 So good afternoon. So now we have the second lecture on algebraic topology. So we had introduced the singular homology. So that means we had to introduce the chain complexes. So that means one has some maps, Cn plus 1 goes with D to Cn, goes with D again to Cn minus 1, and so on. And then we had to define the homology of such a chain complex. So we had h, say n of C star, would be the n cycles divided by the n boundaries, where the Cn of C is the kernel of D from Cn to Cn plus 1, n minus 1. And the Dn is the image. And to make this a chain complex, we need that D composed with D is always 0, which decides we need that this image will always be contained in the kernel. This definition makes sense. And for the singular homology, we had to define this chain complex by making Cn of x, where x is the topological space, to be a formal sum, some sigma, a sigma times sigma. So this is a finite sum. A sigma is an integer. So finally, many of the a sigma are non-zero. And the sigma is a singular n simplex, so a map from the standard n simplex to x, a continuous map. And then we had to find some differential on that. And the homology is the homology of that. And we were now considering, I mean, trying to prove that this is invariant so that this behaves nicely for continuous maps. And for this, we had first looked at it more abstractly again in this context of chain complexes. So if C star D and D star D are chain complexes, then we had defined a chain map, f star from C star to D star is a map at all levels, so is given by, say, fn from Cn to Dn for all n, such that it commutes with a differential. So if we have this map, so this is fn, this is fn minus 1, this commutes. So if you go this way, this way, you get the same as if you go the other way around. And we had seen that if we have such a chain map, then we get an induced map on homology. So if f star from C star to D star is a chain map, then we have a map which I call also f star from Hn of C star to Hn of D star is induced map, which is given by, if I take the class, the homology of the cycle alpha in DnC, then this is just mapped to the class of f star, fn alpha, if you want. OK, this is where we were. And now we want to apply this to show that homology is functorial, so compatible with continuous maps. So this is the following result. So well, first, we want to define the push forward map to let f from x to y be a continuous map. Well, there's an obvious way how you can make out of a singular simplex in x, a singular simplex in y, just by composing with f. So if sigma from delta n to x is an n simplex in x, then f, I mean, I don't know whether it's definition, f composed with sigma. Obviously, from delta n to y is an n simplex in y. And obviously, we can use this to define the push forward map on cycles, on the singular chain complex. But just so we define fn from c star of x to c star of y in the obvious way. Namely, if we have a cycle here, this is of the form sum sigma, where sigma runs through the n simplices, the singular n simplices in x, a sigma, which is an integer times sigma. Well, if you want to apply fn to it, the only thing we can do is we apply it in some sense to sigma, and the way we apply to sigma is by composing r sigma times f composed with sigma. And this, obviously, is a map, and it's straightforward to see. So if you want an easy exercise, why it's not even an exercise, it's trivial, if we have this f star given by the fn is a chain map, c star of x to c star of y. Because obviously, it's compatible with taking difference, if you remember the definition of d. So thus, we have a push forward homomorphism. So this is a chain map. So thus, there is a map, a group homomorphism, fn, f star, from the nth homology of x to the nth homology of y, therefore, the continuous map, which is given precisely as we had the induced map on homology for chain homomorphism, because we apply. Namely, if you have a cycle alpha, this is mapped to fn of alpha. So the class of the cycle alpha is mapped to the class of fn of alpha. So if you want to write explicitly, the class of some a sigma times sigma is mapped to the class of some sigma a sigma times f composed with sigma. And this f star, and this is true for all n, f star is called the push forward. So thus, we have now associated to every topological space sequence of groups, and to each continuous map between topological spaces a homomorphism of these groups. So we somehow see that things start translating into algebra. So we can see that also by definition. If f from x to y and g from y to z are continuous maps, then if I take the push forward by the composition, this is the composition of the push forwards. And this is completely obvious from the definition, because this is just given by composing with the map. So if you compose with the composition, you compose twice. So this is real, and it's also obvious. If we take the identity on x, with this 1x and just the map, it sends every point of x to itself. So let 1x from x to x be the identity on x. Map on x. Then clearly, the push forward by the identity is the identity. Now this is also if you compose with the identity, you get the same thing. So this is usually what one says that if one knows the language, that the homology is a functor from topological spaces from the category of topological spaces to the category of groups, being group, whatever. But we will not use this language, but we can use a simple corollary that homology is invariant under homeomorphism. So if f from x to y is a homeomorphism, so a continuous objective map, which has a continuous inverse, then it follows that the push forward by f from hn of x to hn of x is an isomorphism of groups for all n. And this is obvious from what I wrote here. I spell it out. So f is a homomorphism, so we have the inverse homomorphism f to the minus 1. So we have the identity on the enthromology of x is, as we know, as I just wrote there, the push forward by the identity on the enthromology. Now we have f of f to the minus 1, but if we first apply f and f to the minus 1, we get the identity. So this is the same as the push forward by f to the minus 1 imposed with f, which is f to the minus 1 star composed, f star. So what this says is that we find that this composition is the identity on enthromology. And obviously, the other way around, we also get that the identity on the homology of y on the homology of y will be f star composed with f to the minus 1 star. So we find that indeed f star is an isomorphism with inverse, the push forward by the inverse. So this is completely standard. Now we want to finally compute the homology in one example. And precisely in the simplest example that exists, which is one point. So it's kind of comfortable. I mean, until now we have this rather complicated definition, but we haven't computed any case at all. So we want to compute the most trivial thing we can imagine, which is a point, for example. So let the point be the space, so point be the space consisting of a single point. It's a set with one element. Obviously, this has a unique topology, in which this point is both open and closed. So now we want to compute the homology of this wonderfully complicated space. So first, there are not very many maps from any space to a point, namely always one, which maps everything to that point. So whichever space you have, the only map from that point is the map which sends every point of the other space to this given point. So for all n, there is a unique map, which I maybe call sigma n from delta n to this space. Maybe I call this b. So which sends, namely, the constant map. Any x here is sent to this point. For all n, this map I call sigma n. It's just a constant map. So from this, we can already compute what the chain complex is, what the singular change is. So we get cn of x is the set of all linear combinations of all possible maps from sigma n to point. So this just means all multiples of this one map. So this is just z times sigma n. So this is cn of x is isomorphic to z. The elements of it are a times sigma n, where a is in z. And so for all n, and what is the differential? So if I take d of sigma n, then we have to recall the definition of e, which I did not at the beginning of the lecture. But how did it go? This is sum i equals 0 to n minus 1 to the i sigma n composed with this map e0. We leave out i, where this is a map from delta n minus 1 to delta n, which sends the linear combination of the e0 to en minus 1 for the corresponding linear combination of the 0 to en, where ei does not figure it. And what is this? Well, this composition, whatever it is, is always sigma n minus 1, because this is the only map that exists from delta n minus 1, so x. So this is just an alternating sum of sigma n minus 1. And so this will be depending on whether this number's even odd. Sometimes plus and minus cancels each other out. Sometimes they don't. So this will be 0 if n is odd. Or if n is equal to 0, because this was just trying to be 0, in case n is equal to minus n. Yeah, it's just not there if n is equal to 0. And so we don't have anything at all. And the other case, it is just if you look it up, sigma n minus 1 if n is even and different from 0, or maybe n bigger than 0. And I could also say, in principle, OK. So here you can just see if it's even there, one remains, all cancels each other out of one. Well, so what does it mean? Well, so if you look bn of point, so I claim that this means, if I take the boundary, the singular boundaries, this is equal to n bigger than 0. Because whenever, so this map will be either in the even case it will be subjective, but then the next map is 0 because all it is all the other around. So in every case, the boundaries are precisely the cycles. If you just look at what this gives. There's always one 0 map, and then basically the identity from z to z alternated. And in the case n equal to 0, we actually find that we start, so c0 is z, but the map is 0. So the zn is z times sigma 0, but it's not the boundary. So we have that, but p0 of point is equal to 0, z0 point is equal to z times sigma 0, so that's it for us. So the result, therefore, is that the homology of a point is equal to 0 if n is different from 0 and z times sigma 0, or just z, if n is equal to 0. So we have succeeded in computing the homology of this very central space. So you just have to look what this means and see the obvious. No, just can you, what does h0 be 0? That doesn't make sense. I mean, what can you just say, maybe you formulate in a different way so that I can understand. Yeah. What does it measure? OK. So the question is, h0 is the 0 homology. What does it measure? And you are very lucky because that's precisely the next result that we are doing because the h0 is always z to the number of path components of the space. So in this particular case, a point has one path component, and so h0, therefore, is z. And that is precisely what we do as a next step. So this is, so I'm kind of lucky. So anyway, so the answer to the question is the following. So first, so this is a bit more general. So if xk, k is the set, so let xk, k be the set of path components of a topological space x. So we say that if you have a topological space, we say that two points are in the same path component if you can connect them by a path. So map from an interval x, and there are different path components if you cannot, like here. Then the first statement is that the homology of x for all x is just the direct sum of all the homologies of the path component. OK, so just have an, if you have a kind of obvious so that elements in the homology of x are the sum of elements in the path components. If you have k, if you have the two path components, x1 and x2, then the nth homology of x is the nth homology of x1 plus the nth homology of x2. But in principle, it also holds if you have infinite many components. But you don't have to worry about that too much. And the second statement is that if x is path connected, then the 0th homology is just z, isomorphic to z. So in particular answering your question, so it follows that if x has, say, k path components, then h0 of x is isomorphic to z to the k. So we can really say that the 0th homology measures precisely the path components, the path components. It just measures how far x is away from being path connected. And so the way how one finds out anything about connectedness properties of the path connectedness of space is by homology is decided by computing the 0th homology. OK, so this has these two parts. So the first one is in some sense trivial. I only have to see whether I can convince you of it. So proof. Well, we have this n-simplex, delta n, which is a simplex in Rn. So certainly it is path connected. It's actually a contractable space. So if sigma from delta n to x is a singular n-simplex, then it's the image of the path connected space by a continuous map. So then the image is path connected. So it's path connected, and therefore it lies in one connected component of x. And so furthermore, if we take d sigma, then all the, then this is a linear combination of n minus 1 simplices of singular n minus 1 simplices which lie in the same path component. As is so where an image where sigma composed of n is a singular simplex in the same path component of x. So if then we write x of xk, where the xk are the path components, then we see that every singular simplex lies in one of the xk. So if you have a chain, it's just a direct sum of change in each of the xk's. And the differential just sends something which lies in xk to something that lies in xk. So we get that c star. So it follows that cn of x is just a direct sum over all k of the cn of xk. The d reflects this summation with something like the cn of xk, then d of it lies in cn of xk. So it follows that also this is true for cn and dn. That would be not so n of xk. And then as the homology is the quotient, we get that the homology of x is equal to the direct sum of the homologies of the path components. So maybe I did it fast, but it's really trivial. After all, a singular change is just a formal sum of simplices. And each of the simplices has to lie in one of the path components. And everything is just a direct sum. So now we come to the second part, which is more interesting. So we want to see that the zero homology is z. So we have to somehow construct an isomorphism. And so how do we do that? So for any point p in x, I write by abuse of notation if you can handle it, if not then I could change it. And I write p also for the map from delta n to x, which sends all points here to the same point p, so the constant map. So I just call the constant map to send everything to p, also just p. So this should not lead to confusion, but if it does, then OK. So if you have a point, say x in x, and now the point is called x, well anyway. So let's say for point p in x. So let's choose a point, say x0 in x. Now we have our x, maybe we have any point p here at x0. Then x is path connected, there will always be a path connecting them. So for p in x, let's say sigma p from delta 1, which is nothing else according to our definition in just the interval 0, 1. So x be a path, so we choose a path from x0 to p. So this path here is sigma p. Then if you remember the definition of the differential, the differential of this thing is p minus x0. So the constant map p minus the constant map x0. So then by definition, if you look what the differential means, in this case where you just have one, there's just two points, the beginning point and the end point, the differential is the end point minus the beginning point. The definition is d from sigma p is equal to p minus x0. So for every path, we get that the differential of the path of this difference, this formal sum. So now we define a homomorphism epsilon from the 0 chains of x to z, which is if we have a 0 chain, this is a formal sum, I may be calling it p. So take the points a p times p. This p is the singular, this map. Ah, yeah, I actually don't need that. It's a very constant map. It's just the map that sends the point to p. So that's the way I want it. I didn't want it for all n, but just for n equals 0. Sorry for that. So maybe I can repeat. So we defined the p to be the map which sends the one point that this thing has, which is the point 0, to p. So somehow it's quite reasonable to identify the map which sends one point, the one point of this thing to one given point with this point, because that's precisely information that it contains. And then so this was our p. And now given a point, we find that we can make a path to any given point as the differential of the path need as the one chain is precisely the different point minus the kind of base point we chose. Now we define the thermomorphism. So if we have this, we just take the sum of the coefficients. Now by definition, the chain is a finite subsum. So only 5 to many of the a p, which are integer and non-zero. So this makes sense. The finite sum of things. So now we want to prove that the homology is isomorphic to z. So we know that every element in c0 is also a cycle, because the differential for the next level is by definition 0, because below 0 everything is 0. This is also the same as c0 of x. So then the claim is that the kernel of epsilon is equal to p0 of x. And note that this proves part 2. It proves 2, because epsilon is obviously subjective. No, we can write down anything. We can get whatever we want. We just take one point and take any coefficient. Then we get that coefficient. So the map is certainly subjective. And if we have a map from v0 of x to z, which is subjective and this kernel, if I see this, certainly is an isomorphism, which is an isomorphism of z of the quotient of v0 by v0. z0 of x, I'll just divide by v0 of x with the map induced by epsilon to z. So this is whatever, the homomorphism theorem. And after all, this thing is h0 of x. OK, so we just have to prove this claim. Well, if you have to prove that two sets are equal, you have to prove both inclusions. So that's what you learn. So let's try both inclusions. So let's take, say, an element equal to some p, a p times p, p an element in the kernel of epsilon. So then we have to show that it is a boundary. That is d of something. Well, that's quite simple because we can write it like that. So this is certainly alpha. But we can also, you know, the sum of all the a p is 0. That's precisely what means it's in the kernel of epsilon. So we can add 0 for this thing. It's still the same thing. So this thing is actually 0. But now we see if we look at this thing, this is just the differential of this. So this is equal to the sum over all p, a p times the differential of sigma p, which is the same as the differential of sum a p sigma p. So we see this is indeed a boundary. And the other direction is even slightly easier. I think so if, say, sigma from, again, delta 1 to x is a singular one syntax, which is the same as a path, then what? Well, obviously, we have that by what we wrote here, that the differential of sigma is sigma of 1 minus sigma of 0. This is precisely what it says, which is the path from sigma of 1 to sigma of 0's differential is this. So in particular, so maybe to be sure, if I write something like this in the other notation, you could, like here, you could also write this as plus minus this. So this is what it means as a linear combination. Also here, I could say this is this plus minus 1 times this. So that's what this notation means, like it means in high school mathematics. So in particular, if we take epsilon of d sigma, then this is the sum of the coefficients here. So this is 1, it's minus 1, so this is 1 minus 1 equal to 0. So if you have any cycle, so if alpha is equal to sum sigma a sigma times sigma is a one cycle, is a one chain, then. And we take the differential, then so epsilon of d alpha is equal to epsilon of sum sigma a sigma times d sigma. And epsilon is also just linear, so it goes in here. So this is sum sigma a sigma times epsilon of d sigma. And we know that for every one simplex, epsilon of d sigma is equal to 0, so this is just the sum of 0. So we see this includes. So we have that indeed, v0 of x is contained in the kernel of epsilon. And so this is for position. So now, so much about this. So we have at least found out what the 0 homology computes. And we have been able to compute the homology of a point. So now we want to come to the first theorem in this lecture. So the first time one really has to make some effort. So until now, I mean, I expect you have to have to make quite a lot of effort to, I mean, because there are so many definitions that one has to deal with. There are lots of definitions. It's difficult to keep them in one's head. But the things that we do with the definitions are completely trivial. It's maybe difficult to notice because there are so many, everything is new. But now, we want to do something where one actually needs an idea. And so this is the homotopy invariance theorem. So it might be that the proof is actually not more difficult to follow than the other proofs. But these, if one gives you all the definitions and you understand them and tells you what the result should be, you can prove it as an exercise. But this one is not an exercise. So the homotopy invariance theorem. So we want to show that homology groups are homotopy invariance. So what does it mean? So first, I recall to you about homotopy. I mean, you have had it very recently, but I need to also fix the notation of the definition. So I write now also I for the interval 0, 1. So let f and g be continuous maps from between two topological spaces. So a homotopy f from, I write it just like this, from f to g is a continuous map from, which we want to wonder, x times i to y, such that if I restrict to the part of i, where if one is 0, I get f. If I restrict to the point here being 1, I get g. So I write it in a slightly different way because I want to use this notation in the proof. So such that if I take the map i0 from x to x times i, which sends x to x, 0, and i1, corresponding thing on the other side, if I have these two maps. So for these maps, we have that f is equal to large f composed with i0, just means the same thing I restrict to x times 0. And g is equal to f composed with i1. So this is a homotopy, and you have learned that before. So if we have a homotopy, we call the two maps homotopy. So if there is a homotopy, f from f to g, we call f and g homotopy. As we're at it, I introduce a few other words that you already know. So to say what it means to be. Homotopy equivalent, and then as a special case, contractable. I expect we have learned all this. I don't know who taught with Zimmermann or Dillegas, but anyway, just one notation on the table. So two topological spaces, x and y, are called homotopy equivalent. And if there are maps which are inverse to each other after a homotopy, so if there are continuous maps, I hope that's correct, f from x to y, g from y to x, such that if I first go from x to y and then go back, this is homotopic to the identity on x. And if I do it the other way around, it's homotopic to the identity on y. And I call if x is homotopy equivalent to a point, I say it's contractable. So to the space which I call point, a point, we call x contractable. There's somehow a homotopy which contracts the whole of x to a point. So now, we want to prove this theorem that if two maps are homotopic, then they induce the same map on homology. We know that if we have a continuous map between topological spaces, it induces the homomorphism on the homology. And now the statement is that which map the induce depends only on the homotopy class of the map. So this is the US data theorem. If x and y are no, so let f and g from x to y be homotopic, then if I take the push forward map by f, then it is equal to the push forward map for g, which has a homomorphism from hn of x to hn of y. So homotopic maps give the same map on homology. You have some easy. So in some sense, it says that the homology is not such a super fine invariant. So it only sees, for instance, it only can distinguish the homology groups up to homotopy equivalents. So if two topological spaces are homotopy equivalent, they will have the same homology, where that comes in a moment. So that tells us that there's one thing that they are not. They don't contain all the information about our spaces. On the other hand, it maybe tells us that they are somewhat easier to compute because they only depend on this kind of data. So corollary, if x and y are homotopy equivalent, then the nth homology of x is isomorphic to the nth homology of y for all n. What? Can we use this film without proof? What does it mean? I mean, we use it without proof. No, I first give the corollaries, and then I give the proof. So I'll give you some kind of a, I mean, you want to know that the film has some applications. So I first give the application, and then I give the proof. So I think you can figure out yourself why this is a corollary. You have this, if it's a homotopy equivalent, so that the composition one direction is the identity and the other it's the identity, so that precisely means that if you do it for, you get the same thing for the push forward, and that's it. I can maybe make this, and if you cannot do this, then maybe you should already do it. So the second one is, which is a special case, that if x is contractable, then we have that hn of x is equal to 0, if n is different from 0, and it's z if n is equal to 0. This is because we know that this is how the homology of a point is, and if it's contractable, it's homotopy equivalent to a point, and therefore it follows from the previous corollary. So and now we want to come to the proof of this theorem. And in order to do this proof, we first, again, need to go back a little bit to the homological algebra. Maybe I start with a new thing here. So we need some small amount of homological algebra. Again, we have to talk about change complexes and so on. And so what we need now is a change homotopy definition. Let say f star and g star, c star d to d star d. This means that at each level, we have fn from tn to dn. Commuting with dn, we change homomorphisms. That means we have cn, both fn and gn, tn for n. Commuting with a d. So a chain homotopy from f to g is another sequence of maps which now do not, which go a little bit diagonally. So is a sequence, maybe I call them sn, from cn to dn plus 1, such that then we have this strange formula that somehow interpolates between, if we also use the differential, it somehow interpolates between f and g, such that if you take d composed with sn minus sn minus 1 composed with d, this is, so now I want like this, gn minus fn. So let me write this also in diagram form so that you have some. Anyway, we can see what our theorem is. So we have here, say, if you have cn minus 1, cn, if you want 1 cn plus 1, each time d. And we have here the same with the d. And now at the, so we can do the following. I'll write it here. So what is this map? So we have here the map sn. Here we have the map sn minus 1. Here we have both the maps fn and gn. And now, so this is supposed to be for all n, obviously. And so now we want that if we do, say, if we first go here, and then here, and we subtract first going here, and then here, we get the difference between these two maps. So the difference between the two maps, I mean, after all, these are homomorphisms of the Boolean group. So I can form the difference. The difference between these two maps is the difference between commuting these maps with a differential in the two different ways. So this is a chain homopop. And why would we be, why would this be of any relevance? The point is that if we have a, the claim obviously is that if you want to use this, that if you have a chain homopopy to chain maps, then they induce the same homomorphism on the homology. And then we will use this to prove our theorem to show that if we have a homotopy between two maps, then we have a chain homotopy between the induced maps. I mean, between the induced things on the chain. So let's see. So we first have to prove this little lemma, which is actually quite trivial. That's one of the features of homological algebra that most arguments are extremely simple. It's only that there are very many very simple arguments. So for position, let f star and g star be chain homotopy. So from, say, c star to d star, the chain homotopic chain maps, chain homomorphisms, then, well, if I take the map on homology, it's equal to the given map of homology. And you'll, OK, in order to prove this, we have to remember how this map was defined, the induced map. So how was the map given? So f star of the class of an n-cycle alpha, so if alpha is in zn of c star, then f star of alpha was just the class of fn alpha. That's how we have to find it. And you'll have to see that this is where you find. So first, if alpha is a cycle, then fn alpha is a cycle 2. And if one changes alpha by boundary, then also this gets changed by boundary. But this is kind of trivial and trivial. So now we can want to see that this also works here. So we know what the map is. So let, we are here to prove. So let, so alpha is here. So we have, if we take alpha in, so is a class in the nth homology of x, then we have that alpha is 0 cycle in n-cycle. So that means d of alpha is equal to 0. So now we can try to form the difference between the push forward by f, the map induced by f homologian, the map induced by g homologian. So if we take f star alpha minus g star alpha, what is it? Well, certainly we can, by definition, this is the same as putting the fn into the stain. And then it's just a homomorphism. So we can also put the sum into it. So this is the class of fn alpha minus gn alpha. And now we look at this thing. We have our chain homotopy. So we write it down in this form. This is d of Sn of alpha minus Sn minus 1 of d of alpha. So alpha is a cycle. So d alpha is 0. So the whole thing is 0. This one is 0. So this is the same as d of Sn of alpha, the class of this. But this is d of something. So it is a boundary. So the class of d of something is always 0. Because after all, the homology is precisely given by dividing by the boundary. So we see that the difference between these two is 0. So the maps are the same on homology. So now let me see. So this is the general quarter of now. Now one comes to the actual proof of the theorem. So we'll somehow start with the reduction, which is easy. And then we'll see what we can still do. And at least tell you what the setup is. So now we come to the proof of the theorem. So recall what the theorem is. That if we have two homotopic maps, f and g from x to y, then the induced map on homology is equal. And we want to prove this by eventually constructing a chain homotopy between the induced map on the singular chains. But first we want to make a reduction to a very special case. Somehow there is a very simple homotopy between two maps. And it's enough to prove it for this one homotopy. So if I take the identity of x times the interval 0, 1, this is a nice map from x and p. So this actually is a homotopy from between the two maps, i0 from x to x times i. If you remember, this was sending x to x comma 0 and between and i1. This is obvious that this is a homotopy. It's a map from this from x times 0, 1 to some space. Such that if I restrict it to x times 0, I get i0. If I restrict it to x times 1, I get i1. It's trivial. So the claim is, however, that it's enough to prove the theorem for this idiotic case. So assume we know that i0 star is equal to i1 star as maps from hn of x to hn of x times i for all i, for all n. So assume we know our theorem just for the kind of trivial homotopy between these two maps. Then I claim it holds for every homotopy because we can get every homotopy by composing this thing with a continuous map. So let f from f to g be a homotopy where f and g are continuous maps from x to y. Well, then it's not very difficult to see that. Obviously, f is just f composed with i0. g is just f composed with i1. And if you want, f is just f composed with the identity. Nobody can doubt that. We don't really need it. So we just obtain our given the homotopy we want just by composing this kind of trivial thing with our given f. And then we can use that homology is functorial so that we have an induced map. So then it follows because of that f star. So the map on homology, induced by f, is just f star composed with i0 star. Now, as i0 star is the same as i1 star, this is the same as f star composed with i1 star. But this is g. So we just reduce to this. So thus, we note that it is enough to prove the theorem for i0 and i1. So we want to deal with this case. And now I will start doing the setup. So somehow we have to construct our chain homotopy. We somehow start with single synthesis on x. And we have to somehow make out of them construct from them synthesis on x times i. And we somehow do this by taking our simplex, making a copy of that same thing, 1 at 0, 1 at 1, and dividing this thing into simplices and doing something with it. So let me set it up. So now we come to this. So on delta n times our interval 0, 1, we have the following points. We write vi equal to ei, 0. Remember that ei were these points in the kind of corners or however we call them, the vertices of delta n. And wi equal to ei, 1. So that means if we do just the two-dimensional case, we have here, I mean, the one-dimensional case and then times i is two-dimensional. So if we have here v0, v1, w0, w1, so in this case we get such a square. But if you have another simplex, you would have n. Let's first see how it goes on. And now we define for i equals 0 to n. We want to define a singular n plus 1 simplex in delta n plus i. So we put a singular n plus 1 simplex in delta n times i as follows. Well, if you remember the notation we used to denote maps. So we take the map w0, wi, then we start with v0 for vi, then we start with wi until wn. So this, if you remember, we take this by definition is the map which sends the linear combination of v0 to n plus 1 for the n plus 1 entries. So we call it one linear combination of these. So that means, so this is a map from delta n plus 1 to delta n times i. Maybe I can just write. So if we have, this is the map which sends sum i equals 0 to n plus 1 ei, so ti times ei, where ti are some numbers which add up to 1, to by definition sum i equals 0. This cannot be i because we have an i, so I call it k. So k equals 0. So i tk vk plus sum k equal, how do I want to do it? I plus 1 to n plus 1 tk wk minus 1. So this is the map. So this is just written out what it is. And so obviously the image of this map is what I had denoted like this. So just a set of all linear combinations of these elements. And so if we look at it here, we have, for instance, in this case, we are n is equal to 1. We get two such things. We have the one map which maps the same text for this thing. And the image is this. And the other one maps the n plus 1, the two syntax for this thing. And the image is this, in the obvious question. And it's more difficult to imagine if you are in higher dimensions. But anyway, so that's how it looks. Now I want to use this to do something with it. Let me think about two minutes. So maybe I just write down the operator. And then we have to show that this gives us the event of a chain homo-copy definition. So define the so-called prism operator. So this is p from cn of x to cn plus 1 of x. So it should have been called pn, but I just write p. So I first define it just on singular synthesis. So first I define it for a simplex, sigma from delta n to x, simplex, singular simplex. How do I define it? Well, I just write down a formula. So p of sigma is equal to sum i equals 0n minus 1 to the i. So when one writes formulas, one always wants to sprinkle some signs in it so that afterwards things cancel. Now we take sigma times the identity on i. And we apply this to w0, no, v0 to vi, wi, wn. So we have these simplices, these maps from delta n plus 1 to this thing to the product. And we apply sigma times the identity to it, which gives me a map from this delta n plus 1 to x times i. So after all, that's what I wanted. And we want to, this should be our, what is not composition? Sigma equals 1. No, it might be there's a misprint in the notes. So this is not the composition, this is the product. So it's sigma on the, so this is after all we are, it's sigma on the delta n, it's the identity on. So this thing is a map, if you look at this. This is a map from delta n plus 1 to delta n times i. And now we first do this map. Now we are in delta n plus i times i. And now we apply the sigma on the delta n and the identity on the i. So this is the thing. If you look at this particular case, we apply, we take this whole thing and we apply to it sigma times the identity on i to go somewhere, to go finally to x times i. And we take here, we take it with a positive sign and here we take it with negative sign. And this has the nice effect that if you look at this thing, we have 1 plus 1 minus, and you can hope that in the middle it somehow will cancel. That's why one has this sign. And so then the claim, which will prove the theorem, is that P from i0 to, so P is a chain homotopy from, say, i0 star from c star of x to c star of x times i to i1 star. So that's the claim. And that's the thing we really have to do. But then if we have proven the claim, it follows. So then the theorem follows. Because if we have a chain homotopy between these two, this means we need the same map on homology with us. Because then i0 star is equal to i1 star on homology for all n. OK, so maybe here I stop. So now next time we will repeat the setup and then we will have to prove this claim. And with this we will have proven the theorem. OK, thank you very much.