 OK, then we can start. So this is the third lecture on algebraic topology, where we're talking about the homotopy invariance theorem. So this was a theorem. If f and g, x to y, are homotopic maps, then it follows that they induce the same map on homology. So then f star is equal to g star on hn x to hn of y, for all n. So we had first introduced the chain homotopies. Maybe I will not, or maybe I will repeat it. So this was, if you have two chain maps, f star and g star from a certain chain complex, c star d to another chain complex, d star d. Then we said they were chain homotopic if one had some kind of diagonal maps which interpolate between the two. So then f and g are chain homotopic. If there exist maps s i from ci to di plus 1, such that we have that d composed with s i minus s minus 1, s i minus 1 composed with d, is equal to g i minus f i for all i. So somehow the difference between these two can somehow be obtained as having to do with something composed with a differential. And this will serve to prove that this lemma, which we proved, if f and g are chain homotopic, when the induced map f star is equal to g star from hn of c to hn. So then they induce the same map on the homology of the chain complex. Now we wanted to use this to prove the theorem. And I have to repeat the setup I expect. So we go to the proof. So we had first seen that one can reduce to the case enough to show the theorem in case, say, f is equal to zero and g is equal to e1, where e0 and e1 are the inclusions of x. So x times i, x goes to x comma zero and similar for i1. We had seen that this is enough to prove it in this case. And then we were making this setup. So we want to, out of a simplex, we somehow make a prism and divide it into smaller simplices. So this was the following. So in delta n, so the standard simplex times i, i is always interval of 0, 1. And we take the kind of end points. So we put vi equal to ei comma 0. So somehow the one's in the bottom. And wi equal to ei comma 1. And so if this one is delta 1, then we get basically another copy of delta 1 here. And we have here this case is square, but in general prism. And we define some maps from, given a map from delta n. So if we have a homotopy, we get from this a map from n plus 1 simplices to x as follows. So first we define, we look at these maps, w0, v0 until vi, wi until wn. This, by definition, is a map from delta n plus 1 to delta n times i, whose image is just the span of these points. So for instance, it's not v0, v1, w0, w1. And so here we would have this thing would be v0, w0, w1. So the image of this map and the image of the map, the second one, would be v0, v1, w1. And then using these maps and our homotopy, we can, OK. So you don't need the homotopy. Using our map, we can define this prism operator. So let's just write it down. So we want to first define it on simplices. So we want to define an operator p from cn of x to cn plus 1 of x times i, which in the end is supposed to give us a chain homotopy between i0 and i1, OK. But we first have, so at least it goes in the correct way. It goes kind of diagonally. And we do this first for a simplex. We define it by this formula. We just use all of these, compose it with sigma times the identity on i, and take the alternating sum. So this is sum i equals 0 to n minus 1 to the i. Sigma composed with the identity on i v0, i, w i, until wn. Yes, indeed, it is cross. It is misplaced in the notes, which now I copied. Here it is the composition. So we just have, so this is a map from delta n plus 1 to x times i, and then we compose it with the identity on i times the map sigma. So this is a very different thing. And we put these signs as usual so that in the end something will cancel. And then for, this is just for a singular simplex, then in general case, we do it by linearity. So for alpha, so for we make p of sum over simplices a sigma times sigma. So this is an integer. This is a singular simplex. We define this as usual to be the sum over all sigma a sigma times p of sigma. So always by linearity. So then we have defined this map, which is a homomorphism between these two ABing groups. And we want to show p is a chain homotopy between the maps induced by i0 and i1. From i0 lower star to i1 lower star, where i0 lower star is just a map from the chain complex c star of x to c star of x times i, which is obtained in the usual way by composing with i0. So sum a sigma a sigma times sigma is mapped to sum over all sigma a sigma times i0 of sigma. i0 composed with sigma. That's how this induced map on the level of change was done. We just take the map to x and we compose it with the i0 into x times i. OK, and we want to end the same obviously for i1. And so now we want to prove that this is indeed the case. And you know, so in some sense, this is it. So the proof is difficult because you have to come up with the idea of using this. Now the rest is you just compute. And it's either true or false. And if it's true, then you have proven it. So how do we prove it? So what does it mean that it's a chain homotopy? It means that dp minus pd, always in the correct degrees. So I could call this, maybe I could call this pn, should be equal to i0 star minus i1 star or the other around. It doesn't matter. So that's what we have to show. So we just compute what we have here. So it's enough to show this on simplices, obviously. So let sigma from delta n to x, singular simplex, then by linearity, it's enough to show that if we apply this thing to sigma, then we get this. So let's compute. What is dp sigma? Now I forget the n because there's. Well, we know the definition of d and we know the definition of p, so we just write it out. So this is d of sum i equals 0 to n minus 1 to the i, sigma times the identity on i v0, v i, w i, w n. And you recall the definition of d, which is, again, an alternating sum where one removes the i's of it and takes this or the j's of the things which are written here and multiplies it by minus 1 to the j. It takes the alternating sum. So this is sum. And now there are obviously either one is among the first or one is among the last. So we distinguish these cases. So first the sum j is smaller than i. Do I want this one first? Then the sum j is bigger equal to i. So minus 1 to the i, minus 1 to the j, sigma times the identity on i v0, v i, forget vj. And then if we are, so this is for the first here, now if the number is bigger so that we get to the last one, one has to remember that this one is actually not the i plus first. This one is the i plus first is the i plus second. So it has an extra sign if one makes the alternating sum. So this is minus 1 to the i, minus 1 to the j plus 1, sigma times 1i v0, v i, w i, wj, wn. Because the sign take minus 1 to the which point in this list it is, starting with 0. So this is the jth one. But this is the j plus first. Because first we go until i and then we go on with i again. So this is what it is. So this is the notation which I use, but maybe it's too much of a short end. I sum with respect both to i and j. So maybe I should write it better. Although it is also standard. So it's sum i equal 0 to n, sum j smaller equal to i, and sum i equal 0 to n, sum j bigger equal to i. It's a double sum. And before I meant just wrote just this, and I meant I'm summing over both. But you know, that's maybe OK. So we just note the terms with j equal to i more or less the same. So if j is equal to i, then it just means we don't have this twice the same thing. Because we leave out one of them. So if you have j equal to i here, we throw away the first one. If you have j equal to i here, we throw away the second one. But the effect is the same. You know, we just get the elements that we have, and there's no doubling. But it's not so they cancel. And so these essentially all occur in both sums. But we see that the sign is different. So they will cancel. But it's not completely that they cancel because we have, so they cancel except for two terms. Namely, the case when i is equal to j is equal to 0 here. Then here, when i is equal to j equal to 0, then in the first sum, we have just from w0 to wn. In the second sum, we have the sum from v0, then w1 to wn. It's not the same thing. So one of them occurs in the second, the other one doesn't. So only one of them survives. So the thing which does not cancel here is in the case i equal j equal to n or i equal j equal to 0 or equal to n. And I claim that the only thing that survives is the following two terms, sigma times the identity composed with w0 to wn minus sigma times the identity composed with what now I'm quite sure I got it right. So I think I have to see. And then the other one is v0. So because all the other ones that we get here occur twice, once here and once here. But this one here will not occur in the second sum because we always start with v0 in the second sum. And this one here will not be in the first sum because the first sum, every element, starts with wn, stops with wn. So these are the only ones which don't cancel. But all the other ones, it's always it comes once here and once here with opposite signs. And so the sum, the terms here with i equal to j are only those. Why do I care about this? Well, we want to compute the difference. So I claim that if I do it the other way around, I get only the terms where i is different from j. So let's see why this would be true. On the other hand, if I take pd sigma, well then we do the same thing. First we apply d. So this is so here sigma is just so this will be as follows. We have p of sum j from 0 to n of minus 1 to the j. Sigma composed with E0, Ej. So this was how the map was defined. Now we have this simplex sigma, which is a map from delta n to this. And this is the composition with this base map. It gives us this thing. And we take p of that. And now we have to do the p. So the p is linear, so we have to apply it to each of them. And we do the same thing. So how does it go? So there are two possibilities. So I claim I can write it as follows. This is sum i equal 0 to n minus 1 to the i. So now we replace the sigma. So we also have this sum. So first we assume, again, the case that j is now smaller than i. Obviously, so let's see. So we take E0, Ej, Ei, Wi, Wn. So we take whatever we have. And we double put in the ith position. We put one up, and the other one we leave down. This is what it is. And I forgot, obviously, the whole thing that was written here. This was minus 1 to the i, sum j equal 0 to n minus 1 to the j sigma times the identity on i composed with this. And then, so I wanted to have j smaller than i. And now, so j is the one we leave out. So obviously, now, j cannot be equal to i, because i is the one we double. We cannot double the thing which is not there. So this is plus sum i equal 0 to n, sum j is bigger than i. So we have again minus 1 to the i. And we have again to see that, so we have to think about the sign anyway, but to this we can do in moment sigma times the identity V0, Vi, Wi. And we leave out the Wj until Wn. And so what about the sign? So you see that now we have not left out the jth one, but the jth plus first again, because this one is doubled. So this is minus 1, so the j plus 1. And so if you look at this, this is the same sum, this is the same thing here as what is written here, except to that we don't have the term j is equal to i. So we see that the difference between these two is precisely the terms with j equal to i, which is this. So we see this is the same as dp sigma, except for the terms with j equals to i, which are only in the first sum and not in the second. And so this improves, so maybe this is what we want to prove, you can leave it there. Over the opposite sign it does not matter. So this proves that this thing dpn of sigma minus pn minus 1p sigma is equal to this minus sigma times the identity v0n. And now you have to just remember how this goes. So I mean how everything was defined. So we had this thing that we have here. So we have that w, so vi is equal to ei comma 0 and wi is equal to ei comma 1. So thus it follows. So this thing here is just the image, just this thing is just the map. So w0 mu n is just the map e1 e0 to en. So this is in delta n times i. So the 0.1 here, and this is just the identity. And so this is just the identity. So in other words, this is identity of delta n times 1. And the other one, v0, is just the identity of delta n times 0. So every point gets mapped to itself times 1 here times 0. And so in other words, this is equal to, so therefore this map sigma times the identity of i composed with w0 to wn is equal to i1 composed with sigma. That's what it says. And in the same way, well, and so the map, so thus we see that this difference is equal to i1 composed with sigma minus i0 composed with sigma. So the sign is precisely the other round, which is the same by definition as i1 lower star of sigma minus i0 lower star of sigma. So we have shown that on simplices, this formula holds, and everything is linear, so it holds in general. So thus p is a chain homotopy from whatever, either e0 to 1 or e1 to e0. So e1 lower star to e0 lower star. And so therefore by the previous theorem, then e1 lower star and e0 lower star induce the same map on homology. And therefore by the case that one could reduce to this, it follows that homotopic maps induce the same map on homology. OK, so this was a slightly tricky proof. And you should maybe, the combinatorics of this sum is maybe also something one has to study with some, take a little bit of time to digest. But at any rate, it is not particularly difficult. It's just many little steps, and you have to always know where you are. But this kind of trick of making all these n plus 1 simplices out of the n simplex in this way, this is actually quite difficult to think of something like that. So now we have learned that if you have two homotopic maps between topological spaces to give the same map on homology, so then we can also ask ourselves, I mean, in what kind of other relations are there between homology and homotopic? For instance, we have this other invariant, which is the fundamental group. So how is it related to the homology? And this we want to briefly investigate now. I will not completely. I will prove a little bit about it and state the precise result. But I will not prove it completely because it's too long. But at least I will not miss out anything which is difficult. So relation between fundamental group and homology. So as before, I will briefly recall the definition of the fundamental group, although you just had it, but so we fix notations. So let x be a topological space. And we fix the base point x0 in x. So a loop in x at x0 is a map from the interval 0, 1, which into x, continuous map, which at both endpoints maps it to x0. So is a continuous map, say sigma from 0, 1 to x, with sigma of 0 is equal to sigma of 1 is equal to x0. So we have our space, have our point x0, and the loop is somehow a loop. And then the fundamental group is about homotopy classes of loops. So two loops at x0. Anyway, maybe I call them sigma and tau from 0, 1 to x are homotopic. So not when they are homotopy, not when they are homotopic, homotopic just as map to x, but also fixing x0, as you know. So they are called homotopic if there is a continuous map, maybe f, from the product of these intervals. So interval was also called i, after all, from i times i to x, such that if I take f restricted to, say, which we want to have wanted, maybe i times 0 is equal to sigma, f restricted to i times 1 is equal to tau. And in addition, at all times, the point x0 is fixed. So f restricted to 0 times i is equal to x0. I mean by this, it's the constant map which sends everything to x0. And f restricted to 1 times i is also equal to x0. So this is this homotopy. So if you want to make some kind of picture, we have this product of these two intervals, i times i, of the 0, 1, 1 here. And we have that, say, here, the map is sigma. In the bottom, in the top, the map is tau. It goes from here to here. Here, it's all the time x0. And here, it's all the time also x0. And this makes sense because, after all, both end points here are mapped also to x0. OK. And we also had some, there was some composition of loops. So if sigma from 0, 1 to x and tau, 0, 1 to x are loops at x, loops at x0, then we can define the composition. Maybe sigma composed tau. So I'll make this as a star. I don't know what the standard notation for you is. This is, again, a loop where we just run through one after the other. And we re-parameterize so that we get along from 0 to 1. So for instance, you say t is mapped to tau of twice t if t is smaller than 1 half and sigma of twice t minus 1 if it is on the other, in the other half of the interval. And as at the end point here, one gets always to x0, this matches and gives us a continuous map. One just does one after the other and re-parameterizes. And so the equivalence classes of loops at x0 in x form a group under homotopy. So we have this homotopy of loops. Look at the equivalence classes. Form a group, p1 of x, x0, the fundamental group. And if I write the equivalence classes, for instance, so I write something like that, gamma is the equivalence class of gamma. Then the group structure is just that if I take gamma times tau is defined to be gamma composed with tau. So this was how this was defined. And this is a group. In general, it will be a non-commutative group. So now we want to relate this to homology. And the reason why this will be related to homology, in fact, to the first homology group is that after all this i, the interval 0, 1 is just the standard one simplex. So a loop in x is a singular one simplex in x. So therefore, we have somehow an obvious maps which relate them. Then we have to see what these maps do. So note our definition delta 1 is just the interval 0, 1. Thus a loop in x at x0 or whatever is a singular one simplex in x. And there's more. So we have a loop, say, at x0, if you call it gamma. So gamma is a map from 0, 1 to x with gamma of 0 is equal to gamma of 1 equal to x0. And now, what is it if we take the differential of gamma as a one chain? So then d gamma is, if you remember, the end point viewed as a constant map, the beginning point, no, the end point minus the beginning point. So this is gamma 1 minus gamma 0, which means x0 minus x0 to 0. So in this case, I didn't have to remember who was first because they're equal, but still one should. So thus gamma is a cycle, c1 of x, so one cycle. So as it's a one cycle, we can compute, we can define its homology class. It's just the equivalence class, modular boundaries. So we have a map, which I may be called chi, from the set of loops in x at x0 to the first homology group of x, which sends gamma to its class in homology, modular boundaries. And we want to, and so we basically have the same space, we have the loops, certain equivalence classes of loops give us the fundamental group with respect to certain equivalence relation. And with respect to another equivalence relation, it gives us the first homology. I want to see how these two equivalence relations are related to see how the groups are related. And also somehow, the group structure should be related. And this is the subject of the following proposition. So if gamma, so there's two parts, if, say, gamma and tau are homotopic loops in x at x0. So we have two loops at x0, which are homotopic as loops. Then it follows that they define the same homology class. And the second statement is that this is also compatible. So this means, in other words, note this means we have a map from pi1 of x at x0 to h1 of x, a well-defined map, by sending the class of a loop to the class of a loop as a singular one chain. Because we know that if they are in the same class, they are also in the same class here. And the second statement is that it's compatible with the group structure. So if, again, so maybe gamma, tau again loops, then if I take the class of the composition, this is a class gamma class, a class of tau. So in other words, it means if we put these two together, it means we get a homomorphism from pi1 to h1 of x, a homomorphism of groups. So there is a homomorphism from the fundamental group to h1 of x. So I want to prove this proposition. And then I will just state what precisely is true. Because you would want to know you have a homomorphism. Is the homomorphism injective? Is it surjective? What is the kernel? And I will state afterwards. But first, I just have this. So in other words, I say it again, chi from pi1 of x at x0 to h1 of x, gamma maps to gamma is a group homomorphism. So the proof is somewhat similar to this homotopy invariance theorem, a little bit simpler. So we have two parts. So assume we have a homotopy between these two. So now these two change miraculously that names into sigma and tau. And then, well, anyway, it doesn't matter. So let f from i times e, i, to x be a homotopy. So this means that f restricted to i times 0 is, say, I don't know which one I want first. Gamma, how are they called? f restricted to i times 1 is tau. And f restricted to 0 times i is equal to x0. And this is also true for f restricted to 1 times i. So this was this thing. And we do the same thing as we did for the homotopy invariance. We, again, make a map from other simplices in the same way. So we define v0 equal to 0, 1, 1 equal to 0, 0, 0, 1. I hope I get it right. Yeah, anyway. So I'm not quite sure whether I might have exchanged these two factors, but I think you can manage that. So w0 equal to 1, 0, w1 equal to 1, 1. So we have this square i times i with v0, v1, w0, w1. And we have this map, f, which somehow maps from this. So x. And if I restrict it to the bottom, if I'm lucky, and I think this precisely means I always have to exchange these, then I get sigma, actually gamma. If I restrict it to the top, I get tau. And here's the constant map x0. And here's the constant map x0. And then we subdivide this thing in the same way as for the homotopy invariance theorem. So we define a two-chain, say alpha in c2 of x, by alpha is equal to f composed with v0, w0, w1, minus f composed with v0, v1, w1. So this means we have, so the map which maps the standard simplex first on this half is this map. The map which maps the standard simplex on this half is this map. And we take this one with plus. And so we take this one with plus, this one with minus, and we apply f to it, and this gives us a two-chain. So this is our two-chain. And we want to claim, obviously, that we want to claim that two things are homologous. So we want to show that the difference between the two, these two, is precisely whatever, tau minus gamma or the other route. And it's kind of obvious that this would be the case, because the thing in the middle cancels out. It's x0 constant. It will also cancel out. So we write it down. So what is d alpha? Where do I want to do this now? So this, I write it out, because anyway then. So what is it, d alpha? This is, how does one do d? One leaves out the i-th one and takes it with minus 1 to the i, where i starts with 0. So the first one is, so here we have this one. So we have f composed with w0, w1 minus f composed with, I leave out the w0, v0, w1 plus f composed with v0, w0. And then minus, and so we have a, so the whole thing minus, because we have a minus here. So we get a plus here and a minus here. And then we do the same thing here. So f composed with v1, w1. f composed with v0, w1. f composed with v0, v1. So these two cancel. And what else do we have? This one here is this w0, w1. So this is tau. So this composition, f composed with this is tau. This map, we are here on this side, is the constant map x0. Here we have w1, that is x0. And here we have v0, v1. That is gamma. So we see that indeed the difference of tau and gamma is a boundary, namely that of alpha. So we find tau is equal to gamma. So this proves the first part. In the second part, I will do slightly more, in a slightly more hand waving way. I just draw your picture, and you would have to turn this into a formula to show that the picture describes something continuous. So we have to find, you know, we want to prove that these two classes are that the class of this and the classes are equal. So we have to find, again, a singular two chain whose boundary is the difference between these. And we do this actually given by just one simplex. So let's say sigma from delta 2 to x be given as follows. And I will now just write it. So we take our simplex. I write it in the, you know, I draw it in this way, although the angles are maybe not correct. So this one is e0, e1, e2. So this is a two simplex. And now I want, on this part, I want the map to be tau. So the map from here, so I have a map from here to x. And on this part, the map should be tau. If I, you know, just, you know, this is basically interval 0, 1, the map should be tau. On this, it should be gamma. And then I draw a line here. And I just, you know, whenever I have a line here connected like this, like this, then, oh, it's correct way around. I will find out that it's precisely wrong way around. But anyway, so maybe I call this gamma and this tau. This will not make a difference. So the tau after all comes first. So then on this part, it should be tau, again, reparameterized. So if you take the map from the interval to here, which goes from here, then composed with the map here, it should be tau. And here it should be gamma. And it's always for all these lines, always until the corner. So I hope you understand what I mean. So here, we take, you know, a map from the interval 0, 1 to this line and compose with the map to x. And this composition should be tau. If we take, instead, this line, until this middle line here, then the composition of a kind of linear map from here to here with x, again, should be tau. And on the other hand, for this part, should be gamma. And then we compose this here by having here just tau, gamma, tau, OK? So you can somehow see that if you just take this composition here, you can really see that this is essentially gamma composed with tau. I just make a picture. So and then, obviously, you see, then if I take, what is d sigma? This is the map which you obtain by leaving out to 0. This is gamma, the one which you get by leaving out d1 minus gamma tau plus, so it follows that, OK? So obviously, so the exercise, so to turn this into a proof, you have to write down a formula for this. So as to show that in this way, I have defined the continuous map. So for sigma, so as to see, it is continuous. And that's kind of trivial. You just have to look at this picture and see what you have to do. And then this proves it. But anyway, here I didn't want to do the details. It's obviously often with homotopies that if you actually want to write down a formula, it gets a bit complicated. But I think, I mean, I hope you can imagine what I mean. No? I don't know, you look, I cannot know. I mean, I just say it and nobody says either yes or no. So what I just mean, I have here, I have the map sigma. If I restrict sigma, so I have a map here from, so I have a map, one map from the standard interval to this. For instance, if I just take the map from a standard interval to this with just maps, this with just maps in the obvious way from this 0, this 1, and then compose it with sigma, then I get tau. If I do it here, again, starting with E1 to E2, then the composition with sigma is gamma. And here, I do it in such a way, the map from E0 to E1. I take a standard interval here, the composition with sigma is this. And then I say, I can extend this to a continuous map on the whole thing by always saying that the map, which is the composition of the embedding of the standard interval into this here, into this part, if the composition of the standard, the embedding here with sigma should be tau, and here it should be gamma. And I claim, and I do this for every such. I have the whole thing is exhausted by these. And this gives us, the claim is this is a continuous map from this whole triangle to x. And it's kind of obvious that it should be so. But obviously, to prove it, you have to write down a formula. If you can write down a formula which is everywhere defined and continuous, then it is a continuous map. This proves this proposition. And then I want to just state what the result is. So without proof, so the precise relation between fundamental group and h1. So you can see, we have this map, chi from p1 of x, x0, to h1 of x is a group homomorphism. So in best case, you could hope that they are isomorphic. But this can in general not be the case because you know that h1 of x is commutative. And p1 of x, x0 does not need to be commutative. So that means that if you have a commutator of elements, it always lies in the kernel of this map. So if you write down something, so if you put k to be the subgroup generated by alpha, beta, alpha to the minus 1, beta to the minus 1. So the alpha and beta are elements in p1 of x, x0. So this is the subgroup generated by all such elements. So this subgroup is the commutator subgroup. Then because this group is abelian, it follows that all elements in k will lie in the kernel of this map. Because they are just mapped to the image of A plus the image of B minus the image of A minus the image of B of beta. So then clearly k is contained in the kernel of this chi. And the theorem is that indeed, k is precisely the kernel and the map is surjective. So chi is surjective and k is equal to the kernel of chi. So thus we have that h1 of x is isomorphic to the fundamental group of x. Maybe I should make some assumptions. Anyway, divided by k. And obviously the fundamental group does not work so wonderfully if the space is not path connected. So we assume x is path connected. So then the statement is this. And so you could also say, so often sometimes you say that if you take a non-commutative group and you divide it by the commutator subgroup, you get the abelianized version of the group. Because you get the biggest quotient, which is abelian. You throw away everything. So the first homology group is obtained from the fundamental group by making it abelian. OK, so this is this result. So the proof is not very difficult. You just have to, it's similar to what we did here. So for every thing in h1, you have to somehow construct a path which maps to it. And that's a loop which maps to it that's not very difficult. And it's a bit more, it's not particularly difficult, but it's a little bit lengthy to see that the kernel of the map precisely consists of those. You can see that this requires making some nice two chains of which you get that the difference has to be the boundary and so on. So that might be a little bit complicated. But it's all quite simple. And you can find it in the books. And at any rate, I just wanted to state what the real result is. So in particular, for instance, we get that the first homology of s1, so s1 is the one sphere. So the first homology of that thing is equal to z, because we know that the fundamental group is z. And if we abelianize it, it doesn't change anything. And we have that h1 of sn, sn is the n sphere, is equal to 0 if n is bigger than 0. And as we already know, to z if n is equal, bigger than 1, is equal to 1. So because again, we know that the fundamental group of this thing is 0, if n is bigger than 0. So this was as much as I wanted to say about this topic. Now I want to, so this was slightly more concrete than what we had before. Now again, we began a little bit more abstract. So we want to introduce relative homology. So if you have a topological space and a subspace, you have the homology of the topological space, the homology of the subspace. And there's actually something which kind of is in between these two, which is relative homology. And we will prove that they are connected by something which is called the long-exact homology sequence. And for that, we will need again some homological algebra. First I introduce the thing, and then we will go to this homological algebra, which is what? OK, so let's see. So we want to talk about relative homology. And the long-exact, I actually didn't give you the notes, no? Anyway, homology sequence. So after these lectures, or maybe tomorrow if I don't manage, I will give you the continuation of the notes. So let x be a topological space. And a in x be a subspace. So it means it's a kerosene-induced topology. We define the relative homology. So that's the definition. So we have to define some chain complex. So if we look at the chain complex, cn of a. So these are maps from the n-simplices into a. This is in a natural way contained in cn of x. This is just the maps from the linear combinations of maps from simplices to x, which happen to map into a. So cn of a is equal to the set of all some sigma a sigma times sigma in cn of x, such that sigma of delta n is contained in a for all sigma that we have here. And we also see that if we look at the map d that we have here, if we have d from cn of x to cn minus 1 of x, if we have a chain which happens to lie in a, so that the map actually goes to a, all these maps go to a instead of to the bigger space x, and obviously the d of it, which is just the restriction to some phases, will also map to a. So maps cn of a to cn of a to cn minus 1 of a. So in some sense, so we have this chain complex and cn of x with the d's and we have another chain complex. And this is really, in some sense, a sub-complex. So we can say it as follows. We say we define cn of xa to be the quotient. We know that cn of a is contained in cn of x. We can take the quotient group. And as the d here maps cn of a to cn minus 1 of a, we get an induced map d on the quotient. So if I write for the moment, say, alpha for the class of this in this quotient, then we can define d from cn of xa to cn minus 1 of xa by sending the class of alpha to, I mean, modulo cn of a, to the class of d alpha modulo cn minus 1 of a. So we get a chain and obviously, d composed with d is 0 because d composed with 0 was d composed with d was 0 to begin with. It certainly doesn't become more by taking the quotient by something. We take after the clause. So the class of d composed with d here is obtained by taking d composed with d applied to some alpha. And that's already 0. And if we take a class of it, it stays 0. So thus, cn of xa, so I say c star of xa with d is a chain complex. And the relative homology is just the homology of this thing. So denote, I don't know whether I need this or whatever, cn of xa to be the kernel of d from cn of xa to cn minus 1 of xa. And so these are the cycles, relative cycles. And bn of xa is defined to be the image of d from cn plus 1 of xa to cn of xa. These are the relative boundaries. And then the homology is, as usual, the cycles divided by the boundaries. So the nth relative homology xa is hn of xa, which is defined to be the quotient. So that was rather simple and relatively formal. So actually, but now we want to see how this is related to the usual homology of x and of a. And this is by means of some exact sequences. I want to show that the hi of x, hi of a, and hi of xa are closely related by a long exact sequence. I have to tell you what that is. Roughly, it says it's not quite true. What this will imply is that if you know two of them, you know the third. If you know two of them, you know the third. But in a somewhat intricate way. And so let's see what this is. So I talk about this long exact sequence. So as I do not assume that you know what exact sequences are, I have to review this first. And then I have to set up the homological algebra to this. I have defined, this was a definition. I just said that zn of xa was defined to be the kernel of d from cn of xa to cn minus 1 of xa. So and I had called in the setting where we don't have these two things, but just one, we had called this zn and I had called them cycles. And now if I have the pair of x and a, I say it's relative homology. I say it's relative cycles and it's relative boundaries. But this is just a definition. This is the name how I want to call it. So yes. OK. So you wanted to know what n cycles and n boundaries are. So I maybe can briefly review this. So you remember we had defined these groups cn of x, which were some sigma, a sigma times sigma, where a sigma is in z and sigma is a map from delta n to x. And the n refers to the dimension of the insimplex from which we map. This we call n chains. Now we have a map d from cn of x to cn minus 1 of x. And so we call zn of x to be the kernel of d. So this is some subgroup of cn of x. So these are also n chains. So therefore I call them n something. And I call them cycles. I mean, this is just the word that people call them. Such things, the kernel of d is called cycles. And as they are n chains, I call them n cycles. The property that it has to satisfy to be called an n cycle and n boundary is that it is a cycle or a boundary. So a cycle means that's in the kernel of d. And that's it. And the boundary means that's the image of d. So bn of x. These are called n cycles. So zn of x is the group of n cycles. And it's just defined like this. It's the kernel of d and nothing more. And bn of x is the image of d. This is also a subgroup of the same n chains. And it is called the group of boundaries. This, again, just a name that people give to it. I mean, it's not a name that I invented. And now we know that d composed with d is equal to 0. So it follows from this that bn of x is contained in zn of x. And so therefore we can take the quotient, which is the homology. So this is, with the cycles and the boundaries, these are just words. You just call it like this. For some reason, oh, I mean, this I said the other time. This I said the first time. But you could have asked then. But now I, so you see, these are formal expressions. So therefore I can take the formal sum. I mean, obviously, this you have to know. Otherwise it cannot do anything at all. So you have an element. So if you have a class alpha in cn of x, then alpha can be written as sum a sigma times sigma. So sigma and where sigma runs through. So each sigma is a map from delta n to x. But it's for different ones. So you just have this that you have a certain finite number. So you have to each map from delta n to x, to each continuous map from delta n to x, you associate an integer, an a sigma is an integer. So in such a way that only finitely many of these integers are non-zero. Because if you write sum, you always mean that the sum is finite. So you have for finitely many of these maps, you associate an integer. But you view this in such a way that you associate an integer to every simplex. And for most of them, the number you associate is 0. That's a simpler way of seeing it. Because then you can sum it very easily. If you have sum sigma alpha sigma times sigma, plus sum sigma beta sigma times sigma, how do you sum them? Well, like you would sum, if these were really sums of things, you just sum the coefficients. This is defined to be the sum over all sigma, a sigma plus beta sigma times sigma. So let's look at the, so we can, and the difference is, you can also take mine. So this is the sum in the group. And the way of doing, so the standard way how you would call this construction is you'd have, this is the free ABN group generated by these singular simplices. Just take the, and the minus the element, the additive inverse, you can easily see. So minus of sum sigma, a sigma times sigma, obviously is sum over all sigma minus a sigma times sigma. So let's look at a very small example, just to make sure. So obviously, if you look at a real topological space, I mean some of real life, then it will have very many points. And there will be very many maps from delta n to x. So the set of the sigma over which you sum is huge. It's some uncountable set. But you only take finite sum. So it's not, there are only finitely many coefficients non-zero. But let's look at a very small case. So assume that x is equal to set of two points. See, it's called, it consists of a point x and a point y. So this is a small case. So in that case, so what are, so if I want to look, what is cn of x? So delta n is connected. So here we have a, these are supposed to be a discrete set of two points. So it means that delta n can only map to x or it can map to y. So that means what? So we have two maps. We have the map from delta n to x, the constant map, which I could call, well, maybe I call it x underlined. So which maps delta n constant to x. Every point of delta n goes to x. And we have y underlined, which is the map delta n to y. So every point t is mapped to x. And here t is mapped to y. So what is then cn of x? Well, this sum, this is so, is a set of all formal sums. We have a times x plus b times y, where a and b are integers. That's what it is. And so it is just, and the addition is the usual one. You are just at the coefficients. So this is isomorphic to z squared. But obviously, if you take something like rn, and you want to say concretely what this is, this is some a being group which has uncountably many generators. So it's not so easy to write concretely what it is then. I mean, the miracle is that if you do the homology, so you take this thing, you take the d, and then you take the kernel of the d divided by the image of d, you usually get some finitely generated a being group. At least if you have a reasonable space. So as we for instance saw that if the space is connected, then h0 is equal to z. This is quite miraculous if you think what a big thing you take of which you take a quotient. So you take something huge, and you divide it by something as huge, and it turns out that the quotient is very small, and also still interesting. Anyway, so that's what we have there. So I hope this is a bit clearer then. Okay, thank you. So next time we talk about relative homology.