 Okay. My name is Fernando Rodriguez Villegas. You probably know that. This is the beginning of algebraic topology. So I'll be given five lectures and my colleague Lothar and Götze will be given the other five. So I'll be writing in this paper that gets projected up there in this sheets. We'll scan them and make them available to you if you want to later look at the lectures. So you're welcome to take notes, but also whatever I write here, you will have a chance to see later. Okay. So what we're going to do is today in next class is discuss the fundamental group of a space, of a topological space. And as a way of motivation, I'll start with some concepts of multivariable calculus. So first of all, what is algebraic topology? Kind of the name says quite a bit. We will associate to spaces, algebraic objects, for example, groups, rings, things of nature, that will allow us to, for example, distinguish a torus from a sphere. So these two topological spaces, I think it should be visible from the drawing that they're not homeomorphic, but to actually prove it in a rigorous way, require some tools. And this algebraic topology will give us many such tools. The fundamental group is an invariant that we're going to associate to a topological space, and it has to do with loops. So to say it intuitively, on this torus we have, for example, this loop that goes around, if you think of this as a tire of a bicycle, you can tie it around and then it seems to be, intuitively, clear that this path cannot be moved around and may disappear. Whereas if you do something similar and try to tie a string around the sphere, you can, again, totally see that you can pull it out and make it disappear. So this loop is not really attached, not tied to the sphere, whereas this loop it is. So what will happen is that the fundamental group of the sphere is trivial, and the fundamental group of the torus is definitely not. So that's sort of, that's just to give you a sense of where we're going. So we're going to start by discussing paths, and I won't be very good at giving you all the precise details. On top of the description that should be on the website about what is the bibliography, I will also be following this book by Fulton, which I recommend you have a look, Algebraic Topology, William Fulton. So he starts with reminder or calculus, and he gives a lot more details than I will. So I will sort of try to make a quick summary. So if we have a path in the plane, by that I mean a map from, say, the interval 0, 1 to R2, we would like this map to, at the very least be continuous, and let me just say smooth in the 0 interval. And one, to do this precisely you also have to worry about what happens at the edges, but I'll skip the details. I want just to give you a sense of what we're going to do in comparison to things I hope you've seen. You must all have seen some form of multivariable calculus before, right? So when you have in multivariable calculus one of the things that you do is you consider line integrals. So if we have an expression of the form p of x, y dx plus q of x, y dy, you can, in a path, you can make sense of the integral of the path of the, this object which is a differential one form on a path. It's defined to be the integral from 0 to 1. You basically replace the x by its parametric form given by what the path is. So the path here for a value of t has a coordinate x of t and y of t. You've seen, you all seen this before? Yeah? You do physics, you, all kinds of things you, you, this is a very fundamental concept. And if you assume the functions p and q are continuous, you have a continuous integrand, then the, the integral makes sense. And one of the things that, that is crucial is that if it happens to be the case that your differential form omega has the form like this, which we abbreviate by saying that this is df for some smooth function f, then what happens? It's a question. So I, but more precisely, the integral doesn't depend on the path as one answer. It only depends on the beginning and end points, but more precisely, yeah, that's also a special case that the, he's saying that the integral over a closed path is zero, but I want a, I want a precise answer. I want, I'm looking for an answer to what is this? If we have such a omega on a path gamma, he's saying is omega of gamma zero, gamma one minus omega gamma zero. That's almost right. Just what is, there's something behind this you know very well, right? What is it that will give us what the answer is? What's this called? Something, something of calculus. Not yet. There's much simpler like this. We're doing computing the integral of something that is d of something, right? It's the fundamental theorem of calculus. If you integrate the derivative of function, you get the function back, right? This is just done in sort of a, is done with a path form, but if you plug in into this definition of the line integral, this expression for f. In other words, if you plug in the fact that p is partial of f with respect to x and q is partial of f with respect to y and you write it out, you'll see that you can apply the fundamental theorem of calculus straight away and you'll find that this is f of the end point minus f of the initial point. And as he was saying, if the end point and the initial point happen to be the same, then this number will be zero, but otherwise it's whatever it is. But the point is, as you had said at the beginning, in particular this integral only depends on the end points, correct? Not on the actual path, but this is the case for this particular kind of differential form omega, which is called exact. Okay? Now, so I'm going to introduce or discuss a particular kind of differential of this sort. So this discussion, I just talked about R2, but we can also just concentrate on any particular open set of R2. So let's consider the open set U to be the right hand, right plane in R2. So let's look at the open set where x is positive. So in the usual coordinates x, y. Okay? And in this open set, so this is x and this is y, I'm looking at x positive. If we have a point of coordinates x, y here, I want to look at the angle measured counterclockwise from the x axis. Okay? So let's look at that function angle. And my points are in the right plane, so they can go from here to there. Right? I don't go past. So that theta angle is a perfectly well-defined function on my open set. So it goes to, let's say, the real numbers is the function theta is the angle. Okay? So it's going to go more precisely between minus pi halves to pi halves. And if you use the standard trigonometry, you can write it down as the inverse tangent of y over x. Okay? Now, I want to compute d of theta with the notation that we have before. So I want to find what is the partial derivative with respect to x and the partial derivative with respect to y. Anybody know? No? Okay, I'll write it down in the, given that we have limited time, but I'll suggest you actually do this calculation, which is not hard. Just out of curiosity, how many of you have seen complex analysis already? Some of you have. Right? So I'm doing things over thinking of R2, but many of these things have, in fact, a simpler description in terms of complex variables. This would be dz over z. No, it's taken. Okay. Now, look at what happened. We started with an open set u, which x is positive. We computed this d of this function, which is suddenly angled. This is a smooth function. We get a smooth differential form. And now the thing to note, the d theta is, I'm going to call it omega. Omega is well-defined where? Where does this expression make sense and give us a differentiable one form? Everywhere except 0, 0. So this is well-defined in R2 minus 0, 0. So now we have a differential form that's defined everywhere except the origin. And in particular, if we take gamma to be the path e to the 2 pi i t, or what's the same cosine of 2 pi t, sine of 2 pi t. So we parameterize in the circle in standard form, starting at this point and going around once with a 0 in the middle. What is the integral of omega on this path? Yeah, it's rotational. Something is rotational. We're moving in a circle. This is our path. And we integrating on that path going around the circle counterclockwise once, this particular differential form. And it has an answer that you must have seen before. There's no complex numbers except I did that as a shorthand. When you do the integral, you have a real differential form and you integrate on a real path. The answer is a real number. So you're off. He said 2 pi i. The answer is 2 pi. I mean, all of this I'm saying is multivariate calculus and it shouldn't be hard to do. I suggest you do. I'm just going quickly because I want to get, this is sort of as a motivation for what we're going to be doing. I don't want to spend a lot of time on it. So why does this not contradict what we said before? So if you recall what we said before, let me bring it up again. If our differential form happens to be d of f for some smooth function, then the integral doesn't depend on the actual path. It only depends on the values of the function at the endpoints. In particular, the endpoints are the same. In other words, if the path is closed, the answer is 0. The circle is certainly a closed path. It starts and ends at the same spot. Why is it that I integrate this and I don't get 0? Because omega, this should tell us omega is not the f of a function in all of r2 minus the origin. It is d of something, as calculated here, but in this particular open set. In fact, it's going to be such in any open set as long as it doesn't completely enclose the origin. So this tells us that, so there's another thing to observe here is that the integral, in fact, if I multiply this by r, so that I have a radius circle of radius r, the integral is also 2 pi. So what I'm trying to get at is that for one thing, the integral not only doesn't depend on the path, but this is true as long as we can sort of deform the path in a continuous way. So for example, if you have something of radius 1 and you move and you change the radius, you can think of this as a deformation of the circle into the bigger one. And that's one of the features of this story is that paths don't really play so much of a role, but what it plays a role is these paths up to the continuous deformations of them. And so what the fundamental group is going to do is capture precisely the nature of paths up to the continuous deformations. But this that we are discussing is something that is happening in the world of things that are being smooth. What we're going to do is transition to something where we just assume that functions are continuous. So the story about the fundamental group is about topological spaces of completely arbitrary with no smoothness involved. So we're going to repeat some of these arguments, but done completely in the setting of continuous functions. So just to continue with these thoughts, what happens if I take the path to be gamma n, which is e to the 2 pi n t, which is cosine 2 pi n t, cosine 2 pi n t. What path is this? It goes n times around the circle. What happens if I integrate the same differential form on this path? 2 pi n. So what this is going to lead to is the following. So let me say it in words and we'll eventually prove or at least give some of the details of the proof is that any path on the plane minus zero that starts at a point say on the circle one. So this is the origin. Any path that you draw on the plane avoiding zero will have a winding number. You will be able to measure how many times it goes around the zero. And what will this translate into is that, let me call this x. So I don't have to write it all the time. The fundamental group of this space is asomorphic to the integers. So every path will have a certain number of turns around the origin, which will be described by an integer. An integer which is positive if you go counterclockwise a number of times and negative if you go the opposite direction. And up to the deformations which I'll make precise in a few seconds. Every path has a attach to it an integer. Which is exactly the integer that measures how many times it goes around the origin. And so up to this deformation of paths every path looks like the one we just drew gamma n for some n. So any path no matter how crazy it is can be continuously deformed to be one of the simple ones that go around the circle in one direction or the other a certain number of times. So this is one of the fundamental examples that we're going to discuss in more detail. But this is one way to describe to see what the fundamental group can do for us as an invariant of spaces. The same thing applies for x the unit circle because any path that you have in the plane minus the origin you can shrink it to fit into the unit circle without changing it's the nature of what this measures. Well the technical term is that is homotopic to a path on the unit circle. And so what we have here is an first example of a case where the fundamental group is not trivial. Alright so this was meant as a motivation and now I'll try to be a bit more precise. There are many definitions and many little things to check to do to progress with this theory. So I'll prove a selective few and I'll leave some for you to as an exercise but you also suggest you actually read the book and do the steps as indicated in the for example the book of hatchers which you can download online. You know that the book you can get free online. It's one of those tedious things that one has to do once in your lifetime. This is the first time you see in a fundamental group this is the time to do it. Of course unless you have to teach it in case you have to do it again. Okay so now back to something a bit more formal. So x is a topological space. A path will be simply a continuous map from the interval 0 1 to x. Okay and notice that I dropped the condition that anything about smoothness because x has no smoothness is intrinsically any sense of smoothness. It's simply a topological space. Okay so I'm going to call i to be the interval 0 1 again so we don't have to write the many symbols so gamma is a map from i to x. Actually I'll take it back and as much as possible I'll try to follow hatches notation so you can compare when you if you read it so I'll call that f. Okay so we say that two paths f 0 and f 1 are homotopic so now we're getting into the fine detail of what a deformation means if there is a homotopy between them. So what does that mean? We want a function capital F from i cross i to the space which is continuous and it has the following properties. So for a hatcher the parameter on the path is s which sort of is the kind of thing you have when you parameterize by arc length if you are in in Rn. So at so t let's think of t as time so what we're going to think of maybe I'll do the picture first. We have a starting point which is going to be fixed at all times. We have some path f 0 and some path f 1 and the homotopy will give us a deformation so think of this as a movie the starts of s 0 and as time progresses from 0 to 1 ends up in f 1. Okay so at any given time we'll have some path that goes from x 0 to x 1. Okay and so what I want to we'll write down are precisely those statements so we're going to think of t as time t at time 0 we have f 0 and at times 1 we have f 1 and the fundamental thing that we want is that at any given time the path has initial point x 0 and final point x 1. So at times t equals 0 we want to have f 0 at times t equals to 1 we want to have f 1 and this will not be such a great notion we also as I said want to fix the fact that at for all time the the initial point is x 0 and the final point is x 1. Is that clear? So this is a precise way to define what it means that f 0 and f 1 can be sort of the form from 1 to the other the existence of this function capital F this homotopy is the precise definition of what it means sort of deformation but this is sort of done in an abstract way this space is completely x is a completely abstract topological space. So if you want to think of i times i as a square which is helpful for for some of the things we're going to do with this notion in in a little bit and you will want to do that in your in the homework. So think of this axis as s and this axis axis as t so we start at time 0 with f 0 so as you move s along this path you see f 0 and as you move up in time and you get to time t equals to 1 you move over on this segment and you see f 1 at at all times you see x 0 here and x 1 there so if I take some time t like a half you might see this curve that I had dotted there okay so this is a formalizing the idea of a deformation of a path which as we were saying before it's a natural notion for example in multivariable calculus or just calculus in the plane and this notion of deformation is important because typically things like line integrals don't are not sensitive to deformations of paths so the path itself is not that important what is important is its class up to homotopy so the next thing is to precisely say that homotopy of paths is an equivalence relation so that it makes sense to talk about the class the homotopy class of a path okay so let's prove this we need to prove the conditions of equivalence so what's the first thing we need to prove that a path is equivalent to itself okay so this is the kind of thing that I saying is sort of a tedious work that hopefully you'll see that it's worth our time because it will pile up and create a fairly powerful theory so how do you what is a homotopy that goes from a path f to itself just you don't do anything at all times you have f okay so that's easy so I won't even write it if f0 is equivalent to f1 already I'm abusing notation because I'm writing that is equivalent we still don't know but anyway so there's a homotopy that links f0 with f1 what is the homotopy that links f1 with f0 you reverse the order and just reverse time and you start a one and you finish a zero okay so I won't also write that either and this one the next one requires a bit of more work so let's see how we can do that and as I said I'm gonna do do this one to give you the very a little bit of a hint of the kinds of arguments that one does for many of the things that we'll do later which I'm going to skip the precise details of okay so a picture always helps in this business so here we have f0 here we have f1 and here we have f2 okay so there is a deformation that goes from f0 to f1 and then it's a deformation that goes from f1 for f2 and now we want to find one deformation that simply goes from f0 to f2 okay so let's take a little time to do this carefully and then after that I won't be as careful as that so what is the actual homotopy that we should use by the way how to use it does that this and I think is a useful so we have this capital F function and it's convenient to think of the capital F function as a collection of f sub t's which are the path at time t okay and this notation is consistent when t is zero you get f0 and t is one you get f1 okay so f and a half is this for example this dotted line that we had before so we have so then I should be a little bit more careful and maybe what I'll do yeah this let me do this let's say g0 equals to g1 is homotopic to g1 so we have a homotopy capital F which gives us as sort of a family of paths f of sub t and we have a family of path g sub t so what I want to construct now is a family of paths that go from f0 times t equals to zero and g1 at times t equals one okay so how do we do that so I want an h sub t such that h of zero is f0 and h of one is g1 okay so what do I do okay right so basically what we want to do is put these two things two families two together one after the other one okay so let me write that down as was suggested we're going to define a sub t in two pieces from time t equals one half zero to one half we use f sub t but we want to go twice as fast so by the time we go to the middle we are already done okay so we want to have f of two t right this will tell us that at zero we have f0 and at one half we actually have f1 correct because we are going twice as fast and between a half and one we want to start at g0 and end with g1 so if you think about that is what he was saying we do to t minus one if I'm not mistaken right so t equals a half gives us g0 and t equals one gives us g1 and this is the homotopy nobody complains am I done no I'm not done what's to this hb besides doing that it starts at at the right place and ends at the right place it should be continuous otherwise is this completely useless okay so how do we know this function is continuous was that pasting them you know how to do that okay good that's what I wanted to do so this is the kind of thing you have to do okay you will have to find various homotopies that do the kind of thing you want to prove these results but of course you have to make sure you done it right and that the resulting function is continuous and I won't actually every time point this out or try to do it okay so the arguments are all fairly similar of what we want to do so I'll just account on you that you know how to do it and some of them will be in exercises that I'll give okay so now we have equivalences of paths any path determines a unique homotopy class it's an equivalence relation and now comes the fundamental idea is that we can multiply paths in fact we can multiply homotopy classes and for this to make sense we have a path going from x0 to x1 followed by a path that goes from x1 to x2 say f and g or again to be maybe better I'll write this y0 and this y1 if the path f ends where the path g begins then we can define f times g to be the concatenation of the paths and again we need to verify that we can do this in a continuous way in the sense that the path is a continuous map from the interval to the space so the intervals we could have chosen any any close interval that you liked but we stick to 0 1 so this is parameterized from 0 to 1 this is parameterized to 0 to 1 so to put them together we shrink again the interval to half to get the beginning to be f and then the second half to be g and then we put them one after the other and so what we're gonna say is that f times g is f of t now t sorry it should be s s is the variable that is moving in the parameterization and then after that sorry I guess I should have put f of 2s q and g of 2s minus 1 pass that and again you have to do a little thinking to convince yourself that this is a continuous map and moreover this wouldn't be necessarily that good of a notion because the path there's just too many paths homotopic classes are the thing we are after so what we want to verify is that the homotopic class of f times g depends only on the homotopic class of f and that of g so I'm so here this I'm defining to be homotopic class okay and so finally we get to to this thing we want in particular if f and g are loops so that is f of 0 equals f of 1 equals g of 0 equals g of 1 equals the same point x 0 then f times g so we define the product of the class of f times the class of g to be the class of f times g since the both all everybody starts and begins at the same point we can concatenate them in this construction before requires that one ends with the other one begins this gives us so the main point of this whole discussion is that this product gives the set of homotopic classes of loops at x 0 the structure of a group okay and this group is denoted by pi 1 of x comma x 0 and it's called the fundamental group and what I was saying before is that for example the fundamental group of the circle or what's the same the fundamental group of the plane minus the origin is isomorphic to the integers okay so we are going to prove this fact there's a few things that we need to verify we define an operation in homotopic classes we I'm claiming that our operation gives rights to a group structure so there's a couple of things to verify but let's do one simple thing first okay suppose x inside our n is convex okay what does that mean you know what it means sorry sorry I can't quite hear you right so what you saying is if you take two points in x x and y say and you join you take the segment that joins them it's which makes sense in our end that whole entire segment is included in your set x okay so if that's the case if f and 0 and f 1 are paths from so are from i to to x that and start and end in the same the same points then f 0 is homotopic to f 1 okay and why is that how would you define a homotopic between any two paths whatsoever so here's our convex set say in R2 not quite that sorry so this is our x so here are two paths right so we can simply take a linear homotopic we take t do I want to you do I want one minus t I want one minus t times f 0 plus t f 1 so at any value of s f 0 of s and f 1 of s are points in our set x this linear combination for any t gives you some point in between the two and the line between the two which by assumption because x is convex is a point of x correct and so this is a path in x that starts at f 0 starts of f 1 and what are the values of this at 0 and at 1 f t of 0 is 1 minus t times x 0 plus t times x 0 so that's x 0 and the same for one since the points are the same the segment between them is simply the point itself and so this is a homotopy and certainly a continuous function and so this is a homotopy between the two paths so in particular if x 0 happens to be equal to x 1 if we're talking about loops then every every loop in x is homotopic to any other if we talked about the homotopic classes of maps or loops being a forming a group what would be the identity of this group is a class we can what would be an element that the term is the class go ahead the there is a completely trivial loop which is the one that never leaves the starting point the constant loop so for example then f 0 or any f is homotopic to let me abbreviate by just saying the point or maybe more precisely x 0 x 0 as an abuse of the path that is always equal to x 0 okay so if given the statement and the fact that this particular element is the element the trivial element of the group what is the fundamental group of a convex set trivial everybody is homotopic to did to the trivial path so particular pi 1 of x x 0 is trivial okay so for example x equals the unit disk has trivial fundamental group and if you were listening keenly and you worried about these things you maybe should be complaining why do I call it the fundamental group well somewhat tacitly in this statement I'm saying the fundamental group is trivial but this is a little bit of a pedantic point but it's worth being careful what is necessary to define the fundamental group you're talking about a space and a base point you need to pick a base point to talk about loops they have to start and finish somewhere okay so a priori there isn't just one fundamental group there is as many fundamental groups as points that you could choose okay so let's be slightly careful about that so in many circumstances the base point doesn't play much of a role as we'll see in a second but you have to remember that is there okay the the invariant is not associated to just the space it's associated to the space together with a choice of a base point okay so before and we still we still left aside the proof the proof that this thing was a group but let me before we do that let me mention this statement so just to get it out of the way so if x is path connected what would that mean any two points can be joined by a path that is to say there's a continuous function from the interval 0 1 to the space that starts at x 0 start innocent one so if x is path connected then pi 1 of x x 0 is is this is unique up to isomorphism that's not a good way to say it but let me try and be more precise so more precisely if x 0 x 1 are two points of x then there is an isomorphism between these two groups the groups are physically distinct okay so there are two separate objects but they happen to be naturally isomorphic to each other okay and then in in algebra and many situations in math it's is worth being careful about that not because it can lead to easily to some confusion so how do we do this well we have x 0 and x 1 the the space we're assuming is path connected okay so there is a path that joins them and I want to define my I want to define a map from the fundamental group with base point x 1 to the fundamental group with base point x 0 so if I have a loop starting and ending at x 1 what would be the natural loop that you associate to it that has a base point x 0 well you go from with this path to x 1 then you follow the path and in the loop and you come back okay so what you do is you take a path you have a path gamma I went so path F and this is H then you first do H then you then you do F and then you come back and that come back is the inverse path which is defined to be the path that goes backwards okay and so this will give us the isomorphism between the two groups so with a some abuse of notation one can talk about the fundamental group because oftentimes you this is the kind of spaces you can look like of course if you have a space that looks like a circle over here union the disk if you pick your base point here you're gonna get z if you take the base point here you're gonna get trivial so you know you have that to kind of trivial trivial situations often you we're just gonna stick to one path connected component otherwise things are clear what what the trouble is okay so with that provides so one can speak of the the fundamental group okay so now let's go back to proving the proposition so we would have to prove that we have an identity element that the there is an inverse okay so I leave those to you so trivial element is the path that is always equal to x0 the inverse of a path F is not of the path of the homotopic class is the class of F inverse which as I said is defined to be the path going backwards and these two things of course need a proof but is the proofs are not difficult I'm gonna skip them and just give the the fact that the the product is associated so we have three paths we should see that we have a homotopy between F times G first followed by product with H and the product done in the other way and I'll sketch the proof of this fact and leave the other ones to you as I said they're all sort of very similar in spirit and they're worth doing at least once in your life okay so how do we do this well let's see they we want to write this homotopy so we want to write a homotopy that relates the left hand side with the right hand side so we do F times G first we take the intervals 0 to 1 and divide it by 2 and the first half would be F the second half would be G and now with that going to multiply that by H so we take this interval and divide it by 2 and then follow it by H so what will happen is that if we do it we look at what we're doing in the left hand side here we're gonna do F here we're gonna do G and then after that we're gonna do H right that's the sort of division and pasting of the interval that you do to do the product on the left hand side is that is that clear whereas on the on the other side we first do F all the way halfway and G and H like that so what we can do is simply take a linear map that takes the interval divided this way into the interval divided that way so what we want to do is that if we go from 0 to a quarter we actually want to stretch it so that it goes from 0 to a half and then from here to here it goes from a quarter to a half that's moving a step of one quarter we want to keep it to be a quarter but then this one that goes a step of a half we want to shrink it so that it goes as a step of a quarter so what we can do is simply write a linear map that does that from the interval to itself and so the picture is this we start with with this from and we want to reach on this side this okay so from 0 to a quarter we want to move like that so that when we get to the end of the quarter we're actually at the half then from here to here we want to move straight okay and then from here to there we want to move half as fast so that we get to one okay so this is typical of this type of game that often you the best way to approach this is to try to think in this terms and do a little picture and then you know you have to write it out it's a little tedious but this already is pretty much the proof okay and so we call this function phi okay and we just compose so take the path f times g times h compose with phi of t and this will give us the homotopy okay so again I'll skip the details you have to do convince yourself make sure that this is a continuous map etc okay so then so this is a very powerful fact that homotopy classes of paths can be multiplied and these multiplication turns into take turns the set of homotopy classes into a group so this is the paradigm of algebraic topology you start with a space and you attach to it some algebraic object like a group or a ring something of that nature and then you will like this object to allow you to for example differentiate two spaces or prove things understand things about the space you started with okay so when you have a construction of this sort that you start with a an object of one name some nature and you end up with an object of a different nature you sure you already know by now that you don't just want to convert objects you also want to convert maps so what we expect to see is really something called a functor something functorial so we have a map between two topological spaces that is continuous is to translate into a corresponding map between the groups okay so that's indeed what happens so this construction is completely functorial and you end up with a functorial map between topological spaces and groups okay so if you have fee so we have two two topological spaces x and y and a continuous map that takes x 0 to y 0 remember that you have a base point so it's not just the maps in this category that we are considering it's not just the map continuous maps of map space to another you also have to keep track of the base point so let's do it before we talk about loops about paths that go between two different points a priori so we have a map I so we have a map a path that goes from the interval so we have a map that goes from the interval to x then we can continue that with a map fee the we have to y to obtain a map from I to y and if it so this fee of sorry if f of 0 is x 0 and f of 1 is x 1 then by construction this new path f composed with fee takes 0 to y 0 and 1 to y 0 so and it suddenly continues it's a composition of two continuous functions excuse me I want thank you and this so it takes path to paths and this this is compatible with homotopies so it's again a simple matter to verify that if you have two homotopic maps f 0 and f 1 on going to x then after you compose with fee you're gonna get two homotopic maps going to y by composition with fee and so this leads to the fact that then fee gives the map go fee lower star that goes from the fundamental group x 0 to pi 1 of y y 0 so if x 0 is equal to x 1 and y 0 is equal to y 1 so in particular spaces that are homeomorphic will give rise to isomorphic fundamental groups again keep being careful with the base points so the fundamental group is a invariant let me be more precise the isomorphism class of the fundamental group is a invariant of the space and I don't want to completely start maybe I'll do a little bit of the non the first non trivial calculation that we'll do which is to in fact compute the fundamental group of the circle and prove why we somewhat argued before that that group is the group of integers and not only that but that the the integer that a homotopy class of paths in the circle is is its winding number okay so it's not just a completely abstract isomorphism is something we can understand what is the meaning of the that isomorphism okay so I want to mention an application assuming that we already we have already done this this proof that the fundamental group of the circle is z let's do a very simple application of the deal the fundamental group but it proves something that it would be complicated or difficult to prove and somewhat surprising if you didn't have these tools so this is a theorem due to Brower in where's the U in here 1910 so we have a continuous map from the disk to itself so D here is the unit disk in the plane okay so this is a continuous map then H has a fixed point in other words there is an X in D such that H of X is equal to X okay so that's kind of a surprising fact and we'll see that if follows relatively easy with the tools that we have at the moment and just as an illustration of the kind of things that one can do with these tools okay so this let me sketch how the proof goes this is explained carefully in the book so you can also look at looking at there so suppose it's not the case okay suppose that we never have that H of X is equal to X so let's suppose that H of X is not equal to X for all X in the disk so we're going to define a map from the disk to the unit circle as follows so it's better to see it through a picture so here's X say and here's H of X the value of the function and because we assume that H of X is not equal to X it makes sense to talk about the segment that joined them and I'll do that so I'll draw the line that goes from H X to X I mean the joints so in this direction and it hits the circle somewhere okay and we call that R of X okay so this map R is continuous I guess we would have to be a bit careful and to verify this but if you slightly move divide the X the function H is continuous so H of X is not far from it from what the original value and the sec so the whole segment is not it's a bit of a wavy hands argument but hopefully that should be enough what else can we say about this what happens if X is in the unit circle what should be the value of R X itself okay that I guess is part of doing this carefully because we have to define what it means the value but clearly the value of R if we want this to be continuous has to be that the value is itself X okay so what we have then is a map so we have a subspace topological subspace of the space D and a map back to S1 which is the identity when you compose okay so this such a map is called a retraction so you can think of this as sort of pushing the D down back to the subspace okay so you pull the subspace up and you then you know this kind of pushes it down keeping the subspace as it is unmoved and now I want to show that such a retraction cannot exist and let me say first in words and then we'll describe the proof the point is that if there were such a retraction we would get that the fundamental group of the circle is trivial because the fundamental group of D is trivial so in other words given that we have this map we're going to be able to retract any map any path in the circle using the fact that we have D or this puzzle and we know that that cannot be true because the fundamental group of the circle is not trivial again assuming that fact that which I haven't yet completely proved okay so let F0 be a loop in S1 and just pick your favorite base point at this point it doesn't really matter so in D there is a homotopy FT between F0 and say let's say base point X0 between that path and the and the constant loop X0 why is that so D is convex we talked about this already okay so if we allow ourselves to the path to well is sitting in S1 but if we view it as if it were sitting in and D and D then we can shrink it to a point within D okay I mean and we could write it out I mean we can write it out as we did before with just a linear homotopy so now we consider the composition of R with FT right R goes from D so we take this homotopy and we push it down to S1 what we're gonna get is a homotopy in S1 between that path and the point and we know that is not possible so the point is then that this homotopy is a homotopy between F0 and X0 in S1 right so R composed with F0 is F0 because because it's the identity restricted to S1 and R composed with the point is so that's a simple consequence of the fact that the disc has trivial fundamental group which allows us to retract any path to a point whereas I mean sort of shrink a path to a point whereas the circle we have been arguing is not does not have a trivial fundamental group okay so I start maybe a little bit with the proof of the of the fact that the circle has Z fundamental group but before we do this so so far we haven't done a whole lot I mean we have two examples either the fundamentalist reveal for example if the space is convex convex subset of Rn or we are claiming is Z if it's the circle so for all you know you can be thinking well this is kind of a boring invariant it's always a billion let's see two examples we have do you have I don't know you know you're back and very well so you may have seen a much of this and I'm repeating myself but just if you have seen it then don't answer but if you if you this is the first time you see in the fundamental group make a guess just give a suggestion for a space that has fundamental group that is not a billion what's that sounds like you answer too quickly maybe you knew this in advance or no okay he's saying you take the eight so if you take this space so the two circles joined at at the hip this will have fundamental group that is not a billion I mean we don't have all the tools to be able to show this yet but we will so but let's just argue intuitively I mean much of this requires sort of both you need to have a technical tools to be able to prove things but you also have intuition is important to get a sense of what is happening before you actually improve it so what will you think is the group the fundamental group of this space let's take that to be x0 that looks like a natural point you seem to be trying to say something yes Z plus Z he's suggesting that because the fundamental group of the circle is Z maybe where we have a Z maybe Z cross Z a Z plus Z that's a group any thoughts yeah well it will still be a billion but that's a guess we I'm assuming you haven't seen this and we're trying to and you know trying to understand this intuitively first so why do you say Z plus Z you we have at least two paths that look like we cannot shrink right if you go around one circle and we go another another circle that looks like paths that you know you cannot homotopy that to to be trivial right this is which is the case because each one separately that's what happens if we just had one circle so that looks natural but we have therefore looks like two really independent generators but in fact it's not C cross Z because they're in fact they do not commute okay but this will will need to to get into this so it's in fact the free group in two generators okay so there are two generators with no relation between them but that do not commute okay so in fact it will be a non-abillion group but again this is for later so let me just state what I should be starting with on the next lecture which is a fundamental fact that we needed to prove the fact that the fundamental group of the circle is Z but it's something that is in itself important okay so so excuse me I'll start with this and time will run out and and I'll start again one next time so we are now we're going to concentrate on the circle and trying to understand it so up an important fact about the circle is that you can view it as well we can the language we're going to use the language of covering so clearly the trouble and this is what we were trying we're struggling with when we started with circles and winding numbers and so on is the notional angle and we cannot we argued define a continuous angle all the way around right at some point the thing is going to not match so what we're going to do is embrace this problem and take all possible angles okay so instead of thinking of the circle we're going to think of a cylinder and so we are going to keep track of all the possible angles so this is I'm going to do just an intuitive discussion first so instead of trying to have a continuous function of angles we just going to so remember how many times we went around so instead of thinking of just the circle down here we're going to have an elex so what we are going to need is the fact that if you have let's just think of a simply the path that we talked about going around n times on the circle so what we're going to do is try to define an angle all through that path okay so think about is you start here you go around but instead of just keeping the angle you you also go up okay so by a time you turn you come around to the beginning you're not at the same spot anymore but you're on top of it you did one turn okay and then so if you if the path went around five times you will have five of these little turns on the ellipse another elex and you will end up exactly on top of the same spot you started with but five steps on top okay so this this map is going to be the kind of thing that will clarify for us this idea of winding number and which is the same ultimately as the statement that well as a more refined form of the statement that the fundamental group is z okay so the the map that we want then is the map that takes s to e to the 2 pi s in other words cosine 2 pi s sine 2 pi s so let's think about how this map what are the properties of this map so here think of s as the this height right so we have height zero we are here so if we simultaneously keep track of s and the turning we have you know this turn and this goes like this like that so this helix sort of unravels the circle so this will be height 1 height 2 so on height minus 1 so on so what can we say about this map I'm going to call it P so P what can we say about the inverse image of a point Z in the circle say this point here what does these image look like yeah it's in the circle so all we need is the angle yeah go ahead there was an answer so it's some value of s maybe I call it z zero some angle some z zero s zero then plus what what are the possible ambiguities to this this is the angle well sort of up to factor to P to pi is up to an integer so and also we can be a bit more precise if we take a little open set around this z zero what is the pre-image of that well they're going to be the pre-images look like this and there are all going to be little segments going up so the pre image let me call this subset u is at this joint union of v n with n and z so this would be v zero v 1 v 2 etc v minus 1 okay this is the disjoint union and if I look at P restricted to one of these subspaces these open subsets it maps to you and it does it in a homeomorphic way okay so this is a kind of a fancy way to say that we can define a perfectly one-to-one angle as long as we don't try to do that too far away okay and in fact is that's this statement but also that they're the ambiguity what four angles are integers up to of course a multiple of two pi which I am factoring it out so this type of map is what's called a covering map and what we'll see is that the fundamental group is absolutely tied with the idea of coverings and which is yet another angle to this story that in some sense is even more important than the case of the understanding of the fundamental group so this map this stretching out of the circle into this helix is what is going to allow us to prove in a sort of rigorous way that the fundamental group of the circle is the integers so basically let me say in words and I'll end there if you have a completely arbitrary path loop on the circle you will be able to unravel it and the ending point minus the beginning point whatever you want to start it's going to differ by an integer that integer is the thing you are trying to capture that integer will completely describe the homotopy class of this path all right we'll continue next time