? There's one here and one here, no, two circles, okay? So you have a Z here and Z here. But they don't commute these two Zs. They don't commute, okay? These two copies. So it's not Z times Z, no? It's not a product. It's called the free product. That's a non-commutative version, okay? It is a free product of Z times Z, okay? So this is called Z, that's a free product. A non-commutative version of the, of the abelian product, okay? Of the product. A non-commutative version. And this is called also the free non-abelian group of rank 2. So this is, this is the free, well, I don't know. Free non-abelian group of rank 2. That's what it's called, the fundamental group. Free non-abelian group of rank 2. There's also the free abelian group of rank 2, no? What is that? That's Z times Z, okay? The free, I don't know, you did group theory. The free abelian of rank 2 is Z times Z. This is not times. This is star, okay? That's a free product. This is a direct product, okay? So it's a different product. This is a non-abelian version if you want of this, okay? And so this is the free abelian group of rank 2 and the fundamental group of this is the free non-abelian group of rank 2, okay? It has two generators. The generators are this path, of course, no, for the right-hand side and this path for the left-hand side. One generates this and the other generates this, but they don't commute. This is the fundamental group of the tools. Here they commute, okay? This is also pi 1 of tau. This is a free abelian group of rank 2, okay? The figure 8 is a free non-abelian group. So this is the beginning of group theory, also infinite group theory, okay? Free groups, free non-abelian groups. These are the first groups of infinite group theory. I don't know, did you do that in algebra? Just finite groups, no infinite groups. So this is the starting of infinite groups. Free abelian groups, free non-abelian groups, okay? These are the first examples, important examples of infinite groups here, okay?