 Okay, so are there any questions? I'll give you some exercises for you to work on at the end of the class, okay? So before we get into finally proving the statement about the fundamental group of the circle, let me let's recall the definition of covering, okay? Everybody happy with the definition of covering? Now, I forgot to ask or to discuss the simplest possible type of covering there is. Okay, so let's review the concept. We have a space X, a map P from another space Y. And each point down in X has a neighborhood that is evenly covered, which means that the pullback by the map P consists of sort of slices that look, let's say in words, basically the same as the one below or more precisely the map P is a homeomorphism. What's the simplest possible example of a covering which I fail to say? Identity, what do you mean by that? Okay, okay, so he's saying let's take this space Y to be the same space X and the map P to be the identity. That's kind of a pretty sort of simple kind of covering. I'm thinking it's something a little more elaborate, not a whole lot more. We can elaborate on that same idea but instead of just having one sort of think of the space as one big pancake below, instead of just having one copy on top, you can have many copies on top. Okay, sorry, I didn't interrupt. What was you gonna say in your shirt? That's all right, something like that or you have something else to add, no? Okay, so if you take, so for natural reasons, this is called a trivial covering. So you would have X, Y, P, and Y is X crossed with a set T, a completely arbitrary set T with the discrete topology. So that's a precise way to describe something like what we were saying in words. So here T, say for example, are the integers with just the discrete topology. So every point is isolated, so it's itself an open set. So we would have the map P, which is simply going down and mapping to the point in X corresponding. So this is kind of a boring type of covering, but it certainly is a covering. And with this notion, you can think of a actual covering as something that looks like this locally. So another way to say the same thing of a definition of covering is that if you take a small enough neighborhood of your space X below, and you look what happens on top of that set, and forget the rest, it looks like this. It looks like a trivial covering. And so another way to say what a covering is, is that it's locally a trivial covering. Okay, so if you concentrate on a small enough open set around a given point, it looks like a trivial covering. So this is a concept that appears in math, of course, in many different contexts where the notion of locally could be defined in some more abstract way. You often are interested in things that are locally trivial, but not overall trivial, and you want to understand the abstraction from how much they differ from being trivial. Okay, so back to the fundamental group of the circle. We finally can discuss it more precisely. So we take, let's say, the point one, and the map to Z, which we claim is an isomorphism, is the winding number, which we can define it. Although I didn't do it, sorry, I don't know why I put it till. I didn't actually do it, but you can find it in Fulton's book, for example. You can do this simply with calculus, but we also can do it more abstractly with our covering idea and with the lifting of paths that we discussed last time. So how do I define the winding number? Well, we know there is a lift of this map, of this path to the covering. So recall that we have R and S1 and the map P. And so any path in the space below can be lifted uniquely to a path upstairs in the space Y of the covering, if you fix where it begins. And the difference between where it ends and where it begins is in the case of discovering an integer. And this integer we call it the winding number. And this number doesn't depend on the path, the loop gamma itself. It only depends on the homotopy class, because if we had another path, gamma prime that is homotopic to gamma, they both lift two paths in the covering. But also does the homotopy between them. So the homotopy below lifts also. And so in particular, the ending points of the lifts are the same. So if you fix, for example, to fix ideas, you say gamma twiddle of 0 is 0, then whatever gamma twiddle ends will be uniquely determined by the homotopy class, not by the actual path that you used to represent it, okay? So this is a well-defined map or the concept of the number w of gamma is well-defined because of the lifting properties of coverings that we proved last time. So both the fact that there are lifts of paths, unique lifts of paths and unique lifts of homotopies, okay? So let's prove this statement. So already I elaborated a bit about why this map is itself well-defined, which is part of the proof. But let's say, and I think we argued this already, but just to sink this concept in better, let's convince ourselves that the map is subjective. So there is a path of any given winding number. And what do we say? How do we do see that? I give you an integer n. What is a path in the circle that has winding number n? It goes n times around the circle. We discussed this, so let's see how this goes. Say, let's call it omega n and is the path that goes from s. Let me just say e to the 2 pi i n s to save some ink. That's the same as cosine 2 pi n s and sine 2 pi n s. Now, why don't we actually compute completely explicitly because we can do it? What is the lift of this omega n? If anything else, to practice this concept and get it clear as clear as possible in our heads. So what is omega n? Twiddle, a lift to r. And just to fix ideas, let's just say that the omega n of 0 is 0. So our base point is the number 1 in the complex numbers in the circle. So a picture would be this. Let's say that this is our base point. And so the covering is from r to s1. And suddenly 0 is on top of 1 because it's e to the 2 pi i 0. So 0 is in the pre-image of 1 by the map p. And let's agree that our lifts start there. So what is this omega n? As we proved last time, there's a unique path that lifts omega n and starts at 0. So what is it? n times t. So let me try to keep my notation straight and call it s as I was doing in Hartzler-Lass. So it would be s goes to ns. So this is a path from the interval. This goes from the interval i to r. Certainly is a continuous map. And it projects down to omega n because in this case, projecting is taking the exponential e to pi i of the argument. And that's exactly what our path omega n is. And because this is a lift, we know it's the lift because we know it's a unique one such. OK, so in particular, omega n twiddle of 1 minus omega n twiddle of 0, which is 0, but I write it for completeness, is n minus 0. That is n. So this path omega n, indeed, as we knew all along, and we already mentioned, has winding number n. What we're doing here is proving things rigorously. And this is an example of this. And we've done it without calculus. I mean, this is now pure algebraic topology. Now, suppose you have a path, a homotopic class of loops in the fundamental group. And let's say that its winding number is n. So how do we see that omega is homotopic to omega n? This will show that the map that we have defined is injective. That any path, any class in pi 1 of the circle is accounted for by omega n and only one such n. So how do we see this? And let me do it slowly, because the arguments that we're going to use only use the properties of coverings and one property of the reels. The thing to focus here is that the way historically and the way we're learning mathematics is you learn calculus, you get used to certain things. And then slowly you abstract the notions that are relevant, and then you can expand your theory to include things where you don't have the real numbers, places where you don't have the crutches of being able to differentiate, for example. So let's just try to reason this not by using any calculus, but by using the tools of topology that we have so far. And so the same tools would then apply. And you'll see that the only properties that we need are one that the map is a covering and one very simple property of the real numbers. So let's consider a lift. So we know it exists. And let's just, again, fix ideas and say that it starts at 0. And we also have omega n starting at the same place. How can we see that gamma and omega n are homotopic? We lifted both paths. What happens with these paths upstairs? Same endpoints. Everybody agree with that? Why is that? Well, by definition, gamma lifts to something that ends at n. That's the winding number. And we just checked that omega n total ends at n. So they both start at 0 and end at n. So they're homotopic because r is simply connected, because r has trivial fundamental group. So all we're going to use of the real numbers is the fact that it's simply connected. Is that clear? So we use the covering map to lift both paths. We have chosen so that they start and end exactly the same places. And they are paths in a simply connected space. So that means that they're both going to be homotopic. So in fact, how do you prove that? We have a simply connected space and two maps that go from one point to another. Why are they homotopic? So what he's saying is that you take the first one going one way and take the other one coming back. And this will give you now a loop because they end at the same spot. And that loop because the space is simply connected contracts, it's a trivial, has a homotopic class of a point. And this will give you a homotopy between the two paths. You would have to fill in a few of the details about that. So we use the fact that we have a covering and the fact that the space upstairs is simply connected. That's all we're going to use of the topology of the situation. So we can apply exactly this argument in a similar situation. OK, so if you like concretely because we are in the reels, you can actually write down the homotopy, which we've done before. This is the linear homotopy. But that is not relevant, the fact that you have it explicitly or that is linear. OK, so because r is simply connected, omega twiddle is homotopic to omega n twiddle. And then applying the covering map down, we get the gamma not equal but homotopic. This is omega n. And so gamma and omega n are homotopic. OK, so that means that this map is subjectively so that. And it's also injective because any loop in the circle in the base is homotopic to one of these omega n's. And it has to be only one possible omega n because n is where the lift ends. If it ended at a different number m, then it cannot be homotopic to the one you started with. So the point is omega n and omega m are not homotopic if n and m are distinct because n and m are where the lifts finish. And if n and m are distinct, then they end at different places and their classes cannot be the same because the classes what we saw before determine where it ends determines what its class is. OK, so we would have to check that the map that we constructed is a homomorphism. But this will follow if we check this. And so how would you check that? Property of multiplication? OK, I think it's along the right lines. You're saying that if I understood correctly, you were saying you take the paths and you multiply them. Is that what you're saying? But you have to be careful. And this is a point that we may come back to. As it happens, s1 is a group with multiplication of complex numbers, for example, or just angles. So there are two things you could do with two paths. You can take the paths and actually multiply them in the group. And that will give you a third path. But this is something you cannot do if your space is not a group. So I think I'll postpone discussing that for later. But let's stick to the notion of multiplication of paths that we have discussed. I think is what you're referring to, but just to clarify it. So how would you do this? We didn't really do it in detail, and I'm asking you to do it in a homework. But basically, in words, what is it we do? We have two paths, and we want to multiply them, or rather the homotopy classes. But we start with a path from one class, and then in general, you go from one point to another. And the next path has to start where the other one ends in order to be able to multiply them, and then you basically follow it out. So you do that with this. And so what happens? Omega n, when you lift it, goes from 0 to n. And omega n goes from 0 to m. When you multiply, then what you're going to do is you're going to shift the second one. The first one goes from 0 to m. And then you follow that by going from 0 to n from that point on. And so where it's going to end? It's going to end in n plus m. But that we know means that its class is the class of omega n plus m. So I'll leave the details to be worked out by you. It's not difficult, but it's one of those things that I think we can save a little tediousness by moving forward. So this will show us that the map is a homomorphism, is bijective, and it will prove what we were saying. So just to summarize, what turned out to be useful facts that we used were one that R2S1 is a covering. We have a covering map. And two that R is simply connected. So this situation where you have a space and a cover of it where the top space is simply connected is a very useful situation. And you use it very much in the same way we just did. And what we can say is that R is called the universal cover of the circle. So it's a simply connected covering of the space. So that's a very important situation. And I don't know how much I will be able to fill out this statement. But if you know a bit of Galois theory and you like that point of view of algebras, this you should think of a field. And then the top thing is like the Galois closure. And the fundamental group will play the role of the Galois group of the algebraic closure over the field itself. But I'm not sure I'll be able to fill out many more details about this relation than just what I just said. OK, so that's good. We now have done something fairly substantial. That is a result that we talked about already and even in the first class. But now we actually have all the tools to do it, to prove it rigorously. And we did it with just algebraic topology, not with any other notions that don't easily extend like metric spaces or differentiation or anything relating to the real numbers. OK, so let's do some application. This is a fun application of the things that we know. And as Hasser says, typically we use algebra to inform topology. Now we're going to go back. We're going to use topology to prove something in algebra. So we're going to prove the fundamental theorem of algebra, which says that any polynomial of degree bigger than 0, so non-constant, has a root in the complex numbers. So I don't know how long ago you've seen this result. And you probably take it for granted, sort of incorporate it, and it just becomes part of what you know. You don't stop to think about it, but it needs a proof. You start with the real numbers. You may be fairly comfortable with the real numbers. Suddenly you add one thing, i, and then out of a sudden you get the roots of everything. i is the root of one polynomial. x squared plus 1. But you join one root of one polynomial, and suddenly every other polynomial there is also has a root. That's a pretty strong fact. So if the degree of p is bigger than 0, p has a root in c. OK, so let me first break this up into steps. The first step is sort of a result of independent interest, which is something that you may want to incorporate in the tools that you have, which will tell us a bound of where we are going to be sure there is a root. So let's fix ideas. Let's say the polynomial p of x is x to the n plus a1 x to the n minus 1, a1 x plus a0. So divide out the leading coefficient. So it's monic. It doesn't hurt any of the arguments. So n here is the degree of the polynomial. And so the first point I want to make is that if r is a number, real positive number, bigger than the absolute value of a1 up to a0 and bigger than 1, so bigger than the two numbers, bigger than the max. And then p does not vanish on the circle of radius of any point of norm bigger than r. So in symbols, it would be that p of x does not vanish on x bigger equal to r in the complex numbers. So this is a little bit of calculus. So let's take a point x with sitting in a circle of radius r in the origin, around the origin. Then a subvalue of x to the n is r to the n. r to the n is r times r to the n minus 1. The first factor by assumption is bigger than the absolute value of the sum of the a's. Sum of the absolute values of the a's. And the rn minus 1 is x to the n minus 1. But since r is bigger than 1, this xn to the n minus 1, we can distribute it like this. And so that means that if p of x was 0, we would get that r to the n is minus a1 x to the n minus 1, a1 x plus a0. And if you, this is x, if we take absolute values of this, we would get a contradiction. So r to the n is equal to the absolute value of this. And this is a contradiction. So this was assuming that we had a root on the circle exactly of radius r. But this is true for any r was anything bigger than that bound of the sum of the absolute values of the a's and 1. So this is true for all such circles. So this is good to know because usually we can think of we use this the other way around. So since you cannot vanish outside this disk and we are going to prove it vanishes somewhere, it has to vanish inside. So it will have a root inside. In fact, all the roots are going to be inside this big circle. If you ever wonder how big can the root of a polynomial be, if you have one in your hands, this is a simple-minded, not too bad, bound for where the zeroes of your polynomial must be. So that's 0.1. So now let's look at the path. Loop, to be more precise. Done. So we're going to have two loops. Let's look at the map from x to px over the absolute value of px. Actually, let me do, yeah. So I'll do this for 2.0, just p0 and p1. So p0, I'm taking it to be x to the n. p1 of x to be the rest of the polynomial that make up p. Oops, I think I've been writing the indices wrong. So please correct them if you wrote them. So I think I keep calling both of them a1. OK, so p is by definition p0 of x plus p1 of x. And if we assume that p1, let me go back a second, confused. Sorry, let me write this correctly. So p0 is what I wrote, but I want p1 just to be p. So let me write this again. So p0, I take it to be x to the n and p1 of x to be just p, our polynomial p. And let's have x move in the circle of radius r, the same r that we picked before. So we know that the polynomial does not vanish on that circle. So if we look at p0 of x over the absolute value of p0 of x, well, that's just x to the n is on the circle is x to the n divided by r to the n. The absolute value of x is r on that circle. And then so this will be our gamma 0. And our gamma 1 is going to be p of x divided by absolute value of p of x. And because we're assuming it doesn't vanish on the circle, well, we proved it doesn't vanish with our choice of r, then these paths are well-defined and gives us maps from the circle to the complex numbers. In fact, because we're divided by the absolute value, they map to the unit circle. So this is not quite the unit circle. It's a circle of radius r, but easily we can shrink it to scale it so that it is. So we get maps from S1 to S1. And these are loops with base point, say, x equals 1. So a loop, we said, is a continuous map from the interval 0, 1 to the space that ends where it begins. But that's the same as saying that we have a continuous map from the circle to the space where a given point in the circle maps to your chosen base point. So these are two loops in S1. Is that right? Are you happy with this? OK. So what is the path gamma 0? We've seen it before. So ignore the r. The r is just a scaling factor. Essentially, x goes to x to the n. So what kind of path does that give? What's the path in the unit circle given by the map x goes to x to the n if you think of x as a complex number in the unit circle? Come on, guys. We've seen it before. It goes around n times. So is the path gamma n? Omega n, right? And so second. OK. So gamma 0 is our friend omega n. And this gamma 1 is some other path. But we can consider p sub t to be x to the n plus t a1 x to the n minus 1. Sorry, a n minus 1 up to a1 x plus a0, right? So for t between 0 and 1. So this is a continuous map from the interval across the circle to the, well, a priori to the complex numbers. And p of 0 is when t is 0. So that's x to the n. And p of 1, that's when t is 1, is the sum of these two things. That's our polynomial p. So this is a map sort of homotopy between these two polynomial functions. But you can see by the same argument that we did before that p sub t does not vanish on the circle of radius r, the same r that we chose before. Just exactly the same inequality. If you stare at it, it will tell you that it also tells us that this polynomial does not vanish on that same circle. So that means that p sub t divided by the absolute value of p sub t, so the fine omega sub t to be p sub t of x divided by p sub t of x absolute value. So this will be a map from the circle of radius r, so s1 to s1. And it will be a homotopy between gamma 0 and gamma 1. So gamma 0 is omega n, and gamma 1 is this new map. Now, if you have seen complex analysis, you may recall something called the Ruches theorem. And the argument that we are doing is pretty much the same as Ruches theorem applied to the polynomial x to the n and the polynomial p. OK, finally, point 3. Now what I'm going to do is show that the path gamma 1 is actually homotopic to a point. And this will prove the theorem. Why is that? So we'll prove gamma 1 is homotopic to a point, to the base point, to the constant map equal to the base point. So suppose we do this, why would this imply the theorem? Well, what do we have? We had two paths, gamma 0 and gamma 1. So gamma 0 is homotopic to omega n. Gamma 1 is this path that was built out of the polynomial p, which we're trying to prove has a root. And now we are saying that, on one hand, omega 0 is homotopic to omega n. Gamma 0 is homotopic to gamma 1, which is so that. And now, in claiming that, on the other hand, gamma 1 is homotopic to a point. So the conclusion is that n has to be 0, which is to say the degree of the polynomial was 0. And that is to say the polynomial is a non-zero constant. Or to put it another way, if the degree is bigger than 0, this is a contradiction. And therefore, the polynomial must have a root. OK, so the last thing we need to do, so this implies that n must be 0. And this is a contradiction, because the degree was assumed to be bigger than 0. Now, so how do we show that we actually have a homotopy between gamma 1 and a point? So this is yet one last trick. So consider p, maybe I'll call it q sub t, to be the polynomial p sub p of tx. And so actually, maybe we'll do p times t rx. So if t goes from 0 to 1, OK, so think of our radius r circle. If this is x, we are going to be considered the values of the polynomial in smaller and smaller circles as t goes from 1 to 0. So if t is 1, we get actually, I think this is not what I want. What do I want? Let me just define it like this. 0 to r. And so if p is equal to 1, I think it was right. If t is equal to 1, I have q of 1x is p. And if t is equal to 0, I have p of 0. So this is a homotopy. And to look at things that are the final in the circle, I have to divide by the absolute value. So I can say because p does not vanish anywhere by assumption, if p doesn't vanish anywhere, then we can define the homotopy sigma sub t to be q sub t of x divided by the absolute value of qt of x. This is a well-defined homotopy of the path gamma 1, which is what we get if t is 1 and a point 1 for t equal to 0. So if t is 1, the polynomial is p of x. And we get p of x divided by the absolute value of p. That's how we define our path gamma 1. And if t is 0, q of 0 is the constant p0, which is assumed not to be 0, since p doesn't vanish anywhere by assumption. So we get this over that as a constant divided by the absolute value. Well, maybe it's a sign, so it's a point. I don't care what it is. And so what we get is that gamma 1 is indeed homotopic to a point. And the conclusion is then omega n is homotopic to gamma 1, which is homotopic to a point, which implies that n is 0. OK, so I think I. Is that clear? Is there questions of the argument? So we did it in three steps. First, we showed that the poinomic cannot vanish in a big enough circle. And that's a useful fact to know regardless of proving this theorem or not. And we have a precise description of the radius. Then we showed that we construct two paths by using the polynomials x to the n on one hand and p on the other. And we showed that the two paths constructed from these polynomials are homotopic, and the first one is omega n. Finally, the path constructed with the polynomial p, we showed it assuming it doesn't vanish anywhere. That is homotopic to a point. And the conclusion then is that the degree had to be 0. So I hope I didn't make too much of a mess of it, but it's a nice application of algebraic topology that we get as a conclusion from the fact that we know things about the fundamental group of the circle. We get the conclusion that the polynomial must have a root if it's not constant. OK, another thing I wanted to discuss is a bit more about the torus. So the two torus, we said that this is isomorphic to the Cartesian product of two circles. I argued that the fundamental group of the torus, some point is isomorphic to z cross z. And graphically, we can do this in two different ways. On one hand, we can look at the torus in the standard picture. And then we can describe a generators for the fundamental group that correspond to these two loops, a and b, that go around each one of these circles in this Cartesian product, starting going through one of the two perfect base points. And so this isomorphism is a bit more precise than just saying that it's isomorphic. We can see that a, in this choice, say goes to 1, 0, and b goes to 0, 1. So it might be worth thinking about what these paths look like. I mentioned a little bit of this last time, but for example, let's try to look at what's the homotopy class of, say, the pair 2, 3. So this is sort of an abstract thing that we've done. This is a statement about any Cartesian product of spaces. And the fact that we now know, we have proved that the fundamental group of the circle is the integer. So the conclusion is that the fundamental group of the torus is the integers cross the integers. And we can be a bit more precise as to how that map works. But that means that if I pick a pair of integers over this side, it corresponds to some homotopy class of paths on the torus. So let's try to get a sense of how that works by trying to draw what the path corresponding, say, to the element 2, 3 looks like. So maybe you can tell me what you think it looks like. So once you decide which direction is a and when the direction is b, it will go twice in this direction. That's the a like that. And then it has to go three times in this direction. It's hard, and I'm not sure I'm going to try, but you can see a picture in Hatcher to actually give you a sort of 3D version of the path, which I think is useful for the intuitions. But there's also another way to deal with the torus that is helpful, and is to think of it as a identification of sides of a square. So we've seen, I think, this before. If you call this the a side, this is the b side, what this means is that the b is identified with orientations with itself. So you cannot do this with a sheet of paper because it doesn't fit flat in 3D. So you can do one of the foldings, but not both. So if you take this sheet of paper, you can identify these two sides and make a cylinder. Now, what we have to do is take this disk and identify it with that one, with that circle over here. And that will give us the torus. But of course, you cannot do this with sheet of paper without breaking it because the torus is just not flat. But I can fold it this way, which would be the other identification, or I can fold it this way. So the two identifications actually are what we need to do, but I cannot do them physically with a sheet of paper at the same time. But with this, then, some things have become a little easier to digest. So if I were to draw the simplest possible path that corresponds to the class, sorry, whose class corresponds to the pair 2, 3, what would be, if I were to picture it here, what would it look like? I mean, of course, the homotopy class does not determine the path, but we can think of the simplest possible path with that feature. Just like with a circle, there are plenty of paths that have homotopy class of omega n, but the simplest one is omega n. So one thing I should stress, I mean, when you think of these problems, is one typically draws a path that looks very well behaved. Omega n just goes around two circles. But the path, of course, could be completely crazy, right? You could go back a zillion times and then go zillion back, and it could still be homotopically trivial. So paths themselves are very flabby things. I mean, they could be all over the place. When we talk about the fundamental group, we talk about the homotopy class, which is something a lot more rigid, more restricted. Anyway, so what would be a path whose homotopy class mapped through this isomorphism corresponds to the number the pair 2, 3? Any thoughts? OK. So he's saying we take this picture, and we do it twice in one direction and three in a different direction. So in the A, we're going to do it twice. Well, there's A is 2, right? So A goes to 1, 0, am I right? Let's see. I either get it right or I get exactly backwards, which is possible. So what does the path look like? I've got a picture done myself. I can find it. Anybody else? He's saying that we can take any path that goes from 0, 0 to 2, 3 in these coordinates. Any people agree with that? Yes? OK, good. We're using lifting property. OK, so let me take the simplest possible path that goes from one point to another, namely a line. Is that the loop? So this point is equivalent to this point, which is equivalent to this point, which is equivalent to that point, that point, that point. So if you do all the identifications, this will be the same point. Now if we were to draw this in only one circle, in only one square, well, we can look at this square. This is going to be 2 thirds of the way. So this is the picture here. Now this continues into this picture, which if I bring it down, it means that I identify this point with that, and then it goes to probably half. If I want to take that picture and put it in one square, I identify this square with that one. And so what I see in this square is this little thing. And so this is the same as this point, which will go here. So if I want to keep track of this, I'm going like this, like this, then moving over to there. And then from here, the next one is from here to there. So if we, and I think we may have done this backwards, now that I look at it. So if this, in one square, this loop looks like that. So let's see. It crosses this horizontal axis at this point, this point, and this point. That one is the same as that one, so it cuts it three times. And in the vertical axis, it cuts it only here and there, and that one is the same as that. So I think we got exactly 3-2 with this. So in fact, this class, the class of this path, if I'm not mistaken, should correspond under this isomorphism to the integers 3-2, because it goes three times in the A direction and two times in the B direction. Am I right? It's a small chance that I'm backwards. OK, I find myself having a hard time thinking on my feet, so I leave that to you. And if I'm wrong, please let me know next time. OK, so I suggest you play with this type of things and get familiarized with what's going on, because this captures a lot of what it's necessary to understand the concept we're discussing. These examples capture a lot of the details, and it's a good idea to be familiar with these simple but basic examples to make sure the concepts are incorporated as natural things you don't have to think every time. So one last thing I want to do with the torus is that if A, if I've used the notation a little bit, A is the class of one of the circles, and B is the class of the other circle, then in particular, a consequence of this statement that the fundamental group is z cross z says that A and B commute, because the group is commutative, z cross z. So that means that the class of A, B, and A inverse B inverse should be what? Should be trivial. So that means that if I take the circle in one direction followed by the circle in the other direction and then I do the two things backwards, I should get a path that's homotopic to the identity. So it might be a good thing to actually see that. So let's go back to our square picture. So this is A and A and B and B. How would the commutator, A, B, A, and A inverse B inverse look like as a path? Well, we do A first, and the equation, of course, is a loop, but here I've got to unfold that, because we are seeing some of this lifting property. So A we do from here to here, then we do B, then we do A inverse, which is going through A backwards, and then we do B inverse. So in this picture, the whole commutator path can be described as simply going around the circle in a loop, as he was saying. Now, how do we see what the homotopy is? What is a possible homotopy to show that, indeed, it is the trivial class? If you had to actually write down the homotopy, which is not what you typically do, but if somebody tells you, now, I actually want to see the function, write it down for me. Abstractly, we know this is the case, but I actually want to see the homotopy. So we want to take this path and slowly deform it until it looks like a point. It's homotopic to the constant path, whatever base point you like, because that's the trivial class in the fundamental group. So let's say this is our base point just to fix ideas. It's much simpler than you may think. So what we could just do is slightly deform the path like this. So that thing that I drew is a path in the torus and it's homotopic. We would have to, if you really wanted to write this down, we have to sit down and write some formulas down. And we just keep doing this. So you basically take the path and you see it shrinking until it becomes a point. And in fact, if you search for animation torus, I can give you a link if you like. Somebody actually wrote a little Java applet that illustrates this. So on one hand, you see the square. So what happens in these coordinates? And then the other, you actually see it on the donut. What is the path at each time? OK. Now let's move on to yet another application or another calculation, which is useful. And that is a statement about the fundamental group of the sphere in n dimensions. And if the dimension is 1, that's a circle. And we computed that to be z if dimension is 0. Well, what's the sphere of dimensions 0? How many points? Two points. Two points. So that is not even connected. When you go to a circle, it's connected, but it's not simply connected. It has one loop that you cannot, an essential loop you cannot retract. What happens if you go S2? Any loop should be homotopic to identity. And you can imagine what it is. You just draw it. If you have a little thing that you can see, what you do is basically think of this is a lasso, and you sort of pull the string, and you collapse it to a point. That should be at least intuitively clear that the fundamental group of the two sphere should be trivial. And in fact, it's trivial for any other n. So if you think of the fundamental group as an invariant, it allows you to distinguish the circle from any other bigger sphere. They cannot be homeomorphic, because one has fundamental group Z, and the other ones don't. So if you wanted to distinguish spheres, if you wanted to prove that two spheres of different dimension are not homeomorphic, you would have to have a bigger tool, because this one doesn't see. For the pi1, they just all of the spheres look the same. They look like trivial, so have trivial fundamental group. So the proof of this, in a sense, is simple, but there's a little technicality one has to go around to completely prove it. Well, let me ask you, what would you do to prove this? Think of it as just a regular two sphere. How would you actually prove that the two sphere is simply connected that has trivial fundamental group? Again, intuitively, it's fairly clear, but if you actually want to write a proof, you want us to do something. OK, we have in an R3, fine, is there, projected to R2. And what happens then? So what he's suggesting is that we do a stereographic projection to R2. You're all familiar with that notion. So we take the sphere and we put a plane through the middle and we look at the sphere from, say, up here. And we draw lines and they will hit this sphere somewhere and correspond to points in the plane. And there's only one point that doesn't have a thing corresponding to it, which is this point itself. OK, so the conclusion we want to get from this is that if we take this sphere and remove a point, then we can homeomorphically map it to R2. And the same thing works for any sphere of a higher dimension. You can do precisely the same type of construction and see that the sphere minus a point is homeomorphic to Rn in general. OK, does that prove it? Let's go back. What are we trying to do? We're trying to prove that the fundamental group of the sphere is trivial for dimensions two and higher. So let's concentrate. The idea will be essentially the same for all cases. So let's just concentrate on two for now. How would you prove that the two sphere is simply connected? We just saw that if we miss a point, then it looks like R2 and R2 we know is simply connected. Is this enough to show that the sphere is simply connected? Yes? No? Maybe? Depends on the weather? OK, you say yes. What do you say yes? So what he's saying is we take a loop, we project it on to R2, find a homotopy there to a point, and then bring that back. And it's all right. They say for one little problem, what if the loop always goes to that point? I mean, what if you just unlucky and I take a path that actually goes to this point that I excluded? Yeah, so the base point is somewhere, but my path, I actually make it go through the point you use for your projection. OK, so you're suggesting, well, take a point that is not on the loop. Everybody happy with that? No, you should not be happy because there are paths that go through every point of the sphere. So you cannot do this argument yet because the path just might be completely filled out of the sphere. It could be completely subjective. This is counterintuitive, but there are such space-filling curves. So if somebody brings you that path, then you're stuck. Your argument doesn't work. But what's the saving grace? We don't care about paths. We care about homotopy classes of paths. So somebody brings to you this nasty path, and you change it. So what we would have to show is that if you have any path, you can always homotopy it so that it misses at least one point. OK, now that's what we're getting to. But OK, this at least is at strategy. But it's also hopefully driving in the point that the paths are not the thing we really care about. What we care about are their homotopy classes. So if the path misses a point, use it point p, use p to project to Rn, and use Rn is simply connected. If gamma does not miss a point, find another path in its class that does. And this looks hard to do, but in fact, it isn't too bad. So how to do this? OK, so take your favorite point x. That is not the base point, because the base point path definitely is going to go with the base point, because it's a base point. Take a little ball B around x, center at x. So the picture is something like this. Here's our x. Here's a little ball. And in Rn, yes. No, no, in this sphere. In this sphere. So ball meaning think of it a little disk. And so now let's think of the path. It's just all kinds of branches of the path are going in and out of that sphere. And say lots of them are going through x. What we want to do is take the path and slightly move it inside B so that it misses x. And then everything else on the path will leave it as it is. So the homotopy is only going to do a little damage to the path in this ball. And what we're going to do is move the strings that go through x a little bit, so they miss x. And a priority difficulty is, well, there could be infinitely many things that go through x, in which case it's going to be hard to argue that you can move them all within a homotopy. But in fact, there's only finally many one that do. And so we can sort of iteratively move the little branches one at a time. And that would result in a homotopy that will, in the end, give us a path that misses x altogether. OK, so here we do have to use, after all, we have an interval that always ultimately is going to play a role. The paths are mapped from the interval to the space. So let's take gamma is our path. Take gamma inverse of B. So this is a little open ball. So this is some open set of the interval. And it's, in fact, the open interval, 0, 1, because I can take the ball small enough so that it doesn't contain x0. x0 is outside the ball. And so the values of the path gamma at 0 and 1 are x0. OK, so this open set is a disjoint union of open intervals, A i to B i. And they contain gamma inverse of x. So what kind of set is gamma inverse of x? Let's be careful. What is gamma inverse of x? Is all the values of the parameter s which landed you into x? There could be many. But what's the nature of the set, the gamma inverse of x? x itself is a closed set. Gamma inverse of x is compact. It's bounded. It's in the interval 0, 1, and it's closed. Yes? We're going in steps. Gamma inverse of x is compact. Everybody happy with that? And this sort of rather than I mislead you by writing it begones to, it sort of contains. This is a set. So it's a compact set covered by these open sets. So it has to be covered by finally many. Hence meets only finally many of the A i, B i. Now we're basically out of the woods because now what we just showed is what we were saying would work. That is to say that all these branches of gamma that could possibly go through B meet our point x in only finally many times. And so what we could do is just take more precisely if we let gamma i be the path gamma restricted to the interval i, i, B i. This goes from A i, B i to the closure of B. And the endpoints are where? Well, these open interval maps over to the B. So if we add the endpoints of the interval, they are going to map two points on the boundary of the B. So if we try to think of this as a picture, here's our x down there. Let's say this is a point gamma i. Let's maybe give some names. A i is x i and gamma i of B i is y i. Or maybe x i 0 and x 1 i. So here's x 0 i and there's x i 1. So the path does something and goes through x. It goes from one point of the boundary, goes through x, and then comes back and hits the boundary again. So now what we can do is simply move this path up all the way to the boundary of the ball. And still having a continuous map. But now it just doesn't go through x. OK, so we can push this little strand of gamma. Excuse me? I was going to write it now. Yeah, I was going to write it here. So push this strand of gamma to the boundary of B and repeat so finally many times. And that's the crucial thing that this process will give you trouble if you try to do this infinitely often. So repeat finally many times to get a homotopic path avoiding x. So that's a clever argument. And now once we know that we have a path that doesn't go through x, we? Yes? Sorry? Sorry, this is a gamma. Excuse me. So take the previous path that we had before. 2, 2, 2, 2, the English were 2. So just take the path, sort of go through x inside the ball. So push it all. Sorry, is that clear now? Sorry. No, it wasn't mathematics, it was English. OK, is that clear then? So it's a nice clever argument and avoids this pitfall of having a path that covers the entire ball and would give us the trouble as the entire sphere. So what we did then is make a homotopy from the path that is annoying to a path that we can deal with and then use the argument that we started with. So this shows that the spheres of higher dimension are simply connected. Now we should be careful. Where did we use the n was at least 2? Because if this were, if we did this with S1, what would be the problem? We would have a, well, there we certainly have plenty of paths that are completely subjective, right? Well, where would we fail when we try to avoid a point? We take this little ball, try to avoid the x, but everything else will work. But this, we couldn't be able to do this deformation to the boundary, as he's saying, because the boundary consists of two points, OK? So yeah, so that wants to be careful in stating, I just basically wrote down in words, but if you want to write this more carefully, you have to, of course, verify that this can be done. And that's where you need that n is at least 2. OK, so the fundamental group of spheres, the higher spheres, is trivial. We'll continue next week, and I'll give you some homework for you to work on. Not sure what's the standard about grading, but at least, again, this should be for you, not for me. So try to think of that. And if you have questions and you want to come to talk to me in my office, I'll be happy to discuss them. OK, thank you.