 Okay, so, are there any questions? Yes, is it about the homework? Okay, maybe we can leave that for later for the end. Let me just go back to the example, one example I was doing with the torus that I was getting slightly confused and so let me try to clarify it. We had discussed this already several times, this covering map and we can picture this like this and if we think of this point in the standard way of doing it, this would be zero, these dots would be the integers and so on. So if we wanted to look at the torus, just s1 cross s1, okay, what would be a covering of it? Would be the product of these two covers. So we would have the plane, so it would be covered by r2 where each copy of r goes to each copy of s1, so it would be sort of a product covering and the integers now would become a lattice. So if we use the same maps, the product of the maps, say this is the 00, the origin of the lattice, the plane, and this would map to the torus. So each one of the rectangles in the lattice maps to the torus in the way we discussed the identification of the sides. So if I lift a map, a loop in the circle here, it lifts up, say if I started at zero, something goes back and forth, whatever it does, and it ends in some other integer. In that integer we call it the fundamental, I mean the winding number, and we use that to identify the fundamental group of the circle with z. So we do the same here. We know, we argued already that the fundamental group is z cross z, but this covering map actually allows us to do this more precisely, and this is what he was suggesting last time, so let me repeat it. So if you want to have a path gamma in pi1 of the torus, say a base at 00, then we know the group is isomorphic to z cross z. If we want to find what maps to the say, to give an example, the pair is 23, let me put the origin here. So we just look at 23 in these coordinates, and any path that goes from 00 and ends in 23 when you push it down will be a class, so this would be the lift of gamma, which is the path that has class 23. So this would be 23. And the picture I was trying to find last time, and I didn't show it to, would be this, so I did it like that. It doesn't fit in the picture, so this is the origin down here. And if I take this path that goes across several rectangles and map each piece of the rectangle that it crosses down to one rectangle, what we see is this, this is what I was trying to show you last time. So we start here, go up till it hits the boundary, but this point is the same as that point, so it continues here, then this point is the same as that point, so it continues there, and then this point is the same as that point, and finally it reaches this point which is the same as that. So that's the closed loop in the rectangle form of the torus, and if you want to see it a little more, you can fold it, and you can see how the path is supposed to go like this, like this, second, and then here, and there, or I did it wrong. This here, here, there, sorry. Now, fortunately we cannot fold this both ways without breaking the paper, so it's hard to present the final picture on the torus, but I suggest you play with something like this to be completely familiar with this story, and the thing I was a little confused and I think I may have said it wrong is the following, if we look, if this is called our A path, and this is our B path, then the path in blue crosses this horizontal segment here, there, and there, and that's the same as that, so it crosses this three times. So which is correct in the sense that this is the 2-3, not the 3-2, because if we think of the torus, and this is our A, and this is our B, and this is the base point, then a loop, like if we do B three times, it's going to cross the A segment three times. So the coordinate 3B of a path will translate in crossing the A path three times, and the coordinate 2 of a path will translate in crossing the B path two times. So it crosses the horizontal axis three times, and it crosses the vertical axis here and here, and that is the same as that, so it crosses it two times. So this path indeed is the path 2-3, which is what we were looking for last time. Is that clear, I want to try to say? So you can, so we did this thing in two different ways. Right now, I looked at the universal cover, well, the concept that we'll come back to, but the plane is simply connected and a map is a covering of the space of the torus, and so when we lift the path, we already argued that if you start at a given point, it ends in a point that is uniquely determined, and this is how this isomorphism is achieved by sending a homotopy class to the ending point of a lift of the path, starting, say, at zero-zero. Okay, hopefully this was some help in clarifying this point, but topology I find, especially algebraic topology, is actually fairly hard to write rigorous proves and detailed proves. So you have to combine trying to do that with a lot of intuition that is hard to transmit, and the best way to get that intuition is with just doing the stuff with your hands yourself. So you should try as much as possible to do pictures, use pencil and paper, scissors, just actually physically think of what's going on. I think it will help a lot the intuition. Then it's sometimes hard to translate those things into detail and rigorous proves, but you have to have a bit of both. I think it's very hard to go purely formal, but some of that, of course, is necessary. Okay, so what we're going to do is continue with the fundamental group, and I'm going to do a little bit of examples with the figure eight. So why don't we start there, and then I'll do a bit of the theory. Okay, so the figure eight is two circles joined at a point, and if we have two clear loops here, which I'll call A and B, and hopefully by the end of the class I've done enough to convince you that the fundamental group of this space is freely generated by A and B. Okay, so do you know what a free group is? Can somebody tell me what the free group in A and B would be? Maybe if you've seen it. Have you seen the notion of free groups? No? He's saying the group generated by A and B, where does the word free, you think, come in? He's saying that the position of A and B is free. That's where free come in. Any other thoughts? Yeah? There's no relation between A and B. Right, what the free group means is that the elements of the group are going to be identified with strings of symbols, which are at any given slot, either an A or an A inverse, and a B or a B inverse, and the only relation is that A inverse, of course, is 1, and BB inverse is 1. So you've probably seen, well, we discussed the free abelian group generated by two elements, WBZ cross Z. Right? So we have... So Z squared is the free abelian group in two generators. Okay? So for example, one generator could be one zero, and the other one could be zero one. Everybody is a unique linear combination of these two elements, but the addition... The group operation is addition, so it's closer to the kind of things you do in linear algebra and so on, but the freeness come because of the uniqueness. No combination of these two vectors is zero, unless there is the zero combination. But... So the analog on the non-abelian situation is where... So the free group in two generators, so it will be simply denoted by this, is generated by NB with no relations whatsoever. So elements are words in A, A inverse, BB inverse, with the only condition is that if you see a B and a B inverse next to each other, or a B inverse and a B, or an A inverse and an A, you can cross that out. Okay? So for example, A squared B cubed A to the minus one B to the minus 15 A, that's a element of the free group, and there's no simplification you can do to it. That's what the element is. You cannot switch the A and the B in any way. Okay? So this is different from, say, A cubed B cubed A inverse B to the minus 15, simply because it's a different word. Okay? The elements of the group are like words in a language of A and B. The only thing to watch out, of course, is that if A tenous A inverse cancels out, and a B and B inverse cancels out. But those are the only relations. So this is, in a sense, a huge group. I mean, elements, and when you multiply, just multiplication simply means to put the words next to each other, to concatenate the words, again with this proviso A, A inverse and B inverse, etc. So this is what the fundamental group of the figure A is. Basically, any path is homotopic to a path that consists of, okay, tell me how many times you went through A, then through B, then through A, B, you know, until you're completely done. So this would be the word, this would correspond to the homotopic class of a path that goes twice in A, then three times in B, then one inverse in A, 15 times backwards in B, and then one A at the end. Okay? And so that path is uniquely determined by this word. There's no way to simplify it, and two paths are going to be the same if the words are the same, again, modular that condition. Okay, but I will slowly try to unravel this, and I want to use this to elaborate further on the relation between the fundamental group and coverings. So far we use the notion of coverings to help us compute the fundamental group of a circle, for example. But we'll see that there is a very, very close connection between coverings and fundamental group. And again, if you've seen Galois theory, fundamental groups and the subgroups are going to play the role of Galois groups and the subgroups, and coverings are going to correspond to field extensions. And the fundamental theorem of Galois theory is that there is a one-to-one correspondence, and a similar thing will happen with coverings and subgroups of the fundamental group. I mean, there's several technical details to fill in, but that's the one-centers summary. So to get into this, let me construct some covering of the figure eight. Okay? And the thing to notice, I mentioned this before, is that the figure eight has one point that is special. We'll take it as a base point, naturally. And in that point there is a strand that comes in that has a label A, and a strand that comes out that has a label A, namely this and that. So if you want to keep things clear, the A sort of rolls around. So one strand of the A comes out, one strand of the A comes in, and then, so this would be, one strand of the B comes out of the point and another strand goes in. So to construct a covering, we can do the following. So think of the A circle. In the covering, what will it look like? Well, below is a closed loop. When you lift it to the cover, which we know we can do because we proved these lifting theorems, it's going to either be a closed loop or, more generally, it's just going to be a little strand that starts somewhere and finishes somewhere. Where it ends and begins has to map down to the point, the central point. That's the way this covering map works. And so the same about B. So basically, to look at a covering, we're going to see a bunch of things that look like A, closed circles, or bunch of things that look like A, but that's been opened up. So let me give you an example and we'll try to argue why this is a covering. So here's my central point, my base point, say. I'm going to have an A, but instead of having a B as below, I'm going to have a B that looks like this. I'm going to do my coverings horizontal so I have more space. So my map is going to go like this. Now, then I'm going to have another A that looks like this and a B that looks like this. So what degree does this cover have? Degree is the number of pre-images of a point. If I haven't said it before, I'm not sure I'll be able to actually prove it, but the cardinality of the pre-image of any point is the same for all points. So if you have a covering that each point has a finite number of pre-images, then we call that number the degree of the cover. I mean, there are covers that, of course, are infinite, like, for example, R to the circle, but some covers are finite, a degree like this one. So what's the degree of this cover? Three. There's one, two, three points that are all going to map to the one point below. So the pre-image of the main point here is these three ones. So this is a degree three cover and what we can check to see that this is indeed a cover is that if we look upstairs at each one of the pre-image points, if we just focus a little disc around the point, it should look exactly like the point below. So they should have a strand of A coming in, a strand of I, A coming out, a strand of B coming out, and a strand of B coming in. So that's, if I did this correctly, this is how that should look. So each one of the three special points. Now, so what is the pre-image of the whole segment A? What are the three lifts of A? There should be three lifts starting from any one of the three special points. I have a unique lift, right? A lift of a loop downstairs is unique and I would say once I decide where I begin it. So tell me what is the lift of A if I start here? Just the loop. That's the lift of A if I start there. What is A if I start here? Up to there. So in this second lift, it opened up. Downstairs is closed, but upstairs it goes from this point to this point. In fact, I'm going to give these points names. One, two, three. And so A twiddle, right? The lift of A is either this going from one to one or from two to three or from three to two. These are the three possibilities. Is that right? These are the three lifts of A depending on the initial point. So again, to write very carefully that this is a covering is a little tedious. So let's at least convince ourselves. So for example, if I pick a little open neighborhood here, it should be covered by what? Three little open sets that are homeomorphic to this, to it. So where are they? If I do sort of the most natural map. One is here, here and down. So this is a evenly covered neighborhood of our point below. And you can see that this is going to work for any point other than the base point. And at the base point, all you need about it, if you pick a little neighborhood of it, it should look exactly the same way, which is what we already argued. So it's hard to sort of perhaps write this in full glory, but hopefully I'll convince you. That's correct. So what are we going to do with this? Well already we can see something here when we talked about lifting A because the lift of A, if we start at 1, it ends in 1. If it starts at 2, it ends in 3. And if it starts in 3, it ends in 2. So what we could do is we could say is that A, through its lifts, what it's doing is permuting these three points. And to get this straight is one of these perversively annoying things that sometimes you have to take the inverse instead of the map that you expect if you want things to go smoothly. So I'm going to consider the permutation of the 1, 2, 3, the numbers 1, 2, 3 which are labeling the pre-images of the base point associated to loops in the base in the figure 8. So this is my base space. This is covering any loop in the base lifts to a map that goes from one of these three points and ends in one of those three points. But to do this again for things to work out correctly, you have to go backwards. And the permutation is that what we're going to do is, so if we have a loop, what is this letter, associated to a loop in the base, say gamma, is you say gamma twiddle of 1, it's taken to gamma twiddle of 0. So you have to go from where it ends backwards. In this case, it's not going to matter too much, but if you have a more general, a bigger situation, it will. It's just because these are transpositions, transpositions are inverses of each other. So let's list what this permutation is associated to a. It takes 1 to 1. And if it starts at, if it ends at 3, it takes you to 2. And if it ends at 2, it takes you to 3. So in the more standard way to write this, you can say that it's associated to the transposition 2, 3. And what about b? What does b get mapped to? If I start at 1, what happens, what does b do? It takes you to 2. So 2 goes to 1. So 2, it adds in 2, starts at 1. If I start at 2 and I follow b, we go to 1. So yeah, end at 1, start at 2. And at 3, it's 3. So this is the permutation, the transposition 1, 2. OK. So what we're going to argue is that this covering p, I'll indicate how to prove this. It defines a map. We already said this, p lower star. That goes from pi 1 of the top, which I'm calling, I'm going to call it y and base and calling x, as we had before. From pi 1 y of any fixed point to pi 1 of x. And the image of this map is some subgroup. In fact, p lower star is injective. We'll show this. So in fact, what we have is an injection of this group of the cover, the fundamental group of the cover into the fundamental group of the base. And this subgroup will, if we insist on keeping track of base points, this subgroup will uniquely determine the cover. So it's similar to the statement that in Galois theory, this is a subgroup of the Galois group, and it corresponds to a unique subfield of the algebraic closure. If you've seen that, this is the parallel. And in fact, we can associate to this subgroup. In fact, if you give yourself a subgroup, you can concretely construct a corresponding covering. And I'll do a few more examples to give you a sense of what this is saying with this figure 8 case in mind. But before we do that, let's check one thing. So what is the lift of A, B, A inverse B inverse? Okay, so let me bring back the picture. Can you tell me what does the lift of A, B, A inverse B inverse looks like? Say starting at 1, because it will depend on where you start. Now, this is what I was hinting at before. We compose paths by, the one on the left is the one that goes first. Okay, so permutations we usually compose in the other way. That's why we have to do this funny inverse thing. Okay, so we do first A. So we lift this by lifting each one of them and then taking the product, which is to say one after the other. So we lift A starting here. That gives this loop, right? Okay, so this is 1. This is our A. Then we do B. Then we do A inverse, which will be what? This is A going this way. So A inverse will be that. And then we do B inverse, which is a loop like that. So this map takes 1. This lift starts at 1 and ends in 3. Right? Why don't we do the 3 lifts? So now let's do the lift starting at 2. We'll do it starting at 3. Okay, so we start at 2 and we do A. We have that, right? Then we do B. Then we do A inverse, which is this thing going back. And then we do B inverse, which is, right? So it starts at 2 and ends in 1. And then finally, if we start at 3, we do A. Then we do B. And then we do A inverse, which is this. And then we do B inverse, which is that. So B and B inverse. So we end at 2, right? So we started at 3 and ended up at 2. Is that clear? Okay, so let's see. So this particular element of the fundamental group has an associated permutation of the fiber of the three points on top of the base point, which is now we have to be careful because now it will matter if you do it right or if you do it backwards. So the way to get this straight, you have to look at where it starts, sorry, where it ends and where it starts backwards. So 1 starts, if it ends at 1, it starts at 2. 2 is 3 and 3 is 1. So this is the permutation that takes 1 to 2, 2 to 3 and 3 back to 1. So just to, sorry? Sorry, you just wanted to see it? Okay, so out of, just to check our sanity, let's just see that this is in fact, yeah. I think I'll just leave it at that. So let me skip what I was going to say. So what is the conclusion? What conclusion how we draw from this little calculation? Right. They found out in the group all the figure 8 is not a billion because this particular element, which if it had been a billion would be the trivial class, trivial homotopic class, it would lift to give you a trivial permutation. Okay, so this implies, okay, so there's several things here to fill in, but in particular, we see that A, B, A inverse, B inverse is not homotopically trivial. Why is that, by the way? How do we lift the homotopy trivial class? If this path were trivial, the class of it were trivial, it would be homotopic to the constant map equal to a point. How do you lift that? We'll just take it to any point that you want to to the constant thing, and the maps are unique. So if it's homotopically trivial, it has to start and end in the same place, and it doesn't. Okay, so that means, pi1 of y, respect to any point, is not a billion. So we claim a lot more, we claim that the group is, the free group is far from being a billion, is as non-abillion as you can make it out of two generators. Okay, but these are all things, so in principle we have not seen a single example of fundamental group that was not a billion before. Okay, now we have a concrete example in our hands. So to continue with a bit more of these examples, let me, as I mentioned before, Hatcher has a long whole page of examples, but just to go quickly through some of them, let's see. Give me a, here's two points. Okay, give me a double cover of this figure eight, not a figure eight space. So it has, the preimage of the base point has two points. So we have to pick for A and for B two different lifts, and at each point it has to look like this cross with the A coming in and out and the B coming in and out. So let me start, pick A to be this, and then there isn't much room, I have to pick what B is pretty much determined. Oh, sorry, A and A, sorry, excuse me. Now otherwise the A and the B have to meet at one of the lift points, so that was wrong. A again? From this point, what do I have to do? It has to be B, because at the crossing it has to look like A, A and B, B. So do B, B. So now this point looks fine. So he's saying that again here we have to put A and A, and if we don't want any other point, then it will have to look like this. Okay, tell me quickly what is the permutation that I associate to A and to B? A is the identity because it takes, yes, A takes one to one and two to two, and B one to two and two to one, and it doesn't matter how we do it because it's its own inverse. Okay, and so you can construct several examples like this. Another example of degree two, so this is degree two, would be if we do A like this, then if I do B here and B there, it would be basically the same as when we drew, so we can have this instead. Okay, so that's another cover of degree two of the figure eight. And what permutations do we get here? They both map to the same thing, to the permutations of transposition swapping one and two. So the way that this works, is that we have the fundamental group of the base, and we look at the cosets by the image of P lower star of the top, which as I mentioned, we'll see is actually a subgroup because P lower star is injective, and these cosets are going to be in bijection with the points above x0. So in this example, the covering y has a fundamental group that injects into the fundamental group of the base with index two. So there's a coset that corresponds to the trivial class and a coset that corresponds to a transposition. So we can identify, say, one with the subgroup itself and two with the other coset. So what is going to be useful to have a space, a covering space y that is simply connected. And that, remember, means that it has trivial fundamental group and is path connected. So I'll say, I'll end with this example in a little bit, and then I'm going to go and try to do some of the theoretical aspects. But before that, give me examples of, even without having seen the existence, we can try to find such things. So if we have such a y, we call it the universal cover of x. And you should be complaining of the use of the word the, but hopefully you'll stop complaining afterwards. Because at the moment all I'm saying is, any space that happens to be a covering of their base, but it's also simply connected, we call it a universal cover, but we'll see that it's essentially unique when it exists. So give me examples. X is as one. What is a universal? What is a cover? What is a cover that is simply connected? R. We don't have a whole lot of examples though. S1 cross S1, the torus squared, and we already used these facts quite a bit to study the spaces x themselves. What else do we know? Well, something really trivial about the, if x is R, R itself is a point, is a point itself. We don't have enough tools to give any more substantial examples yet, but let's give a try to find the fundamental, sorry, the universal cover for the figure eight knot. So what should it be? It should be something that looks like the examples we're doing. We will have a bunch of points. There are a pre-image of our base point, and in each one of them we should see this cross, the A in, A out, and the B in, and the B out. Okay, but it should be simply connected. That's what we're looking for. So if we look at the examples like this we constructed, we have, it's a covering indeed, but this is not simply connected. So if we go A, A like that, it's definitely not a trivial class. So it should be something that has no loops. Okay, so let's give it a shot. Here's one of our, pick one of the base points that are supposed to be on top. It should have this cross picture. So let's just draw the cross. Okay, let's say we put the A's here horizontally, and the B I'm going to put it vertically. Okay, so now each one of these new four points should look again like this cross, but I don't want to ever come back because I don't want to create a loop that will prevent me from the space being simply connected. So basically what we do is we continue this. So we do another B, and I stop writing the labels because it's going to become unreadable, and then you do the same here, and you can imagine that you do this with each time iteration with a slightly smaller size, say half of the size that you had before. So this will fit into this sheet of paper, and it will in the limit give you a space that looks like what we want. Each cross has the right shape, and there's no loop because this is simply a tree. I mean, you never have something that closes back. So this would be the fundamental, sorry, the universal cover of the figure 8-0. Okay, so, and this is not far, now we're not very far from proving that the fundamental group of the figure 8-0 is a free group in two elements because how do lifts of paths look like? Well, a lift will, if you take any word downstairs, that corresponds to a sequence of A's and B's and maybe A inverse and B inverse in the figure 8. When you lift it, well, if you start here and you see an A, you go horizontally. In A inverse, you go to the left. If you see a B, you go up, and you see a B inverse, you go down. And so, if you scan the word from left to right, you can exactly say, well, I do so many steps up, so many steps down, so many steps north, so many steps south, and then at the end of the day, you're going to end in some spot in the plane, and that spot, there's only one way to reach it. This is like having this graph. It gives you a way to have coordinates on each one of these points. Each point is uniquely determined by a sequence of A, A inverse, B, B inverse, and so on. And if you see an A, A, A inverse, for example, you go to the right, but then you do A inverse, you go back, so you might as well not have done it. So if you cancel all of these A, A inverse and B inverse possibilities, you have a unique way to reach any given point in this picture, and this tells you that there can't be any relation because if it were a relation downstairs, it will lift to the trivial homotopy, the trivial path. So it's a bit of a hand wave argument, but it's not far from something one can write carefully approved to show that the fundamental group of this figure eight knot is the free grouping two elements. In fact, we're going to prove it next time, rigorously in a different manner, using Van Kampen's theorem, which is the other next sort of major theorem in the subject that we need to discuss. I'm going to leave that for next time. But hopefully this way of seeing it also sort of gives you a bit of more intuition as to why is the case that this is the free group. Now something I should point out that if you've never seen it, will probably come as a huge surprise. So let me take one of the examples. Let's take this example of degree two. Say this example here. And let's take this example. Take that to be the one point. This is B2. This is A, A, B, B, A, A. Let's try to describe the fundamental group of this. So what are elements, sequences, words in A and B that actually are closed paths here? So we start at one. A is a closed path. So if I do this, I'm going to get a closed path. The picture in Hatcher. So A is a closed loop. Then if I start at one, I could do B and then B. So I'm writing B squared upstairs is something that lifts the loop B squared below. And then I can do B, A, B, or A. Sorry, B, then A, and then B inverse. I can go back to the same spot. So these are closed loops. And so in fact, they freely generate pi1 of y, 1. Meaning that pi1 of y, 1 is free group in these three generators. And pi1 of y, 1 is a subgroup of pi1 of x, 1. So this is the surprise that this is a free group in two generators and it has a subgroup that is free in three generators. This is counter-intuitive because one is used to linear algebra where if you have a space of dimension 2, it can contain a space of the subspace of dimension 3. That's completely ludicrous. But in this game, in this world of non-Abelian groups, that notion just doesn't hold true. So this is a pi1 of this as a free group in two generators and this has pi1 which is a subgroup and is free in three generators. And in fact, you can have a subgroup in as many generators as you like. And moreover, it's a theorem of Nielsen, and let's get this straight, that every subgroup of a free group is free. And it could be of any number of generators. And I spend more than I wanted in the examples because I think they're more illuminating than giving proofs but of course we can't live on just examples. But maybe I'll do more of the proofs next time. I can't find the statement. I can't find it. Well, I'll mention it again next time. So let me end the example since I've already carried away with a question. If what I've been saying is the right understanding here, any subgroup of the fundamental group will correspond to a cover. And if you fix the base points, that's a one-to-one correspondence. So in this example here, this subgroup generated by AB squared BAB inverse a subgroup of the free group in two generators AB corresponds to the cover Y given by that picture. So to illustrate this, let me give you a subgroup and you tell me what the space Y is. Okay, let's give it a try. So take the subgroup H of pi1 x1 to be the group generated by A squared and B squared. No, no, no, I want the subgroup of pi1 of x below. See, below we have the free group of two elements. Any covering will give us a subgroup of the pi1 below. And if you go all the way up to this crazy cross, you get the trivial subgroup. So the universal cover in this... So this has pi1 of Y1 to be trivial. And this injects two pi1 of x1, which is the free group of two elements. Okay, so this is supposed to continue. I'm not writing it, but it's supposed to continue in every possible direction. Okay, so this is an infinity. Right, so the way it goes is that a subgroup of the fundamental group downstairs corresponds to a cover and vice versa. Again, you have to be a little careful with the base points to make this statement precise. I'm somewhat ignorant of that for a minute. Okay, so now I give you the subgroup. And let's try to figure out what is the corresponding cover. Sorry, two points. So he's suggesting that we take two points. Maybe you're thinking of this. Well, so this A, if we do it twice, is a loop. And a loop upstairs corresponds to a trivial element downstairs. So the subgroup in question contains A squared and B squared too, but it has more stuff in it. It has AB. Or yeah, so we can go A and go B. Okay, and well, there are other options, but they're all going to follow from that one, maybe. Or maybe we also need BA. Be careful. We basically have to look at all the little cells that we see that are enclosed and then write the corresponding element. So there's a theorem there that... I think we'd, no, I think we're fine. I think with that we cover all cases. So again, this is a free subgroup in three generators. So it has AB that is trivial in this covering, which we don't want it. We only want A squared and B squared to be trivial. Okay, so maybe I'll leave it to you to think about. Tell me next time. Okay. All right, so let's do a little bit of theory then to try to fill out the things that I was saying mostly in words and pictures and so on. Okay, so the first statement that we would like to see is that if we have a covering map, let's say the y0 maps to x0, then we know we have a map p lower star that goes from pi1 of yy0 to pi1 of xx0. And the statement is that this map is injective. And the beauty of this is that then we are associating to a cover of x, a specific subgroup of pi1 of the base. Okay, because the map is injective so we can see the image is a subgroup, which is isomorphic to the fundamental group of the top. Okay, I think I may skip most of the proofs because I mean many of these proofs are always arguing in exactly the same way and they're not necessarily very illuminating. The statements are a lot more interesting than the proofs. So I think I'll skip that, but that's one ingredient. The next ingredient is a statement about liftings in general. We talked about lifting paths to coverings and lifting homotopies to coverings. We want to decide when can we lift an arbitrary map from a space to the base. Okay, and so the situation is this. We have z and a function f to x covering p to y. And we're interesting in the existence of a lift, f twiddle. And we'll fix base points so we have a y0 that maps to x0 and a z0 that maps to x0 and the map f twiddle should also preserve the base points. So our map f twiddle is a lifting so p of f to o should be f and f twiddle of z0 should be y0. That's in symbols the same as those two pictures. So what is a necessary condition? And tell me what the necessary condition is and we'll prove that in fact is also sufficient. So he's saying that we need z to be simply connected if we want the f tilde to be unique. We'll worry about uniqueness in a bit. I'm more concerned of the existence of the f twiddle at this point. So we prove that we can do this with no condition if z is an interval. And in fact if z is an interval cross an interval because we've shown how to lift paths and how to lift homotopies between paths. So now z is arbitrary but for the statement I want to assume the z is connected and in fact we need that is locally path connected. And as Fulton says in the introduction of one of the chapters we'll see yet another condition that is very technical that is required for various statements and these things kind of can easily get in the way. So if you want to concentrate on the issues minus we'll take z to be as many properties as you need but being simply connected is not one of them. No we don't actually need z to be simply connected. Well okay let's get to the statement and then I'll explain what I mean. There's a necessary condition and it helps to think in this functorial way. The fundamental group is not just a group you associate to a space. It's a group you associate to a space and maps between the groups associated to maps between the spaces. With all the properties that if you compose maps on the spaces you get composition of maps on the groups. It's a functor. So that is actually very crucial for this construction of fundamental group. You not just sort of randomly associate a group to a space but it has these properties that if you do things to the space corresponding things happen to the groups themselves. Okay so let me start with the proof of the phantom statement and then we'll... So I'm going to show you what is the necessary condition and then I'll argue that actually is sufficient. Well suppose we have the lift. Okay? So what do we conclude from this? If we now take... So this is a picture of spaces. There are continuous maps between spaces. Sorry. The proposition is still not finished or is halfway through. I'm just trying to... Rather than dumping down the condition I would like to find it with you. Okay? No you don't. Hold on a sec. No you don't know what we're proving. But let me try to find a necessary condition for the existence of such a lift. So suppose we had the f total. So now we have a picture of spaces, maps between spaces. Now I apply to that picture the fundamental group. It will convert this picture into a picture of groups and maps between them. Okay? So we will have pi1 of zz0 maps by f lower star to pi1 of xx0 p lower star pi1 of yy0 and then this map f total lower star. So if we have such a picture of groups what can we conclude about the relation between f lower star and p lower star? What he's saying is that the image of f lower star should be a subset but more precisely a subgroup because these are homomorphisms. The image of f lower star should be a subgroup of the image of p lower star. Yes? Because what we have here is that f lower star is the composition of p lower star with f total lower star. So let me write that down as a necessary condition. If this condition does not hold then we cannot expect to have a lifting of such a map f. So I'm going to write down that the image of f lower star should be a subgroup of the image of p lower star. Am I writing this right? So the statement now is full. I claim that that is actually sufficient. So what we need to prove now is that if you have a map f whose image of f lower star at the level of fundamental groups lands in a subgroup of the image of p lower star then you can lift the map f. And in fact if the map exists it will in fact be unique. So the lift is unique when it exists. So now let's revise this statement with the statements we actually already proved which were the case of a path and a homotopy. What happens in those cases? What is the space z? Is an interval or an interval across itself? So what is the image of f lower star in the case that z is an interval? Well if z is an interval, pi1 of it is trivial. The image is trivial and is included in any subgroup no matter what p star is. So in that case this condition is sort of vacuous. It's always satisfied. So that's why we never saw it. Okay? So if you like then a corollary of this statement is that if you have z simply connected as you were suggesting then we will always have a lift. So your statement was correct if the question that we're trying to answer was when can we always lift the map? Yes. Yeah, path... I believe it has to be connected and locally path connected which might imply... No, it's defined for anything but I'm choosing a base point. So in fact it will only matter for the path connected component that contains my point z0. So those are technical issues that are not sort of at the center of the matter. You need some hypotheses but you definitely don't need simply connected. I mean that's kind of the point here that we're trying to lift things from a space that needs some properties but not that is fundamental Goobish trivial. Okay, so I'm going to leave the proof of the other direction to you because it's more or less you follow your nose with the statements if you look at the proof of the case of lifting paths and you apply into this. At some point you will have to use the statement of course but the ideas are pretty much the same all the time. Okay, so as a corollary of this suppose you have two coverings y and y' with two base points y0 the map to the same base point downstairs y0, x0. So what we would like to say is a notion of when these two coverings are isomorphic coverings and for that we want a map from y to y' homomorphism that preserves the map p so this is p and p' or rather it takes p to p' so we want a homomorphism that fits in this commuted triangle in other words p is p' composed with h so we have such an isomorphism of coverings if and only if, what do you think? So look at this way, go ahead so I think what you're trying to say is the image of p1 of p should be equal to image first we want it to be a subgroup so that we have a map one way but we also want to map the other way so we want it to be also contained the other way around and so they have to be equal so again there's a little bit of thinking some few little things to check to complete in this argument but the statement is that these two should be the same so this is going some ways into what we were saying before if we have two coverings they give us subgroups of p1 of x namely p lower star of p1 of y and p prime lower star of p1 of y and if they happen to have the same subgroup then the covers are not necessarily the same but they're isomorphic so this goes some ways into filling out this story between the bijection between subgroups of p1 of x and coverings so rather than coverings I should have said isomorphism classes of coverings so here I guess x has to have the same hypothesis as before this is a corollary of the theorem and a further corollary what can we say about coverings of x which is in the hypothesis of the theorem x as in proposition which is simply connected on top so this is the corollary first corollary, the second corollary I want a conclusion about coverings so a covering of a simply connected space which is locally simply connected in the hypothesis of the theorem has to look how it must be trivial what does that mean we're talking about the space I'm asking I have a covering y to x where y with x is simply connected so I think you're in the right direction but maybe let's make precise a statement go ahead sorry what is the universal covering so x is its own universal covering because it's simply connected the map x to x is naturally a covering but let's be a little careful I give you r give me many covers of r so we have r what are possible covers of r? r two r's three r's any number of r's okay so the fact that it's simply connected doesn't imply that a covering has to be just r just itself but it implies that it's trivial which is to say that we have r with a trivial covering so there are several copies of it which could be whatever cardinal number of copies you like so the conclusion that I like to get I want to claim is that y is the trivial covering or rather is a trivial covering i.e. y is isomorphic to x cross some subset t which is discreet in fact t we can take to be pi inverse of x0 okay so let me just give you a sketch of the proof so we have x going to x and we take the identity map because by the proposition the pi1 of x is trivial so the image of p lower star is well it's just because the map so this is our f the identity it maps the pi1 of x to pi1 of x it's just the trivial group to the trivial group it always satisfies the hypothesis so we can lift this map okay and let me call it phi phi we can lift it so that phi of x0 is any point y0 in the inverse image of x0 so x0 is down here and it has a whole bunch of pre-images if I pick what I like the value of x0 to be then there is a unique lift like that so phi of p is the identity of x so the identity of x has a lift for every possible choice of a pre-image of x0 and so when you lift that map you are getting a well you have to do a little check but this the idea would be that this phi gives you a map to this slice at that point that looks like x if you choose a different point y0 you are going to get another slice and so with this way you can sort of slice your y and more precisely what you have to check is that you look at x partition product x inverse of x0 and you map that to y by taking x y0 to phi of x sorry, this phi so this phi is attached to the y0 and we have to check that this map is actually an isomorphism, is a homeomorphism and now I think now going back that I forgot to say something very very important which is the correspondence between subgroups of the fundamental group downstairs and coverings has to include the word connected okay, I did not, I don't think I mentioned that before in the hurry to make a statement so already here we see that if we have the fundamental group is trivial it has no subgroups other than the thing itself so it must have only one cover what I said before was correct but that's only true if you add the word connected to the word cover there's only one connected cover of a simply connected space, namely itself and we can just argue that if we don't include the connectedness of course you can have all kinds of covers for example you can have, you just have as many slices as you like okay, so this goes some ways improving what we were doing before and all the covers we were doing of the figure 8 knot for example were all implicitly it was only looking at connected covers and that's why we only worried about subgroups, anyway I'll continue with this and there'll be my last lecture so I'll do a bit more of this and find campus theorem and that's about all the time we're going to have so we'll stop here