 Okay, so we defined homotopy, path-homotopy, product of pathes, and then also product of, so product of pathes was written as this, f times g, okay, and then also the product of homotopy classes of pathes, so this is well defined, okay, and now we want to construct this group, fundamental group, so we need the first lemma, three lemma, three lemma term, lemma, first one, lemma one, so this is associativity, so well f times g, this is notation, times h, it's equal to f times g times h, so this is trivial, so this means maybe this and this, this is associativity, okay, of this product, so what does it mean, f, and the product has to be defined always, okay, this means g starts where f ends in the point, and h starts where g ends, okay, otherwise the product is not defined, then this is all defined, okay, so this is the proof, and I will not write well proof, I will indicate the proof, what do we have to prove, so we have to prove that this is equal to this, this means that f, we have to prove that first this is, this is f times g times h, and the class of this is the same as here, this is first f, and then g times h, and so they represent the same class, so they are homotopic, that we have to prove, okay, so we have to define a homotopy between this and this, okay, and this I will do schematically, so I make by picture, I will not write the formula, in the book you find the formulas, but this is not so interesting, I mean it's clear, let's see what happens, so this product here, okay, we, this is the interval, 0, 1, okay, the homotopy, so the first one, f times g times h, what does it mean, so it means we have h here, and we have f times g here, half interval, half interval, f times g however, is also a product, okay, so we go f with f, with four times the origin velocity, okay, between 0 and 1 over 4, then we go g, also four times velocity between 1 over 4, 1 over 2, and h twice, two times velocity, okay, this is the parameterization, right, between 0, 1 over 4, 1 over half, instead on this side we have just another parameterization, the path doesn't change, okay, the parameterization changes, the image doesn't change, so what we have here, we have first f, this is twice velocity, okay, and then we have g, h, so it's this one, the parameterization, right, and now we have to define h here, everywhere, okay, on all levels, okay, so we want to define this h, so our space, what is it, x, okay, homotopy, where are we in x, wherever we are, there's no one name, so these are paths in x, okay, f, g, h paths in some space, which is almost x, so what would you suggest, what to do, well we have to define h, it's defined here, by the product is defined here, here it should be constant anyway, right, the path homotopy means constant here, so in which point, f of 0, in the initial point of f, this is always the point, this is fixed, okay, this path homotopy, here it's constant path in the final point of h, no, everything else here, okay, so constant here, constant here, here this here, we have to define it everywhere, okay, so what you do is you make some thing in this way, and then you want to do, say, what on an intermediate level t, what should be this, so this is t between 0 and 1, this is level 0, this is level 1 in t, what we do, okay, how we define the map, okay, and it's clear that we just change parametrizations, okay, from this parametrization to this, so we want to take f here also, okay, but this is a parametrization between 4 times the velocity and 2 times the velocity, okay, so we decrease the velocity from 4 times to 2 times here, right, this interval is 0, 1 over 4, this is 0, 1 over half, okay, so just a question of parametrization, and then we have g here, this is 4 times velocity, okay, between, the interval is of length 1 over 4, this one, okay, and here it's also 1 over 4, I didn't write the points, so it's the same velocity, that doesn't change now, g, but the interval, okay, changes, no, here we are from what, 1 over 4 to 1 over half, here we are from 1 over half to 3 over 4, okay, and here we have, we go back, okay, so the velocity is the same, but the interval that's defined, this g, which changes, okay, and then we take here h, this we have to reparameterize also, okay, so we have to, all these three things have to reparameterize, okay, the velocity and also the interval, reparameterization, and now you can write formulas, okay, if you want, you have to find this equation of this, and then in this area, this, in this area, this, in this area, and then you have for some, some algebraic, okay, but I'm not interested in that, okay, that's in the book, but it's clear that if you sit down one hour, then in some way you find, okay, an explicit formula, you have two coordinates, this and this, right, both are intervals, and then for, but this is not very interesting, I mean, it's clear that you can do that, okay, in a continuous way, and I don't know the formula, I didn't, I mean, it's in the book, but I'm not interested, okay, and so this defines h, the homotopy between this and this, okay, anyway, there's a small lemma, but this, the image of this path and this path is the same, the image is the same, no, but it's a reparameterization, okay, this path upstairs, and they are not equal here, that's important, okay, that's not, we don't have associativity for the path itself, this path upstairs here, this one downstairs and upstairs, they are not equal, okay, why are they not equal, the image is the same, but it's a different parameterization, so two paths are the same, if everything is the same as maps, okay, not just the image, but also the parameterization, so they differ by a parameterization, all these paths, okay, it's always the same image, but different parameterizations, right, that's a general fact by the way, so this is an exercise, if you want, if you reparameterize, what is a reparameterization, so f is the path, f from i to x, some paths, and now I want a reparameterization of this path, okay, what does that mean, well in general you take the Dithiomorphism, homeomorphism of the interval, okay, and compose, so let f pass and r, reprameterize, mapping from i to i, in general Dithiomorphism, homeomorphism, whatever, but that's not so important, continuous, even continuous, such that map, such that, such that, so the only thing which I want is that the initial point goes to the initial point, the end point to the end point, I don't want something straight with this point, such that r of zero is zero, and r of one is one, okay, they preserve the end point of the end point, the rest only continuous, then the exercise is then f, and now I take first r and then f, so I want to call this reparameterization of f, okay, reparameterization of f, maybe you want r to be a homeomorphism then, okay, but that doesn't matter, continuous is sufficient, then this r homotopic, a path homotopic of course, that's an exercise for next week, it's easy, okay, this exercise, the first exercise, okay, so if you reparameterize the path in this sense, then the homotopic class is the same, okay, this is easy, we are in the interval, okay, in the interval, in the interval we have some, you know, this is straight line, homotopy type, okay, so this works, then I give the second exercise, so yesterday I made a special case, which also this is very easy, so exercise one next week, this is exercise one, so exercise two, because yesterday I wanted to do this, but I made a special case, so, so this exercise, these are reparameterizations clearly, okay, of the path, no, and then you can define this r, which makes this to this, this to this, and this h to this, in a linear way, okay, that's a reparameterization, so if you want, this implies this, okay, so exercise two, that's what I wanted to do yesterday, so I give an exercise, a very easy exercise, two for next week, so this says that if f and g from x to rn are continuous maps, so we have two continuous maps, to rn again, okay, then f and g are homotopy, not path, not path, okay, continuous maps, then f and g are homotopy, what I showed yesterday, I wanted to show this yesterday, and then what I showed is that any map from x to rn is homotopic to a constant map, okay, but don't use this, you prove directly this, okay, any two maps to rn are homotopic, from space x to rn are homotopic, now, so how is the homotopy called, of course, what is the name of this homotopy? Yes, of course, okay, straight line, we are in rn and then we use straight lines, okay, okay, so this is also, that's what I wanted to do, and then I made some special cases, well that's the first, these are two exercises, okay, and now we have associativity of this product, and now we, f, we need what else, lemma, two, we need a unit element, okay, if we want a group, so if f from i to x is a path, a continuous mapping, then we need something which looks like a unit element, and f times, so this would be a constant path, okay, so this has to be the constant path in the endpoint of f, right, otherwise the product is not defined, so this goes from f of 0 to f of 1, and then we have the constant in f of 1, okay, it's equal to f, this is a homotopy class, and also the other way, only that I have to write now f of 0, the constant path in f of 0 times f is also f, now I have to write f of 0, no, constant path, and then this starts where this ends, okay, otherwise the product is not defined, so this is the second lemma, and here we have, in principle we may apply this again, this, so proof, but I make the, what you have to, let's prove the first one, f times, we have to prove, so the first one is f times constant path f of 1 is homotopic, this means homotopic to f, okay, that's what one says, this one, right, that we have to prove, and we have to define again homotopy here, and now we look what we do the same kind of picture, so I want to define this h to x, so let's look at this, so we have f times constant path, so how is the parametrization of this, half of the time interval for the first, the other half for the second, so here we have f, right, and here we have all this constant in the end point of f, f of 1, that's this product, okay, here, here instead we want f, okay, the normal f, okay, so you may apply this exercise, no, it's not exactly, you have a map from here to, I mean the image is the same, right, what happens here is you go f, okay, what happens here, you go f with twice velocity and then you stay in the end point for the rest of the time, right, that's what happens, here you just have f, okay, some f, that's f, what happens here, you go f also but you go twice in velocity, okay, twice as fast, so you go twice as fast the same path and then for the second half of the time interval from one half to one you just stay in this point, you relax, okay, here you go with the normal velocity, here you go twice velocity and then from, from between zero one half and between one half and one you relax and stay in this point, okay, that's the second passing, this one, right, no parametrization or not, so I said only a continuous map, right, we need this continuous map from here to here to reparameter us, okay, by homomorphism this is difficult because, but we may map this to this by mapping all this to this point, okay, and this to this, in the sense of the first exercise, this is a reparameterization, it's not a homomorphism, no, because this interval goes to one point, okay, so here we want constant map in which point f of one, here we want constant map in f of zero, no, on this side, so in the sense of exercise one, also this is a reparameterization, it's not a homomorphism, no, because the whole interval gets to one point, okay, but it's no problem, so either you apply this or you again, at the time t you say what you want to do, okay, and then there are the formulas again which I don't know, so this is time t, and so what you do, what you do, well it's clear, you do this for example, this, okay, and then you take f here, but reparameterize, now here you have twice velocity, here is the original, and here you pass from twice to the original velocity, okay, and the time interval changes, and here on this part you take everything constant, so here everything is constant here, so here it's constant on this part, constant in this point f of one, in this whole area here, so this is, and you see they are not equal again, okay, they are not equal because they have the same image, but different parametrizations, okay, so they are not equal as passes, you have to pass to homotopy classes of passes to have equality, okay, it's not true for the passes itself, okay, this pass and this pass in general are very different passes, okay, the images are the same, but they are different, two maps are equal, a pass is a map, right, a continuous map, two maps are equal if they are equal for each t in the interval, right, and these are not equal obviously, so it's the equality we have for the homotopy classes of passes, not for the passes itself, okay, here we have only homotopic, but not equal, is it clear, we have to pass to homotopy classes, and then there's a third lemma, and the second is the same, okay, obviously, this is almost, so the third lemma is lemma three, we need what, we need inverse elements, okay, so let f from i to x be a path again, so we have a candidate for the inverse, no, so define f minus one from i to x by, well, if you have f in this direction, and now you just go back, the same pass, okay, so you change, sorry, f minus one, so you change, okay, the direction, so f minus one by f minus one of t is equal to what, one minus t, no, so you go back, yes, then f times f minus one, sorry, I'm lost, what, what, this is not correct, right, f of one minus t, yeah, of course, what did I write, one minus t just, no, f of one minus t, yes, right, f of one minus t, okay, f of course, no, one minus t is, okay, then this is, so what should it be, constant pass, so this is passed from f of zero to f of one, this goes back from f of one to f of zero, so f of zero, the end points of the product are f of zero on both sides, so, and f minus one times f is now what should I write, f of one, okay, we start in f of one and we finish in f of one, so f of one, okay, well, you see the homotopy here, okay, let me make the picture, so proof, we have to fix only the end points, right, so proof again, proof, we have to prove, what do we have to prove, the first one is f times f minus one, not equal in general, homotopic to g f zero by a homotopy h, okay, and then the second one in a similar way, let's look at this, so this means I have to define this h again, this homotopy h, okay, and between this, here we have f and then f minus one, okay, so this is f minus one and here we have the constant map at the time for the homotopy, okay, and here we have what, the constant map, and here we have again the same, here we have the same constant on all sides, okay, and now the homotopy is the following, so again, so what can you do, well, well, the one thing which you may try this one, okay, and then again I say at a level t what do I want to do, how do I want to define this map, okay, and here we are in the point f of one, right, so here's the constant, this is the point f of one, here we have this f of zero constant, so here f of what, one minus t, no, then we go between f of one and f of zero, okay, one minus t, and here on this part we take f but not the whole f now, okay, so what we do is at time t we go just to a certain point, okay, then we remain at this point and then we go back, f minus one but also here we have a restriction, okay, so I have to write maybe the restriction to what, and then we have to reparameterize of course, okay, so we stay here, so this is f restriction and then we go back, this f minus one restricted to, okay, and reparameterized and then at time, this is at time zero and at time one, we just stay all the time in this point, okay, so here in a minute we go just a little bit, we stay a long time here and then we go back and finally we stay all the time, this is a homotopy intuitively, okay, which I'm describing here, so f restriction to what, to zero, t, t depends on t, one minus t, because here is zero, so if t is equal to one, this is zero zero, so this is okay, one minus t, yeah, one minus t, that's okay, and then f one minus t, okay, anyway, and then we return f minus one restricted to, so what is the interval here? One, it's t, okay, I think this is okay, okay, well, anyway, this is a homotopy, okay, at time t you go just, you start in zero and you go to one minus t, okay, then you stay here and then you go back with f minus one, okay, that's it, you can have a look in the book and then you find the formulas, okay, if you don't like this, but they are sort of more technical to write down, so now we don't define a group, fundamental group, so the product of course has to be defined always in a group, okay, now we have a path and we have a product of paths, but we need that the second path starts in the endpoint of the first path, otherwise, so we take closed paths, okay, so let x be the definition, so x is a topological space and x zero is a fixed point in x, so pi one of x, x zero is equal to, so you take all closed paths, f from i to x such that f of zero is equal to f of one is equal to x zero, so this is a closed path, you say a closed path, same endpoint and initial point, a loop sometimes, okay, a loop, so f zero, f is a path of course here, f is a path, no sorry, I want to have a homotopy class of paths, so I take f, so homotopy class of f, where f from i to x is a path, a closed path, anyway it must be a path, continuous, the homotopy class of all loops which start and finish in this point, no, so the pictures, this is called, this is called the fundamental group of x with base point x zero, base at x zero, with base point x zero, this is the base point, we have to choose the base point, fix the base point, so this x zero is a fixed and then we have this, why is it a group now, what is the product, of course the product is, now the product is obviously fine, okay, so we have the product is our product of paths, so f times g, so these are closed passes now, this is f times g of course, this is always defined, that's the only remark here, always defined now, because all paths start and finish in the same point, okay, so it's always defined, it's associative, okay, that's the special case of lemma 1, by lemma 1 the product is, what is the unit element cx0, okay, cx0 is a unit element by lemma 2, this is lemma 2 and inverse, so f inverse element is the class of the inverse paths, f minus 1, so this is f minus 1 and this is lemma 3, okay, and so what else, so it's a group, that's it, we don't, that's the three things we need, no so, now this implies pi 1 x x0 is a group and this is the fundamental group of x with x0, that's the definition, example, pi 1 is a fundamental group of rn with some base point, whatever, that is 0, well that's the next question, the role of the base point, okay, does it depend on the base point or what happens with the base point, so let's take 0 here, is a trivial group, what means trivial group, so this means it's just one element and that's the unit element, okay, so cx0, that's the only element, okay, it's a trivial group, just the unit element, well proof, that's again here you have x0 and we have to prove that any path, any loop is homotopic to this constant loop, okay, that's the only homotopic class, so we have some f here in the fundamental group, right, represented by f, no, by the path, and now we have to prove f is homotopic to the constant map in the point 0, by some homotopy h, okay, now we are on rn, it's always straight line homotopy, but what will you do, so you just let these points go back, right, so what is the definition of this homotopy, so h of, the only thing we have f, no, f of, so what should I write, f of s is the original path, no, s times t or 1 minus t times s, doesn't matter too much, let me try this, 1 minus t times s, so that means at time t equal to 0, we have f, okay, so this means we have f, and at time t equal to 1, we have f of 0, always f of 0, which is x0, okay, this is the point, sorry, around x0, 0, here x0 is 0, okay, so this is a constant map in the point 0, this means that we have just one class, no, f, each class is, so this is a triangle, that's again straight line homotopy, okay, that's what happens here, so you see how it gets smaller and smaller, so we retract here and then we retract here and then it gets smaller and smaller at the ends, that's the homotopy, you see how it, the loop goes to one point, okay, so this is one example, not interesting because trivial fundamental loop, no, that's not very useful, so different passes, so definition, it's very formal, still, so definition, it's a formalism and then we have to find something which is not trivial, no, but then we have to work and then we have to have some applications, okay, then it might be useful, this is a formalism which is for the moment, it's not clear, obviously it's interesting or not, so what, let alpha be a pass, so we have two different, two different base points, okay, in x from x0 to x1 and then we define alpha hat from pi1 of x, x0 to pi1 of x, x1, so we have fundamental loop of the same space, but with different base points from maybe, okay, x0 and x1 and how do we find this, well alpha hat of a class F here, so we have again this picture, so we have x0, x1, we have alpha and now we have F here, right, and it's clear how to define this now, now we want something based here, okay, what do we do, we take alpha minus 1 times alpha, we go back, then we take F and then we go back again, so alpha, or if you want to write in different way, it's alpha minus 1 times F times alpha, that's the same, and this minus 1 you may also write up, okay, out, okay, we find alpha, then alpha, so the definition is, this is lemma and definition, definition, well anyway, then F, then alpha hat is an isomorphism of groups, this is an isomorphism of groups, so proof, well first you have to prove it's a homomorphism of groups, okay, so what does it mean, it means that alpha, so let's see what you have to prove, alpha hat of F times G, yes, this is alpha hat of F times alpha hat of G, okay, that is homomorphism of groups, so what is this side, this is alpha hat of F times G, this is a product, and alpha hat obviously fine, this is alpha minus 1 times, well I should have left this even, times F times G times alpha, that's the definition of alpha hat, no, first alpha minus 1, then the path, then alpha, this on the other hand, what is this, this is alpha minus 1 times F times alpha, sorry, I need more space, so this, I will go on here, okay, here's the definition, yeah, so see, well I should write small, so this is alpha minus 1 times F times alpha, this is this, that's the definition, times, and no, it's the same for G, alpha minus 1 times G times alpha, again the definition, what, two times the definition, the parenthesis in principle are these, okay, that's the first, that's the second, well alpha is not closed in general, no, no, alpha goes for, alpha is not closed, no, no, not closed, it might be closed, but in general it's not closed, that's not a closed path, well the class, the homotopy class of alpha, okay, of a non-closed path, that's redefined, we have to fix, okay, so I go on here, right on this side, working on this side, the lemma 1, 2 and 3 are proved for arbitrary passes, not for closed passes, okay, so it's associative to the product, for arbitrary classes, okay, the product must be defined, but then it's associative, if it's defined, okay, so I can put the parenthesis in a different way, as I want, by associativity, so this is alpha minus 1 times, and now times F, I want to do it this way, of course, this alpha alpha minus 1, okay, so times alpha times alpha minus 1, I'm not in a good shape here with the blackboard, and then we have what is the remaining g times alpha, maybe this one, okay, but this alpha alpha minus 1 by the lemma, what is this? This is equal to C, which point, alpha starts, x0, okay, that's lemma 3, and now I have something times C x0, okay, and then I can forget this C x0 by lemma 2, right, this is like, behaves like a unit element, so this means I can forget, okay, so this means I can forget this part, right, by lemma 1, by lemma 2 and lemma 3, and associativity, so we use the 3 lemmas, which are not only for closed passes, okay, they are for arbitrary classes, so what remains is alpha minus 1 times F times g times alpha, and this is now finally equal to this, no, that's the other side, this alpha cancels in the mid, okay, alpha and alpha minus 1, they cancel, it's not written nice because the blackboard, okay, but so using lemma, we use alpha 3, 1, 2, 3, 1 for associativity, we don't care about parentheses, 2 for this unit element, which we can forget, and 3 for inverse, okay, so this is, so it's a homomorphism, so this implies alpha hat is a homomorphism of course, isomorphism, so now you can guess, isomorphism is inverse 1, okay, and what will be the inverse, isomorphism is, so what do you suggest, okay, and you can check this, I don't know what, okay, but the same, you use all this, okay, easy to check, I don't want to write that not also, because it's the same as this, okay, you just apply definition, definition, and then you cancel, but you can cancel, and then you see that you get what you want, okay, so that's it, that's the definition of alpha hat, and alpha hat is an, this is important, alpha hat is an isomorphism of course, so this means that if x is path connected, okay, so observation, if x is path, no we need path connected here, then for all pairs, for all x 0, x 1 in x, pi 1 of x, x 0 is always isomorphic, so this means isomorphism to pi 1, x x 1, it doesn't depend on the base part, the isomorphism okay, if it's path connected, they are all isomorphic, however there's a big, small problem here, the isomorphism in general is not canonical, it's not canonical, okay, in general you don't have a canonical isomorphism between, okay, the isomorphism is not canonical, what does it mean, natural, I don't know, it's not a natural isomorphism, natural or canonical, where is he saying, for me, it's not a canonical, that is it, in general it depends on the path alpha, okay, alpha other, it depends on the path alpha, if you take another path, in general you have a different isomorphism, okay, in general the isomorphism is not canonical, but it depends on the choice of the path alpha, that's what I say, but you have to believe this, of the path alpha, for the moment, I cannot prove this, I don't want to prove this, I'll say it's not canonical, okay, then you cannot identify these groups in a canonical way, okay, it's not the isomorphic, but you cannot say it's the same group, there's a famous example from linear algebra, where you have isomorphism not canonical and canonical isomorphism, somebody remember, remember what that is, between vectors, you have a vector space and then you have a canonical isomorphism with another vector space, you have isomorphism, in any case isomorphism for vector space means they have the same dimension, okay, but then you go one step further and then you get the canonical isomorphism, what is this, so you go to the dual, okay, if you have a vector space of finite dimension, v, okay, then you have the dual, v star, and this has the same dimension, so v is, so remark, just to give you a feeling what is canonical and not canonical, okay, so v has a vector space now, finite dimension, okay, finite dimension, then v star, the dual has the same dimension, okay, same dimension, the dual has the same dimension, if it's finite dimension, okay, so they are isomorphic, two vectors spaces are isomorphic, if and only if they have the same dimension, okay, so they are isomorphic, isomorphic, because they have the same dimension, but this isomorphism in general, you cannot choose in a canonical way, okay, you cannot identify this, okay, but what can you identify, what? Yes, exactly, so I go on one further step, I take the dual of the dual, no, this is the dual of the dual, okay, b by dual, I don't know English, b by, by dual, okay, how do you pronounce by, I suppose now, by dual, and now this and this isomorphic, you can identify, they are isomorphic in a canonical way, v isomorphic to v is a by dual, so here you have a canonical isomorphism, so you canonical isomorphism, here you have a canonical isomorphism, and then you can identify using this, okay, so then you say almost that it's the same, you don't distinguish, equal, they're not equal, but you can identify in a canonical way, so it's the same, okay, that you cannot do here, you need some choice here, you need the choice of the basis, and then you have the dual basis, and then you have the isomorphism, you choose another basis, then you get a different isomorphism, okay, so you don't know how to fix, okay, so now we're doing linear algebra, okay, what is the canonical isomorphism here, what, well that's not our theme, so you have, this is conceptually a little bit, the by dual, okay, you apply, so if you have something, how to call this, v in the by dual, okay, the dual of the dual, and you have a phi in the dual, then you have v, you can apply this, well let's forget this, I don't want to do linear algebra, I just want to say that isomorphisms and canonical isomorphism, okay, and the most famous example is that this is not canonical in general, but this is canonical, so you identify these, okay, this is the same, and here's the same, you have an isomorphism, but in general you cannot choose it in a canonical way, okay, because you have to choose this pass alpha, and if you choose another alpha, you have another isomorphism, so you cannot identify everything, okay, so this is a similar situation, okay, what is the next definition, well that's the way of talking, x is simply connected, x is pass, is pass connected, and the fundamental rule of x, x is trivial, and this does not depend on the choice of x zero, okay, because you have pass connected, okay, does not depend the triviality on the choice of x zero, so this is simply connected, pass connected and trivial fundamental, okay, simply connected is pass connected and trivial fundamental, if it's not pass connected, this is also maybe useful remark, if it's not pass connected, then let's say we have two pass components, okay, that's a remark, trivial remark, but sorry mark, to give you a feeling maybe, if x has, for example, has two pass components, so here's one pass component, here's the other one, so this is a, b, pass components a and b, okay, and now you choose the base point here, a zero in the one component, and then you choose the base point b zero, okay, then these two fundamental components are completely independent, then it's a fundamental group of, then pi one of x with base point a zero, this by the way is then pi one of a with base point a zero, no, because you pass in x which starts here, never goes to the other, okay, you don't have a pass from here to here, so also if you have a close pass here, you cannot pass to this side, okay, because otherwise you have to pass from here to here, okay, then pi one of x a and pi one of x b zero, this is pi one of b b zero are completely independent, okay, are independent, one has nothing to do with the other, independent, so this means one has absolutely nothing to do with the other, okay, so you cannot suppose that, we have to prove non-trivial fundamental groups, but if we have, then we take a space with one fundamental group and here a space with another fundamental group, okay, then you take the disjoint union and then it depends on the base point, okay, so it's useful for, it does not depend if it's pass connected, pass connected is important, okay, if it's pass connected then the fundamental group up to isomorphism is independent of the base point, okay, now to finish all this, so here's a last exercise, an x exercise, exercise three, if x is simply connected, this is all easy, if x is simply connected, any two paths with the same initial point and also the same final point are pass-homotopic, then any two paths with the same initial point and the same final point also are pass-homotopic, are homotopic, pass-homotopic, so we have two passes, now we have two passes here which have the same initial point and the same final point, okay, F and G, and if it's simply connected with this definition here, then any two passes, these are not closed now, okay, F and G, not closed, it doesn't matter, okay, this is exercise three from next week, so to finish all this now I have to, this is a functor, okay fundamental group, so it goes on with these formalities, but well, of course I would like to see a no trivial fundamental group, but maybe I should write this theory which is the first theory with the proof but not now, pi one of s one, this is a unit sphere, okay, with any base point x zero, what would you, so this is s one, here's a base point and now what would you suspect, what is this fundamental group, yes, and why is that, so you have a path, okay, and you start running along, okay, and what you count is the number of, how many times, the number, the rotation, okay, and this is this integer, okay, maybe this is one, this is minus one, this is two, okay, this is minus two, and so on, however a path can be really very very complicated, no, so this is not trivial, so it goes in this direction, then in this direction, and in this direction, then it returns, and that's a very strange thing, okay, so how do you find the rotation number, you have doing complex analysis, no, there's also some kind of rotation number, okay, by some interval or something, right, now we are coming to this anthropology, so this is, we prove, okay, but not today, tomorrow, not tomorrow, whenever, next week, so definition, other definition, I want to factor, let h from x0, from x to y be continuous, we have a continuous map, so it has to preserve base points, so I choose base points x0 and base points y0, so define h star, the induced map, from pi1 of xx0 to pi1 of yy0, I want to define a map, okay, so how to define h star of f, so what do we suggest, we have a path in x, and we apply h, and we have a path in, so we take f times h, it's a homomorphism of groups, it's a homomorphism of groups, the homomorphism induced by h, okay, the homomorphism induced by a continuous map, well the proof is, this is, it's a homomorphism, first it's well defined, okay, maybe you think, well I use representative, okay, h is well defined, that's maybe the first even, it's well defined, but why is it well defined, so if f is homotopic to g, to f prime, okay, then fh, which is this definition, okay, and f prime h, which is, there should be homotopic, also, it doesn't depend on the representatives, okay, and what is the homotopic, so we have homotopy here, what is this homotopy, homotopy, h, very clear, now this is always h, okay, so you just compose everything with h, this means it's well defined, this is a homotopy between these two, okay, so it's well defined, that is homomorphism, h star of, homomorphism of groups, so f times g is equal to what, h star of f times h star of g, that's what we have to prove, right, so let's work on this, so this is what, h star of f is hf, h star of g is, first g then h, that's the definition, two times, okay, and here we have the product, and I say this is equal to h times f, of course, this is equal to hf, and then hg, what is the definition, no, you take representatives and you take them, but this is equal to, and now you see that this is h times f times g, also without homotopy, okay, so this is a triviality, okay, h times f times h times g, composition, composition, composition, this is the product, this is the composition, okay, it's equal to h, composition f times g, they're equal even, without the parenthesis, okay, even without the parenthesis, they're equal, here you take first the product, and then you compose with h, here you first compose with h both, and then take the product, the result is the same, so this is equal to this, and this is of course this, h star from, this is by definition, okay, so they are equal, yes, it's a homomorphism of groups, this is a group homomorphism, what else, it's a homomorphism of groups, so that's what we say, okay, so it's a homomorphism of groups, and now we have the factorial properties, factorial properties of the fundamental group, so we have this situation, we have h, we have mapping from, so this is the first one, we have continuous map f from x to y, and g from y to z, we have the composition now, and we have the base points, f of x0 is y0, and g of y0 is z0, okay, the base points are preserved, and then f, first f and then g star, the induced map is equal to f star, so this says that the induced map of a composition is the composition of the induced maps, okay, right, the induced map of a composition is the composition of the induced maps, okay, that's the first property, very simple, and the second property is even more simple, that says if you take the identity map of a space, okay, identity map of x, so this is the identity map, this means every point goes to itself, okay, the identity of x star, the induced map is the identity map of the fundamental group, okay, pi1 xx0, this is the second function, these are the two function points, the second one is trivial, if you take a path and you compose with the identity map of x, you have the same path, okay, so it's the identity of this, if you take a path and compose with the identity h before, okay, then you don't change the path, no, you have the same path again, so it's the identity also of the fundamental group, okay, and this, well, proof, how much time do I have, 10 minutes, proof, well, it's also trivial, no, what is it really, so g times f star of the shape, no, the path was always f, no, now we have a continuous map f here, okay, and g, so now I have to change notation, the path, w, okay, right, that's the path, these, sorry, are continuous maps, of course, I forgot, so these are continuous maps, everything is continuous here, so how is this defined, this is w and then we compose with this, no, that's the definition of the induced map, however, the composition of maps is associative, no, not of classes, of maps, so this is wf and then g, in fact, this is equal to this, without the parenthesis also, okay, we don't need to parenthesis, so what is this, this is g star of f times w, what, by the definition of g star and this is g star of, so this is equal to g star of f star of w, by the definition of f star, and so this is first f star and then g star of w, and that's what we want to prove, ah yes, now, I started with w, okay, and this implies for each w, okay, so this means this map and this map are the same, so this implies, for each argument they are the same, so this is equal to this f star, since for each argument they are the same, they are the same, two maps are the same if for each argument they have the same value, okay, so this is i, two i is even more trivial, now I have still a little bit of time, but this is, so identity map of x of w, how is this defined, this is w identity map of x, okay, that's the definition, but the identity map doesn't change anything, so it's w again, okay, so this means that, sorry, identity map star of, so this means identity map x, induced map is the identity of a map of pi 1 of x, okay, if you compose with identity nothing changes in other words, okay, if you compose with identity nothing changes, and in the first one, if you compose with the composition you can do it step by step, okay, by associativity of composition, right, so these are almost trivial things, okay, and nevertheless they are important, these two factorial properties, so I'll give the last thing for today, yes, I give, so these are the two factorial properties, the composition, the induced map of a composition is equal to the composition of the induced maps, and the induced map of the identity map of a space is the identity map of its fundamental, okay, very simple stuff, however now we have the first short application, so theory, if f from x to y is a homomorphism, so we have, I need base points always, this f of x0 is equal to y0, if you have homomorphism, then what you suspect is that the induced map from pi 1 x, x0 to pi 1 y, y0, they have to preserve base points, okay, it's an isomorphism, okay, a homomorphism induces, a continuous map induces a homomorphism, a homeomorphism, homeomorphism induces an isomorphism, okay, proof, very simple, so what means homeomorphism, we have an inverse continuous map, okay, let f g from y to x be the inverse map g of f, also continuous, that's not automatic, okay, also continuous, by definition, this is definition, for some spaces we don't, we have a bijection and then the inverse map is also prove what spaces, compact spaces, but for general spaces the inverse of a bijective continuous map is not continuous, okay, but if it's compact, yes, so we have to ask, this inverse map of f, also continuous by definition, okay, so this means that first f and then g is what, is the identity map of x and first g and then f is the identity map of y, right, and now it applies to, so this implies first f and then g star, this is trivial, is equal to the identity map of x star, now if the two maps are equal, the induced maps are equal, right, but now this is f star and this is this, this is the property, okay, the induced map of a composition is the composition of the induced map and the induced map of the identity map is the identity of the fundamental, this is a star, okay, similarly, now you do the other one, you have g star and then f star is the identity map of pi1 of y, y0 and that means that f star is an isomorphism with inverse g star, okay, so this says f star is an isomorphism with inverse isomorphism g star, so in some sense you can write this way if you want, you say f star minus one is f minus one star, no, the inverse map of f star is the, okay, the inverse, okay this, the inverse map of f star is, well whatever, okay, so last remark, so this means, so last remark, if x and y are path connected, so this means the fundamental group does not depend on the base points, okay, if they are path connected and pi1 of x, x0 and I take any base point, it's not isomorphic to pi1 of y, y0, then x and y are not homeomorphic, okay, then x and y are not homeomorphic, they're no homeomorphism, in the other say, if these, if x and y are homeomorphic, then the fundamental groups are heteromorphic, that's the same, right, if they are homeomorphic then the fundamental groups are isomorphic, if the fundamental groups are not isomorphic then the spaces are not, so in this sense a fundamental group is an algebraic invariant, is a topological invariant, okay, first example We have topological properties, no? If you have two spaces, one is hostile, the other is not, they are not homomorphic, OK? If you have two spaces, passively, no, and the fundamental groups are not the same, isomorphic, they are not homomorphic, OK? So this is the first example. So the fundamental group is an example, an algebraic invariant, a topological invariant. Not property, no, invariant, OK? Invariant means something algebraic, OK? So the fundamental group is an example of a topological invariant for pass-connected spaces, of pass-connected spaces, OK? For pass-connected spaces. So compare this topological property, OK? Houst of first count to the second count of a pass-connected collective, OK? Now we have the first example of a topological, not property, but invariant, which is a group, the fundamental group, OK? And there are many invariants. I mean, the next one is homology of all time, co-homology, homology. These are all topological invariants, OK? Homology, co-homology. And maybe numbers, but that's more difficult to define. Euler numbers, OK? That's a number, OK? But these are algebraic. So topological invariant of algebraic type, OK? Not a property, it's something algebraic, no. In this sense, fundamental group is a functor. So that's the last. Well, fundamental group, fundamental group. The fundamental group is a functor. So fundamental group means pi 1, x, x, 0, OK? That's a group. But also this h star, OK? The maps, we need also the maps. Induced maps h star, both, OK? The fundamental is a functor from the category. Well, that's a way to say something. It's from the category of topological spaces with base points to the category of groups. From the category of topology with base points and continuous base points preserving maps, OK? And continuous base point preserving maps to the category of what? Groups and group homomorphisms. So the category of groups and group homomorphisms. That's a way to say it, and group homomorphisms. Or if we have a continuous map x, y, h, h continues and preserves the base point, then we continue h of x0 is y0, base point preserving we need. Then this gives me h star in the other category. And this goes from pi 1, x, x, 0 to pi 1, y. So we go from one category to the other category. The category of topological spaces with base points and continuous map to the category of groups and group homomorphisms, OK? And if this is a homomorphism, then this is an isomorphism. This is just a way of talking. That's not so important. But it's more important what we will do next week. Because now we have to, this is a formalism, right? If you say it's important, but it's also boring maybe because we have no application, right? We just define something and define. Next week, we have to compute some group. The first one is of this circle, OK? And then give some nice applications. So then we see if that is useful, OK? Fundamental, OK? This is, up to this point, it might be that all fundamental groups are trivial now, for example. Then we have to prove first something, OK? If all would be trivial, that's nothing. Then we can forget, OK, fundamental group. But that's not the case. And the last remark, I never said that the fundamental group is abelian, OK? This in general is not abelian. It's a general group, OK? That makes it difficult, fundamental group. Abelian groups are easy, finitely generated. Then they are easy. General groups are very difficult, OK? Non-abelian groups can be very difficult. So we will see examples, OK? In the next two weeks, there are two weeks left. OK.