 Good afternoon. So I don't yet give you back your homework. I was a bit sick. And for the same reason, I don't have new homework. But you will still get homework maybe tomorrow, and you will also get back your homework. OK. One week? What? Well, I don't know. Maybe you will just give some small team. OK, I still have to decide. But to just one, at least one more exercise. OK, I think we can start. OK, so we had been talking about relative homology. So I recall that we had, so A was a subspace of x, the topological spaces. Then we have, but in a natural way, the n chains, single n chains in A are contained in the n chains in x. And we can consider the relative n chains, which is just a quotient. And then in the usual way, one has the differential d. In the usual way, one associates to this, the relative homology. And we had seen that, say, writing i for the embedding from A to x, we have the long x-sector homology sequence. So this was something like you have some hn of A, hn x, hn of xA. And then we have this. This is the map given by the push forward by i. And here we have some boundary map to hn minus 1 of A. And it goes on like this, a long x-sector sequence. And one thing that is useful is to know how to compute these, how to simplify the computation of this, because it's given to us in this rather complicated way. And this is given by the excision theorem, which says that, again, if x is topological space and z and A are subspaces of x, such that the closure of z is contained in the interior of A. So we have here x, we have here A, and we have here z. Then we can cut out z from both x and A and get the same homology. So then the hn of xA is isomorphic, hn of x minus C, A minus C. And this uses this theorem of the cover, which I want to recall. So we had defined cn. So if, say, A and B are subsets of x, such that the closures, such that the interiors cover x. So interior of A union the interior of B is equal to x. Then we had seen, or then I had claimed, that so we can consider this what was called cn of A plus B. Which was supposed to be, was defined to be the set of all n chains, which are linear, such that every simplex is either contained in A or in B. So it's a set of all sum sigma, A sigma times omega, just like some i, ai, sigma i, i equals 1 to n for some n. And the ai are elements in z. And sigma i is a map from delta n to x with either sigma i of delta n is contained in A or sigma i of delta n is contained in B. And by the way, notice that by definition, we have that cn of A is contained in cn of x and cn of B is contained in cn of x. And if you think of what this means, we just have that cn of A plus B is just the sum. No, not the direct sum, the sum. So these are all formal sums, which can be written as sum of a formal sum with things in A and with things in B. And then the statement was that the relative homology. So the homology, so I can just say it like that. If I take the homology of the chain complex of x with d, which is just the homology of x, is equal to the homology of a k. So this was the theorem of the cover. So you can compute the homology of a space by looking at chains such that each of the simplices is contained either in A or in B. So there are no simplices which go over both. OK. And then I wanted to say one more thing, which we had maybe I think briefly said at the end. So if we had introduced an exercise, the reduced homology, I will not repeat. One thing that we had seen is that if x is path connected, the reduced homology is 0. And then it's a remark and not difficult to prove that the long exact homology sequence also holds in reduced homology. So recall that for n bigger than 0, the reduced homology is the same as the usual homology. So there's nothing there. But as I said, if the space is path connected, then the 0's reduced homology is just 0. And that makes it easier to apply the long exact sequence because most of the spaces you look at are path connected and then it starts with 0 in the beginning. So you don't have to have so complicated arguments. OK. And now we want to be more or less, I will talk a bit long. But anyway, now we are at the point of continuing where we were. So I wanted to prove this invariance of dimension for this. I actually will have to use a theorem I will only prove later. So we'll prove below the following that I don't know whether or not. So this is, I don't know, I call it theorem, but that's maybe nonsense, that if I take hk of Sn, this is equal to 0. So Sn is the n-sphere, the usual n-sphere. So this will be 0 if k is different from 0 and n, and z if k is equal to 0 or n. And for the reduced homology, it's clear so that this thing is connected. So reduced homology will be 0 also if k is equal to 0. So we are going in two cases. So this I will prove later. But now I want to use it to prove the invariance of dimension, so which says that if you have open subsets of Rn and Rm, which are homeomorphic, then n must be equal to m. So the dimension of open subsets is preserved by homeomorphism, which is something which maybe seems obvious to you, but it's not so clear how y should be true and you can now prove it using homology. So let u subset Rn and v subset Rm be open and non-empty and u homomorphic to v. Then it follows that n is equal to m. You know, I think there are examples of continuous surjective maps from, say, R to R2 or things like that. So you can see that this is not a, but if they are homeomorphisms, then this cannot happen. The dimension is preserved. And this we do using this relative homology and the excision theorem. So we take a point. So we have assumed that x and that u and v are non-empty. Obviously, otherwise, there's also no statement. You cannot certainly conclude anything from talking about the empty set. So let x be a point in u. And now we look at the relative homology, hk of u, u minus x. We can use excision. What is it? I claim this is equal to hk of Rn. So we were u was in Rn, Rn without x. So we just take, in the excision theorem, we take z equal to Rn minus u. And then you see that the excision theorem applies because this is contained, the closure of this, which is actually closed, is contained in the interior of both sets of Rn minus x. So then we can apply the long x-sector homology sequence. And I will apply it to the reduced homology. I mean, one can also do otherwise, but it gets slightly simpler if one is reduced homology. So I take hk of Rn, maps to hk of Rn comma Rn without x, goes to now the boundary map, hk minus 1 of Rn minus x, goes to hk minus 1 of Rn. And then it goes on. So if you look at this, Rn is contractible. So all the homology, except the 0th homology, is 0. The 0th homology is z. But you can also, if you take the reduced homology, then just all the reduced homology is 0. Because it's path connected and the higher homology is 0. So this is 0 for all k. And this is also 0 for all k. So we find that this relative homology is the same as this. So for all k, we have that hk of the relative homology is equal to the homology of this. But now you have that Sn minus 1 is a deformation retract of Rn minus x. I mean, you could assume you just, or at least is a, I mean, you can assume that the point x is 0. I mean, it's just by shifting. So assume that, then you just have here the Sn minus 1. You have here the point x. And you just can, if you just make the homotopy from the identity on Sn minus 1 to the identity on Rn minus the point, if you are inside, you kind of pull here with a certain speed so that at 0 you are at 1 you arrive here. And if you're outside, you do it here, all along straight lines going to infinity. And you find that you have a, you get in this way a homotopy equivalence between the identity on Sn minus 1 and the identity on Rn minus x. So it follows that this here is actually equal to hk minus 1 of Sn minus 1, which we know. So we find that hk, what? What's the problem? So we find that hk of, say, now I go back to where we were, u without x is equal to hk minus 1 of Sn minus 1, the reduced homology, which is equal to, according to what we have written here, to 0 when k is different from n and z if k is equal to n. So if we take this relative homology, it will remember the number n, because it's the only part where the homology is not n. So if f from u to v is a homomorphism, then it induces an isomorphism from, say, hk of u, u minus x to hk of v, v minus f of x. And as you see, according to what we just proved, this is 0. I mean, both sides are 0 unless n is equal to the, so this one is 0 and z. And this is if k is equal to n. And this one is 0 or z if k is equal to m. And these things are always supposed to be equal, so it means n is equal to m. OK, so this was this application. So I hope that's clear. So now we want to, so this was one small application of the excision theorem and the long exact homology sequence. So now we want to come to a kind of even somewhat more useful tool to directly find an exact sequence for the homology without talking about relative homology. And this is the Meier-Vietore sequence. And this is in some sense maybe the most useful thing that we will introduce in order to compute homology. So this is the Meier-Vietore sequence. So this is after two mathematicians, one called Meier, the other one, Vietores. I don't know very much about Meier, but Vietores died very recently at the age of 105 or something years. And I think he still was publishing papers until very soon before, so he can have a long career. But this I think he did when he was rather young. So I mean, this is the Meier-Vietore sequence. So let x be a topological space. A and B open subsets of x. And we assume, I mean, it's just an open cover. So x is equal to A union B. And I just for security, I mean, to have the notation, I introduce all the inclusions. So I take EA from A intersected B to A, the inclusion, IB from A intersected B to B. And then we also have, say, GA from A to x, the inclusion, and JB from B to x, the inclusion. So that's that we can see what all the maps are. So then we have a long exact sequence. So it somehow goes on in the usual way, Hn of A intersected B. So it's always, so all the maps, except for the boundary map, are given by inclusions. So here this will be, this goes to Hn of A plus Hn of B. And this is just by the two inclusions. So IA, IB. So the map from this to this one is the push forward by IA. And the one to the other one is the push forward by IB. And then we have the map to x. And here we basically take the difference. So we take, again, the push forwards by these two inclusion, but we take the difference. So we take, say, JB star minus GA star. So if you have a class alpha here, we send it to JB star of alpha. So the push forward minus JB star of alpha. And this would be a class here. And then we have the boundary map delta, which will, again, as usual, bring us to Hn minus 1 of A intersected B. And it goes on like this. So now, in this case, we want to give some kind of a proof. This, again, follows from the theorem of the cover that we had. That's why I repeated it. So the proof, we have a short exact sequence of chain complexes. So 0 goes to D star of A intersected B. Goes to C star of A plus C star of B goes to C star of A plus B goes to 0, where this map, I call phi in this psi and phi, is just IA star IB star. So if I have a, if I take class alpha here, so phi of alpha, if I have a chain, so linear combination of singular simplices in A intersected B, I can view them as singular simplices in A and also singular simplices in B. They just take the sum. I mean, I take it here and here. So this is equal to alpha. And here, I can, for psi, it's the same thing. I mean, again, just translate this J. So that means if I take psi of alpha, if I have a singular chain in A, I can view it as a singular chain in X. Or I mean, I can just say I take. So I have now, here are two things, alpha better. And I take the difference, beta minus alpha. And why is that an exact sequence? So it's kind of clear that these are all maps of chain complexes. It's just the D is always the same D as we had before. This is the restriction of the D in X on C star X. And the D on C star B is the restriction of that to B. The D is always the same. So it's certainly, these are all chain maps. So the first step, so phi is certainly injective because we know that according to the definition, we just have that C star of A intersected B is just a subset of C star of A and also of C star of B. And the map here is just the inclusion. I mean, the inclusion is called I A lower star. But it is just the inclusion. And now, what else? It's obviously injective. So the composition is obviously 0. If I take psi composed with phi, so psi was with phi is 0. Because if I take psi composed with phi and apply it to some class alpha. So what do we do? I mean, we first send it to alpha, alpha. And then we subtract it from itself. So it cannot really be more obvious. And now, but we want to say that the image is precisely equal to the kernel. So if we have alpha beta, which lies in C NA plus CNB, if this is in the kernel of psi, what does it mean? It means that beta minus alpha is equal to 0. So it means beta is equal to alpha. But alpha likes in CN of A, beta likes in CN of B. So it lies in the intersection of the two. The section of the two is CN of A intersected B, if you look at the definition. And so we see that then it is equal to phi. And then finally, the subjectivity is by definition. If you look at the, we have just observed that. We had seen that CN of A and CN of B are both sub-modules, I mean, sub-rings of CN of X. These are all the simplices where this is contained here. And by definition, we have CN of A plus B is equal to CN of A plus CN of B, inside CN of X. Or if you want to, and so therefore, and this is precisely the image, which by definition is the image of this map. But to see it more explicitly, we can also say, so let more explicitly, if we take an element alpha in CN of A plus B, then we can write it as alpha is equal to sum over sum i, alpha i sigma i plus sum over sum j, beta j tau j, where alpha i and beta j are some integers. And sigma i is a map from delta n to A. And tau j is a map from delta n to B. Because that's precisely how this CN of A plus B was defined, that it's all n chains, which can be written as where all the simplices map either to A or to B. And I've now just divided them in such a way that I first write those map to A, and then I write those map to B, that I can certainly do. But that just means that alpha is equal to psi of what? Well, what are we supposed to do? We are supposed to write this as a difference of these two. So I could say sum i, alpha i, sigma i, comma minus sum j, beta j, tau j. Because if we take the difference between these two, this is precisely alpha. So the map is also subjective. So thus, we can apply the long. So now apply the snake lemma. And this precisely gives us the long exact homology sequence. Which one? I have not understood the question. Yeah? So I made it the wrong way around, yeah. Okay, so there was a sign around. Anyway, what matters is the principle. But I understand it's a, okay. So I want to also state without proof, but it's not difficult. It would be that the following remark, which is the same as for the exact homology sequence, it also holds in reduced homology. So, and again, this is practical because you have less case distinctions for the low homology groups. So the Meyer-Viertos sequence also reduced homology. And that's essentially an exercise to modify the proof here. Okay, now we want to give some examples. So first we have to prove the statement that I made at the beginning of the lecture that computes the homology of the n-sphere. I was using this for this invariance of dimension. So I should better prove it. So we have the n-sphere. So we have, as you know, have Sn set of all x0 to xn in n plus 1. Such that the sum of the squares of the coordinates is equal to 1. And so we want to compute the homology. So the claim was, after all, that, so I just say it for reduced homology. So hk of Sn is equal to z if k is equal to n and 0 otherwise. Okay, let's do it. So we want to use the Meyer-Viertos sequence. So we write it as a union of Sn as a union of two open subsets, of which the homology is easier to compute. So we write Sn equal to a union b, where a is equal to Sn minus, say, I could call it the north pole. So n is the point 0, 0, 1. And b is Sn minus the south pole. And I think you can imagine which one that is. And so we have this picture that, so we have here the sphere Sn. And so here we somehow have the north pole. And here we somehow have the south pole. And we have for a, we leave out this one point. For b, we leave out this one point. And then we also consider a intersected b, where we just leave out both points. So now I say that a and b are contractible. And a intersected b is homotopy equivalent to Sn minus 1. So maybe you can imagine this I will, so I make this some kind of exercise. That means I don't carry out the details. So check this. So we have, so basically we can, I want to claim. So we can retract. So we have a, so S is a neighborhood retract of a. n is a neighborhood retract of b. And there's n minus 1, which is the equator. It's 0 to xn minus 1 in n plus 1, so 0. Such that sum xi squared is equal to 1 is a neighborhood retract of a intersected b. And it's clear that e is homeomorphic to Sn minus 1. Yeah, a neighborhood retract. So S is a neighborhood retract of a. I introduced the last time what a neighborhood retract is. And OK, so I want to, it's an exercise, I say. But the exercise is just to carry out the detail. If you look at it, for instance, for a, so how does it look like, we have this picture. So here's the south pole, and we kind of throw, we take out the north pole. There's one point missing. So we can just make the kind of the homotopy by pulling it down along great circles. So each point lies on a great circle, and we just shorten it down according to n equals 0. It's the identity of this thing. At the point 1 in the homotopy, it will be here. And in between, for every point, we go half the way from the point until S along the great circle. And obviously, for b the same, and you would have to write down the formula to see that everything is continuous. And for a intersected b, it's basically the same story. We have now this thing. And we are now missing the north pole and the south pole. And again, we can just take, we can again just pull down along great circles, but only until the middle. So if we have any point anywhere here, if you are at the homotopy for t, you go so and so much along the way until you reach here. You can just write down the formula. This gives you a homotopy between the identity on this thing and the identity on the equator. So you have to write down the formula to see that it is a homotopy in a homotopy equivalence. And now, given this, we can apply the Mylvia tour sequence. I haven't done this very symmetrically to apply the Mylvia tour sequence. So what do we have? I wiped it out, but we have, at some point, we have hk of a plus hk of b maps to hk of sn, maps to hk minus 1a intersected b. And maybe one more to hk minus 1 of a plus hk minus 1 of b. So, and now we do everything with reduced homology. So in principle, we only have to put it here because the other ones will be the same. But for any k, we do it with reduced homology to make it easier. So we know that a and p are contractable. So their reduced homology is all 0. So this is 0 plus 0. And this is 0 plus 0. So we find that hk reduced of sn is equal to hk minus 1 reduced of, well, a intersected b, which we know is, which has the same homology groups as sn minus 1. And now we know, so for instance, we have that s0 is two points. The set consisting of the points 1 and minus 1. And you can easily compute that the reduced homology of a set of two points is z. So hk0, if k is different from 0 because it's just points. We know that for a point, all the higher homology vanishes have a union of path components. It's the whole homology is just the sum. And then we take the reduced homology, so we have one sum and less, and z of k is equal to 0. And the statement says that, therefore, by induction, by this formula, the result follows. Thus, by induction, hk of sn is equal to 0. Different from n, z, k is equal to n. I mean, one could have done the whole argument also without reduced homology, but with the usual homology. But then it's always a bit complicated because you do not have the just things are 0, but there's some z's and so on. And you will have some maps of which you have to show the injective or something. So it always is a shortcut to use the reduced homology. Because if something is path connected, the reduced homology, the reduced homology is 0. And that somehow makes things easier. So this was this example. I want to make a couple of small applications of this, which I think you already had done for n equal 1 using the fundamental group of the circle. We can now do it for all. It was the only thing for all spheres. So we have the following corollary. sn minus 1. So we write dn to be the unit sphere. So the unit disk in dimension n. So this is the set of all x0. Well, maybe I write now x1 to xn. Well, maybe I write that. Whatever. It doesn't matter. In rn, the sum of the xi squared is smaller equal to 1. So it's just the unit disk. And sn is the, now with sn minus 1, we take, we also changing the numberation. This is OK. It's the boundary. And the claim is that sn minus 1 is not a retract of dn. And the proof is, I think, something you already had with homo is very similar to what you had for homology groups. So assume we have such a thing. So let f from dn to sn minus 1 be a retraction. So that means it's continuous. And f restricted to sn minus 1 is the identity on sn minus 1. And we also look at the inclusion to dn be the inclusion. Then you actually had an exercise. I mean, you had an exercise which you have already solved, which says that i lower star, say, from hk of sn minus 1 to hk of dn is injective for all k. Now, for simplicity, I maybe will assume, I mean, I either would, again, have to do something with the do-so module. Otherwise, I assume that n is bigger equal to 2. It's an exercise. I mean, in the following, it's an exercise to conclude if n is equal to 1. So in this case, we see that, for instance, if we do this for the n minus 1st homology, we have that i lower star is a map from hn minus 1 of sn minus 1 to hn minus 1 of dn. But this one is equal to z. And this one is obviously equal to 0, because dn is contractable. I mean, everybody can write a homotopy which contracts this thing to a point. And so this is a contradiction. Obviously, it can be no injective map from z to 0. And so we deduce that there can be no such retraction. And from this, one gets in the standard way the Brouwer fixed point theorem that I think you also have seen already in the case n equal to 2. So every continuous map, f from dn to dn has a fixed point. So by this, I mean that is an x in dn, such that f of x is equal to x. And so how do we do that? Well, that's very simple. I first make a drawing. So I will anyway also just say it in words. You have to write down the formula. So let's assume we have such an f without fixed point. We want to use it to make a retraction of dn to sn minus 1. We do this as follows. We define dn to sn minus 1. And now I don't write the formula as follows by letting r of x to be the intersection point of the line from x to f of x with sn minus 1. So there will be two intersection points, no? But I take the intersection point on the side of x. So the picture is as follows. So we have here our sphere, the disk. And here's our point x. Somewhere is the point f of x. We just, you know, we are in rn. We can connect them by line. And the line will intersect the sphere in two points. And we take the one which lies nearer to x. We call it r of x, OK? And obviously, one has to check exercise r is continuous. By definition, r is a retraction. Because if we already are on sn minus 1, then the intersection point, if x is already in sn minus 1, then the intersection point with this line with this n minus 1 will be x. By definition, r is a retraction. And to prove that r is continuous, you have to write down a formula. And you know, if you write down some rational expression in continuous function, where the denominator is not 0, this will be continuous. So this proves r is fixed point 0. So let me see. I have half an hour. OK, I want to see that one more example. Just to show that we are now able to at least compute some homology groups. So say I call it g2, with a union of two circles which touch in a point. So this thing looks like this. And in fact, these two circles are called c1. So c1 and c2 are the two circles. And here we have intersection point p. And we want to compute the homology of this wonderful space. And so I again compute the reduced homology. Then you can compute from it the non-reduced one. So we do this by using, again, the Maya-Vietou sequence. So we said, so we have to write g2. So the thing which would be the easiest thing to do, which doesn't quite work because the assumptions don't apply, would be to take u equal to c1 and v equal to c2. And then apply the Maya-Vietou to this. This doesn't quite work because we have assumed that u and v must be open. And they are not open here. I mean, there is a sharper form of the Maya-Vietou sequence which actually would apply. And this would be OK. But we now have to use what we have. So we take u, so let epsilon be some small positive number. So I mean, really small. So maybe if the radius of this is 1, it's to be smaller than 1. And we take and let, so e epsilon p be a ball of radius epsilon around p. So we assume we are in R2 here. And we just take a very small ball. So we have here our small ball. And so we take u to be c1 union b epsilon of p intersected with g2 and v to be c2. So that is, we have this thing. And we have also this little thing here. And now it's easy to see. So first thing is note, if you take u intersected v, this is just this. And it's not difficult to say to see. I mean, you can do this as an exercise that this is contractable. In fact, p is a neighborhood retract of it. And we just have to write down some coordinates how to pull this together. And in the same way, we have that c1 is homotopy equivalent to u. So you can make a homotopy which pulls this extra part in. And c2 is homotopy equivalent to v. So now we are in the situation as if we had taken just u equal to c1 and v equal to c2. Because now we can again apply the Myovir-Tour sequence. Is it? So we have hn. So again, do it with reduced homology for making it easier. u intersected v goes to hn of u plus hn of v goes to hn of g2. Now that's the usual thing. You have the intersection, the direct sum of the open subsets goes to the union. And then we have the boundary map. We need it, the delta to hn minus 1 of u intersected v. And now u intersected v is contractable. So the reduced homology vanishes. This is 0. This is 0. This u is homotopy equivalent to this circle, c1. So this is equal to hn of s1 plus hn of s1. So we find this. So it's reduced. So which means that these two are isomorphic. We know that for s1, the reduced homology is z if n is equal to 1 and 0, otherwise. So we find hn of g2 is equal to 0 and different from 1, z, n is equal to 1. And in the usual way, we can know as g2 is path connected non-empty for the non-reduced homology, we find that hn of g2 is z if n is equal to 0 or 1. So nobody protested. I've just proven something else than what I said. I mean, I have proved it's z plus z, and then I try to tell you it is z. So you should complain. Anyway, and then for the non-reduced one, you get z also in case n is equal to 0, because it's path connected. OK, so we can see what I can do. One more example. So I do not think I will. So I want to do one more way of one more application, which is about attaching of cells. So what happens to the homology if you make your space more complicated by adding some piece, which looks like a disk, but connecting it wrong, the boundary. So let's see. So let dn, well, we have the dn we had now 1,000 times, but I mean, dn is the same as it always was. d0 to an minus 1 in n, so it's the sum in i squared is equal to 1, and it contains sn minus 1 in the usual way, the unit sphere. And we now want to say what we mean by attaching cells. So definition. So assume, so let y be a topological space. I mean, this is not, and let f from sn minus 1 to y be a continuous map. We want to somehow attach dn to y using this map, so that it's kind of only fixed along the sn minus 1 given by this map f. So we define y union over f dn to be, we take first the disjoint union of these two, model some equivalence relation. And it's with a quotient topology in the usual way. The quotient space where, so obviously, so the equivalence relation is induced by identifying a point in sn minus 1 with its image point in y. And everything else is not identified. So is the equivalence relation induced by p in sn minus 1 is equivalent to f of p in y. So if you want to do it explicitly, what it means is if you have a point in y, which does not lie in the image of sn minus 1, it's only, the equivalence class consists only of the point y itself. If you have a point in dn, which does not lie in sn minus 1, its equivalence class only consists of that point itself. And if you have a point in the image, then it's identified with any point which maps to it. No? And so this gives you, if you then give this thing the quotient topology from the disjoint union model of this thing, you get some nice topological space. And you can somehow think you have y here. And somehow this sn minus 1 gets mapped into y. And then you have to somehow, now this is not going to work, then you somehow have to add this dn just attaching it as this as a boundary. So we cannot really make a picture, but you have this thing here, which kind of sticks out. So OK. And now the statement is that again we get some exact sequence, which so lemma. So assume we are in this situation. So let f from sn minus 1 to y continues. And we have, we write now just yf for this space here. Then we have a long exact sequence. And I again will do it in reduced topology, because it is what works best. So say hk of sn minus 1 goes to, so by the map given f to hk of y goes to hk of yf goes to hk minus 1 of sn minus 1 and so on. And as I said, I wanted to do it with reduced topology. It also holds non-reduced, but with reduced it's better to use. So we have such a long exact sequence. So well, we just have to see that this is in a suitable sense an instance of the Myriviator sequence. But obviously, if it was just the Myriviator sequence, it would be with two trivial, you have to see why. So proof. And again, I will have to cheat a little bit, so to leave some of the more geometric sides as an exercise. So we have to write yf equal to u intersected v. So for instance, if we write, maybe I should write it here, we certainly have a map, or maybe write here, j is the map from dn to y union over f dn, which just sends a point here to its equivalence class. So this is a completely fine map. We see that if we, for instance, now take u. So we write it as union like that with u equal to, say, the image under j of the ball with radius 1 half. So this I just mean, you know, so what do I want? It's 1 to xn in dn, such that the sum of the xi squared is equal to 1 half, smaller equal to 1 half. So this is just some subset, and I think I have to take the open ball, so it's smaller than 1 half, because it's supposed to be open subsets. And v, I just take the whole thing minus the image of the center of the disk. So we have the 0.0 in the disk. We just take away this one point, make a little hole into this thing. Now clearly, yf is the union of these two sets, and they are open. So we can apply the Myriator's sequence, but there are a few things we can see. So j, if we restrict j to this disk, it's a homeomorphism, because there's nothing we identify once we are inside the disk. The identification is only on the boundary. So we have that this is homeomorphic to just the ball with this, and so it is contractable. So we have u is contractable. And what we did with more difficult to see is that v is a homotopy equivalent to y itself. Because if you think of it, once you have taken one point, you can again kind of, after all, I mean, one would have to write down this in formulas, and this would be an exercise. But you have this disk, vn, you have taken away one point. So the origin, you can certainly kind of retract this to the boundary by just moving out along the radii. And you can make a homotopy equivalence of this thing with y by taking the identity on y in this map on the n, and then taking the identification. And I claim you still get a nice homotopy, because you would have to work this out. So exercise proof this. So but once we have this, the statement is clear. Because again, this is a retraction. So nothing happens on the boundary. And the boundary is the only place where we have identified things. So it's in this picture, which I just wiped out, where we have this kind of something like a bubble on this thing. We just make a hole here, and we can contract it. And then we are left only with the thing we had before. So anyway, once we believe this, it's always simple. We apply the My-Vietto sequence, which says that if we take hk of u intersected v, this maps to hk of u plus hk of v, maps to hk of this wonderful space to hk minus 1 of u intersected v. And again, in the same situation, this space is contractable. No, no, no. So it's correct. But so u intersected r. So I haven't said what is u intersected v. Now look at the intersection of these two. We take this ball, and we take out this point. So this just looks precisely, again, like a smaller disk taking out one point. So that's homotopy equivalent to Sn minus 1. Sn minus 1. So this is just the k-thomology of Sn minus 1. u and v are so u is contractable. So it means all the reduced homology is 0. This is 0 plus. And we have seen that v is homotopy equivalent to y. And then we have hk of yf. And then we have hk minus 1 of Sn minus 1. So this is an example. So it somehow tells us that we have a way, if we, for instance, built our space by successively adding such pieces, we have a way to compute the homology in an inductive way. What? Ah, yeah, yeah, certainly. So we write yf as a union of two open subsets. Yes. We want to apply the Myviator sequence, and that is by doing that. And it was a misprint. But u intersected v. Here it is u intersected v is homotopy equivalent to a sphere. So now there is no further time. So one application of this result, which obviously will not be able to give you, because I first also would have to define this thing, is that one can now prove, for instance, if I'm in particular bad mood, I will make this an exercise, but we will have to see. So we can look at p and c. So this is the set of all, say, so this is cn plus 1 without the origin, divided by some equivalence relation, where we have that if we have some n tuple a0 to an is equivalent to p0 bn. So n plus 1 tuple of complex numbers is equivalent to another one, if and only if this vector is just a multiple of the other by a non-zero constant. So there exists a lambda in c without 0, such that, say, a0 to an is equal to lambda v0 until lambda bn. So we have this space. And one thing, there exists a lambda, lambda lambda, in a non-zero complex number, lambda, such that if I multiply this vector by lambda, I get the other vector. This is just the space of complex lines in cn plus 1. And what one can prove with this theorem, or also in other ways, but one can prove, so this can show. And this would be, but it is a bit complicated. I might make it an exercise if I give you enough hints, but I'm not yet sure. Namely, that hn of cpn, also hk of cpn is equal to 0 if k is odd. It's also 0 if k is bigger than 2n. And it's z if k is even and 0k. So here one would have a space where it is a little bit more complicated. So one case where you can check it is for instance, if you take cp, just to see that whether this might be so true, cp0 is just a point. And that's OK. You get just z for k equals 0, otherwise you get 0. And cp1, it's not so obvious, but one can prove that this is isomorphic to s2, not s1. s1 is over, if you have r, we are here over the complex numbers. And then again, you will find that this is true. But in general, you get this. So we have now a space where there are also a few more homology groups which are non-zero. But anyway, a much harder exercise would be to do the real projective space when the answer is also much more complicated. But that, I think, is almost impossible anyway. OK, so that's, I mean, I don't know. So that's maybe all I want to say now. So I have tried to introduce some of the main results and to do some examples. And we had very little times we didn't get very far. But anyway, I hope it is enough to at least get some impression and to be able to compute some examples and some applications. If one, I mean, algebraic topology is a big field. So if one really wants to study it, then one, you know, if you look, for instance, at the book of Hedger, which I partially use, I think I covered about maybe 25 out of 500 pages. So, but anyway, that's what is just the first impression. But I mean, it is, you know, you have seen some of the methods. OK, maybe I stop then, and then we will see. No, no, you? Well, I mean, that's one thing. I don't know, this may not. I mean, algebraic topology is not only about that. So I mean, it's one thing. One can study whether two spaces are homomorphic. Or you can also see, after all, we have proven that homology is invariant under homotopy. So it's not really about homomorphic. But even if they're homotopy equivalent, you get the same homology. So you can decide that. But you know, sometimes you can prove that certain things are impossible, like with this Rauer-Fick's point theorem and so on, because you can do certain things. It's not just about things being homomorphic. But in some sense, you can always think like that. If you want to decide anything in topology or in any other subject, you can also always reformulate this as a statement about maps. So about continuous maps or something like that. And so the homology, if you have a continuous map, it will give you a map on homology. And sometimes it's impossible that such a map exists because the homology groups do not fit. And so you can see that certain things are impossible because you cannot do that. And sometimes you also can find the converse. And then one thing which I did not say is that homology actually has something to do with the fact what the space looks like. I mean, it's not clear here, but I had said in the very beginning something that if you have the zeroth homology tells you how many pieces there are, so you can really see it. The first homology tells you something about certain kind of holes. I mean, you can kind of imagine that if you can make a simplex around the hole in your space, then it is a boundary if you can somehow fill it up in the space. And so you get non-zero homology if there will be a hole. And there's some way how you can make this kind of precise. And there are also other things. There is co-homology, which is dual to homology. It's not much more difficult than homology, but it has some extra features. And one of the things you have with homology, and this then is, for instance, also related to dirham homology. It somehow turns out to be the same as dirham homology, something with differential forms. And dirham homology, for one thing you could say, it's just another way to compute homology, which is most as the same as homology. But in fact, dirham homology is about solving differential equations. So it means homology or homology tells you something about solving differential equations. You can, by computing, doing certain computations in algebraic topology, you can decide whether certain differential equations can be solved or not. Certain partial differential equations are many forms. And there are other things. So often when you have some space, you can make some construction on top of the space. And this construction will give you a certain homology class. And so then maybe there are some reasons why this homology class cannot exist or cannot do what you see. And so you see that this construction that you would like to make cannot be there. So there are many cases. It's not just about two things being homomorphic. That's just the kind of causes thing. But it's kind of a more universal tool to, you always want to associate some groups to your question and some homomorphism between the groups to the question. And then you have translated your problem into a different one.