 Good afternoon. So maybe now I would start. So we were talking about relative homology. So I prefer to recall where we are. I mean, basically, we're just more or less given the definition. And then now we have to do something about relative homology. So x is a topological space. A is a subspace of x. So with induced topology, then, as I had said, I can view, say, the n-chains on A to be a subset of the n-chains of x. It's just the map from the n-symplex to x, whose image lies in A. I mean, the formal linear combinations of such maps. And D will map, obviously, cn of A to itself. So it respects the inclusion. So therefore, we can form cn of xA, which is just a quotient. And we have an induced map. So an induced differential, which I still call D from this to the one with one less. So we still have D, which goes here to cn minus 1 from xA. Just if we have any n-chain in x, alpha, we look at its class, modulo cn of A. And this is sent to the class of D alpha modulo dn minus 1 chains in A. So this was this definition. And then we defined the corresponding homology in the usual way. So we again write, first, the cycles, cn of xA is equal to the kernel of D from cn of xA to cn minus 1 of xA. And the cycles and dn of xA is equal to the image of the previous D. And as the D is just the map induced on equivalence classes by the previous D, we have obviously the D squared is equal to 0. And so one is contained in the other. So bn is contained in zn. And so the homology, relative homology, is hn of xA, which is defined to be the quotient of the cycles by the boundaries. So it's the same thing that we did before. So the reason why one considers these relative things is that they are tied together with the usual homology of x and the homology of A by a long exact homology sequence. And so I want to first talk a bit about exact sequences because I'm not sure you know about them already. So this is some simple algebra, or whatever, homological algebra if you want. So let, so this is somehow maybe the definition. So if I have A, F, B, G, C, so F from A to B and G from B to C, it's a sequence of two of homomorphisms of a billion groups. So we say that this sequence is exact at B. So it's called exact at B. If the image of the first is precisely equal to the kernel of the second, if and only if the image of F is equal to the kernel of G. So if you look at it here, if you take these, if the map would be just, for instance, just as a, so the relation to what we're looking at before, if you have this map, Cn of x goes to Cn minus 1 of x with T goes with, so maybe yes, n plus 1, n, n minus 1, with T. If you look at this thing, then this will be exact at Cn, if and only if the entomology is 0, by definition, because the entomology is the quotient of the kernel of this map by the image of this map. OK. So in particular, if the sequence is exact, then we have that the composition of the two maps is 0. But it's more, the kernel is precisely equal to the image. That it would be that the composition is 0 means that the image is contained in the kernel. OK. So in more generality, we can look at, so if in more generally, if say, n plus 1 to an to an minus 1 to an and so on is a sequence of, so possibly longer, maybe infinite sequence of Bn groups and homomorphisms, we say the sequence is exact if it is exact at every element here. So at each step, the image of the first previous map is equal to the kernel of the next. OK. So in particular case, that one looks at is a so-called short exact sequence. This is a sequence consisting of three things, or five, depending on how you count. So the short exact sequence, 0 or an exact sequence, 0 goes to A, goes to B, goes to C, goes to 0, is called short exact sequence. So what does it mean? So 0 is also in a Bn group. So we have that the kernel of this map of the F is equal to the image of this map. The image of this map is 0. So this is equivalent to the fact that F is injective. This is what this says. Then here, the image of F is equal to the kernel of G. And then here we have the, again, the 0, a Bn group. So the kernel of any map from C to the 0 group is 0, is the whole of C. So it means, and the kernel should be equal to the image of the previous map. So we should have that G is injective. So that's what it means that we have a short exact sequence, like these three properties. And just for later use, I can make some remark. I mean, this is an exercise. It's all somehow trivial, but one has to, so that one gets used to this concept. So first is, this in some sense I just wrote, 0 from a to, say, G F to B is exact if and only if F is injective. I just said it. Second, A goes to F to B, goes to 0, is exact if F is injective. This I also just said. The image of F is the kernel of the 0 map, so the whole of B. And then by putting this together, if you have 0 goes to A to anything, B goes to 0 is exact. So if and only if, F is an isomorphism. Because after all, it's equivalent to being F both being injective and subjective. And then slightly more complicatedly, if I have A alpha to B, it's better to see an exact sequence. And what, say, the second map injective and better injective. So that means that the kernel of better is, do you really want that? Yeah? No, I didn't. No, I think I wanted that. Obviously, if better is injective, then its kernel is 0. And then this is not the case. So if, say, gamma is injective, then alpha is subjective. This is because if gamma is injective, then its kernel is 0, which is equal to the image of better. And so, which is equal to the image of alpha. And so alpha is subjective. So these are all very simple things. And the same way, say, if A goes to alpha, B, better, C, gamma, B. And, say, alpha is subjective, then it follows that gamma is injective. So again, it's an exact sequence. So you can figure this out. This is very straightforward. So as an exercise, you can, if you want to prove it. So one example of an exact sequence would be, for instance, you have 0 goes to Z. And you take, say, the multiplication by 2 goes to Z. So the image are all the even numbers. And then the quotient will be Z, modulo twice Z goes to 0. That's that. And so this is an exact sequence. I should also maybe make another remark, which is also, which one can also see here. So if 0 go A to B to C to FG is exact, then C is isomorphic to B divided F of A, which is, I mean, also standard. I mean, the image, so C, the map is subjective. So it's the image of G. And so the image of the map is isomorphic to the source of the map divided by the kernel. That's some homomorphism theorem. OK. So why do we deal with these exact sequences? So now we want to come to some slightly more serious homological algebra, make the whole thing a little bit more complicated by looking at exact sequences of chain complexes. So you know what chain complexes are. We have some, so let A star D, B star D, so a short exact sequence I want to deal, C star D, chain complexes. So that means we have A n plus 1 goes with D to A n, goes with D to A n minus 1, and so on, and same for B and C. And now we assume we have a map F from A star D to B star D and G from B star D to C star D, the chain homomorphisms. So remember that this meant that we have, after this map, we have A n plus 1, A n, A n minus 1, always the D. And this F is given by a map F n from F n plus 1 to A n plus 1 to B n plus 1. And here we also have the D. So that we have a map of chain complexes, a homomorphism of chain complexes, or a chain homomorphism, means that we have all these maps at all levels and they commute with D. So this diagram is commutative and similar for B and C. And now, so we have these two chain homomorphisms and we call this an exact sequence, a short exact sequence of chain complexes if at all levels we get a short exact sequence of a B in groups. So 0 to A star to F, B star to G, C star is called a short exact sequence of chain complexes if for all n. So we have the corresponding maps at level n. So we have the map A n goes to F to B n, so F n, G n to C n. So if we complete this with 0 here and 0 here, if this thing is exact. So the fact that we have these two chain homomorphisms means that we have always this map F n and G n. And the statement that it's a short exact sequence of chain complexes means that this F n is always injective, the G n is always subjective, and the kernel of G n is always equal to the image of F n. And we introduce this because of the following theorem, which is, yes? No, I mean that's a misprint, B n minus 1. Thanks. So we have introduced this concept because there is a certain theorem, which is maybe the most useful theorem of homological algebra, although it's not very deep, but it is a bit tricky. But anyway, it's very useful. And it's the following theorem. It's called the snake lemma. And it says the following. So let 0 goes to a star with F to b star with G to c star to 0 be a short exact sequence of chain complexes. Then we have a long exact sequence for the homology of the chain complexes. As these are chain complexes, I can form the homology. And it is tied together by a long exact sequence. So then there is a long exact sequence. So we have, at some point, say we have hn of a star. So you remember that's the image of the kernel of d from a n to a n minus 1 divided by the image of d from a n plus 1 to a n, as usual. So we have a map here, which is the map induced by F to hn of p star goes to G star hn of c star. And then we have another map, which I call delta, which then goes to hn minus 1 of a star F star hn minus 1 of p star. And then with G star hn minus 1 of c star, and so on. So here we're always delta. So you have an exact sequence like this, which goes to all the n. And you can imagine why it's called snake lemma. It's just because of the form of this diagram. You write it down this way, and it looks like a snake. And as I said, F star and G star are the maps induced on homology by F and G. So this would mean that F star of the class in homology of some element a will be the class of F star a of n of a, say, if a is in the n. So and delta is called the boundary map. And so actually the most surprising thing about this, so this F and G are somehow F star and G star are the maps which are obviously induced by F and G. So they certainly exist. But the most surprising thing about this theorem is that this map delta actually exists, which goes from the entomology of the last one to the n minus first homology of the first one. And then the whole thing is exact. So at every step, the kernel is equal to the image of the previous map. OK, so this is a statement in homological algebra. And it's proved by a method which is called diagram chasing. And it's some kind of so proved by diagram chasing. So somehow the idea is you take some element in some of these groups, or maybe some of the groups that we will see later. And then at each step, you can either take a pre-image under one of the maps that you have, or you can take the image under one of the maps. And you try to do this. You kind of let you always do one of the two. And somehow you try to chase this element through the diagram by either mapping it or taking the pre-image until it does precisely what you want. And this sounds kind of difficult. And you will see now that the proof is maybe a bit confusing. But as a matter of fact, it's very simple. Because at every single step, you have only very few choices what you can do. You can either take a pre-image under one of the maps or the image under one. So you have only very few possibilities of what you can try to do. You can basically try all of them. And if at all there is a proof, you will find it this way. So it's kind of something which one could easily program a computer to prove things. For the human being, it's actually slightly more complicated because it's not so intuitive. So now let's try to prove this. So we will not prove the whole thing. We will construct delta, which is the most difficult thing. And in the notes, I've proven that it's exact here. But I actually will only prove that the composition of these two maps is zero, which is slightly easier. But I do the most difficult steps. And then the other ones, you have to just, at each time, you have to do something similar to check the exactness. But the most difficult thing is to actually construct the delta and that we do as a first thing. So we want to proof construct delta and show that say delta composed with G star is the zero map, which is part of the exactness here. So as before, I write Zn of A equal to the kernel of D from An to An minus 1. And I don't know what the exactness is. No, I don't think I need this. And similar for B and C. So let's see what we can do. So I write, so at some place, I have to somehow keep track of what I'm doing. So I write one diagram where you can see where we are. So we are here in A, B, and C. So this is, you know, so here we have, say, we are at level N, N minus 1, and N minus 2. So if we are here, for instance, we are in Cn, Cn minus 1, Cn minus 2, and so on. And so the horizontal maps are always given by Fn. So I give them by F. This way, this is G. And the vertical maps are always given by D. So we have this. So now we start. So what do we want to do? We are supposed to make delta from Hn of C to Hn minus 1 of A. You have to construct such a map. So OK, we start. So let C be an element in Hn minus N of C. And so that means it's the class of an element C of an N cycle C. So it is C is an element of Zn of C. It's an element. And then its homology class is this. So we have here, so this is our zero step. So here we have our C. So now we want to, I mean, we have to, from here, go all the way until here. So we have to somehow go backwards. So what can we do? We can apply this. We take an inverse image under this map, Gn. So anyway, just to write it out, so we need to construct delta of C equal to A for A and N minus 1 cycle in A. So the first step is, obviously, there exists an element B in Bn such that C is equal to Gn of B. This is because we had it all level, An to Bn to Cn is an x-axis sequence, so Gn is subjective. So we have here our element B, which maps to it. This is the first step. Now, we can take its, we can apply the differential to it D. And we get Db. So we have Db is an element in Bn minus 1. And note, what happens if we map it here? We can apply here Gn minus 1 to it. And we get to C. What do we get? We get if we take Gn minus 1 of Db, this commutes. These are chain maps, it commutes with it. This is D of GnB, which is the same. Gn of B is C, Dc, and C was a cycle. So this is 0. So this thing actually maps to 0 here. So it means that Db is in the kernel of this map Gn minus 1. So as we have an x-axis sequence, An to B, An minus 1 to Bn minus 1. So Cn minus 1, it is the image of some element here. So let me, now I don't have quite enough space. So by the exactness, so this would be 3, I think, there exists an element, a unique element, say A in An minus 1, such that fn minus 1 from A of A is equal to Db. So we have here found our element A. And now the A is actually supposed to be this A here. So in order to see that it's good, we have to see that A is a cycle. So therefore, we should apply D to it, and it should become 0. Let's see where this is true. So now note that this map, fn minus 2, is injective. Because the first map is always injective. We have a short note there. fn is injective, Gn is surjective. So it's injective map. On the other hand, what is fn minus 2 Da? This is the same thing we just did. This is D of fn minus 1 of A. But fn minus 1 of A is Db. And I know that twice applying D gives 0. So I find that this thing maps to 0 here. And as the map is injective, it's 0 itself. I've said that fn minus 1 is injective. So Da is 0. So thus, Da is equal to 0. So A is an n minus 1 cycle. And we define delta of C to be A. So should be slightly careful. I have not completely constructed because there were some choices. And one has to prove that the definition was independent of the choices. So this I call maybe an exercise. So A is independent of the choices. And there were actually two choices. So one choice was a choice of C. Because C is a homology class that we have chosen representative. First choice of C, just of the representative of C. And the second is that B is B which took some pre-image of our C. But nobody has said that the map gn is injective. So there will be more than one. So it's also independent. Should also be independent of the choice of B. And that's actually both very easy to check. You somehow have to show that if you take a different choice, then what you get changes by boundary. And it's quite easy to do that by limit. So this will show. So modulo this exercise I have constructed this map. And I mean, you maybe really should try to see whether you can see this independence. It's really very simple. So now I want to prove the exact this statement. So where is it? So we want to show that delta composed with g, say, star, is the zero map. So we have this map delta composed with g star. It will be a map from hn of B to hn minus 1 of 8. So let B an element in hn of B star. So given by some cycle B in zn of B. And we first apply g star to it. So let's see. So c g star of B. And I take c, the representative of it, to be gn of B. So now I want to show that if I apply now delta to this c, I will get 0. For this, in principle, I have to go through the whole of the definition of delta to see that I get 0. But there's one thing that one notices. So here in the construction of delta, we had some point chosen a B which maps to c. And I claimed it's independent of which one we chose. And here we do have a B which maps to c. So why don't we choose precisely that B here? So in the construction of delta of c, so this is at step one, we have chosen a B in the end with gn of B is equal to c, like here. So just choose now the B here as that. So maybe I should have given it a different name. But anyway, so choose the B. Yeah, maybe I write it like this. Choose the B to be B bar. So the B that the element that we have here which maps to c, we chose. The B, here we are free to chose an element of bn which maps to c. And we chose the one that we are given here. What happens then? B is a cycle. So we have then dB is equal to 0. So A is an inverse image of B by this injective map. An inverse image of dB by this injective map here. This map is injective. And now dB is 0. So A will be 0. And this shows that this composition is equal to 0. So you see it's, I mean, I do not expect that you can follow it quite as fast in all the details. But you see that in some sense, we always do kind of the same thing. We either take the image of something or we show that the image under something by a map is 0. And then by exactness it comes from the one before. Or we want to show that something under d, something is a cycle. So we map it under d, we get 0. And in this way, we somehow move forward and backwards until we precisely know what everybody does. So in this case, we show that if we take delta of the c, which is the same as delta gn g star of B, this is equal to 0. And so this was as much as I wanted to show. Obviously, you have to show many more things. You don't just have to show that this composition is 0, but that the image of the g star is equal to the kernel of the delta. And you have to show the exactness at all the other places. But this gives you a fair impression of what you have to prove. All the other steps are somehow similar. OK, so this is this snake lemma, which is kind of a standard example of this argumentation by diagram chasing. So now you have seen this once. And actually, there is a book where, I mean, this is a bit excessive. Obviously, I think it's by searching, but I might be wrong. Anyway, so where it's about another topic, and then he says a few things about diagram chasing and maybe prove some easy lemma and says, afterwards it says, exercise. Use the methods used. Take any book in homological algebra. Take any theorem and prove it. Because the claim is that you basically know for many things this is all you need to be able to do. So but it's a bit excessive, obviously, but that is a theorem. Now we get the long exact homology sequence. So we go back to our homology story. So we are back to single homology. So let A in X be a subspace, X a topological space. And I actually want to explicitly write I the inclusion from A to X. Then there is a long exact homology sequence. So at some point, you have this delta. Hn of A goes to via the map induced by the inclusion. Hn of X goes to Hn of Xa by the map induced by the quotient map goes to Hn minus with delta H1 minus 1 of A and so on. So we have a long exact homology sequence which goes to all the possible ends in this way. And this is basically a corollary to this snake lemma. Because what do we have? So by definition, if you look at the map from C star of A to C star of X, which is given by I star by the inclusion, and also the map from C star of X to C star of Xa, which after all is just C star of X divided by C star of A. So at each level, though this means just at each level, we have I n from Cn of A to Cn of X. And we have the map from Cn of X to Cn of Xa, which is equal by definition to Cn of X divided by Cn of A. So these maps are obviously chain maps, chain homomorphisms. I've said it before, the inclusion of A with X commutes with differential. And the differential here has been defined in such a way that we get a chain map. We just have said the class of something, d of the class of something is the class of d of it. By definition, these are chain homomorphisms. And also, obviously, for all n, if you look at the sequence 0 goes to Cn of A, goes to Cn of X, goes to Cn of Xa, which is just defined as Cn of X divided by Cn of A. This sequence is exact. Because I've told you that Cn of A can be viewed as a subset or subgroup of Cn of X in all those chains for which all the simplices, instead of mapping to the whole of X, just map to A. And here, we have just defined this to be the quotient. So certainly, the sequence is exact. So at all levels, it's exact. So we have 0 goes to C star of X, C star of A, obviously, C star of X, to C star of Xa to 0 is an exact sequence of chain complexes. Now one just applies the snake lemma. And the snake lemma says precisely that there is a long exact homology sequence in this form. That's precisely the statement that it makes. So that's it. So we get this sequence. Now I should say, in the moment, one can question the usefulness of such a result. Because somehow, maybe our aim will be to compute the homology of some spaces like X or like A. And now we have an exact sequence where we put in addition these relative homologies, and we don't know what they are. So we have kind of related the things that we want to know to other things that we do not know. So that by itself, one could doubt whether this is a very useful thing to do. So this can only be useful and help us to, for instance, compute the homology of actual spaces if we have some way to simplify the computation of this, to somehow understand how one can compute these. And there is a theorem which serves this purpose, which is the so-called excision theorem, which is the next thing that we are trying to do. But in this case, I mean, the excision theorem is really, the proof is really complicated. And it's so complicated that I cannot possibly cover it in this course. It's not particularly difficult, but it's just very long. So it would take me maybe, I expect, two lectures to do it. And I mean, I don't have two lectures. But I will just state it and give some hints about why it's true. And then we move on. So we have the excision theorem. So the statement is the following. So let x be a topological space. And let, say, z and a be subspaces of x, such that the closure of z is contained in the interior of A. So z is the closure of z in x. And a is the interior of A in x. So this is the smallest closed subset, which contains z to all the intersection of all closed subsets of x, which contains z. And this is the union of all open subsets, which are contained in A. So if you have this, then we can cut out z from both x and a, if you want to compute homology. So hn of xA is isomorphic to hn of x without z A without z. So we can somehow, so we have this space x. And we have in it the subspace A. I want to compute the, maybe I make it a little bit bigger. This is A. We want to compute the relative homology of x and a. Well, if we have some z, which lies still inside this, to compute the relative homology of x and a is the same as if we throw away this. So if this is not there. So we can, whether the z is here or not, we get the same relative homology. And so this is why it's called excision because apparently, although it's not a very standard word, excision means something. To excise something means to cut something out. So we can cut z out of both x and a and get the same homology. OK. And this is true for all n. So one can formulate this for the proof, which I don't give, but which I have to give at least some hint. One has to reformulate it in a slightly different way. A given equivalent reformulation. So let say A and B be subspaces of x, such that x is equal to the interior of A union the interior of B. So we can cover it by two subsets, by the interiors of two subsets. So for instance, if A and B are both open, it just means that x is the union. Then h star of xA is hn of xA is equal to hn of B A intersected B. And this, although it's not completely obvious in the first moment, is precisely the same statement. If I tell you what the relation between z and B is. So the A, the x and the A are the same in both. And the B is, I think, something like z without B. Let's see. So now to make it more interesting, I have actually not written what it is. So I can either try to figure it out myself, or I can be lazy and look it up. So I say for, so in one direction, so maybe this is the first statement, and the equivalent second statement is this. And then I say to go from 1 to 2, I put z equal to x without B. And I don't know whether. And to go the other round from 2 to 1, I do the opposite. I put B equal to x without z. And then you can see. So assume, I don't know whether it's 1 to 2, we know which way around it goes. But so if, for instance, if we assume we know this one, and we put z equal to x without B, then the statement that the interior of B and the interior of A cover it means precisely that the closure of C is contained in the interior of A. And so that the assumption for A and B is fulfilled is equivalent to the assumptions from z and A to be fulfilled. And as z is the complement of B, taking B and A intersected B is the same as taking x without z and A without z. So this is just the same statement formulated by once for A and B, and once for A and the complement of B. So now let's try. So as I said, the proof is amazingly complicated. Although it's not, well, it is also a bit difficult. But anyway, so it follows the main step. And the most difficult part is another theorem, which we will also use for something else, which is the theorem of the cover. So the main step is the theorem of the cover, is the following theorem. And this says the following. So let, so I don't want to use this. Let x be a topological space. And I make the same assumptions as here. So A and B subspaces such that the inter, in the interiors cover x. So then we can do the following. We call Cn of A plus B. We define this to be, so let, this is a set of all cycles, sum sigma A sigma times sigma in Cn of x. So these are singular n chains in x. So just linear combinations as usual of n, of maps from insimplices to x. With the property that when, if the coefficient is non-zero, so when it really is there, then the image of this n cycle, of this n-symplex, either lies in A or in B. So if A sigma is different from 0, then sigma of delta n is a subset of A. Or sigma of delta n is a subset of B. And then, so again we can, and we make the corresponding chain complex. So we have a chain comp, so anyway. So let C star of A plus B mod D be the corresponding chain complex. You can see if you apply the usual D to such a thing, it also is true for the faces. This will also have the same property. Then the inclusion C star of A plus B to C star of x induces an isomorphism on homology for all n's. So if I compute the homology with these chains, where each simplex, the image of each simplex is either contained in A or in B, I get the same homology as if I allow all chains, also those where maybe it's not contained in one of the two. So this is this theorem. And it's not very difficult to prove from this the excision theorem. In the notes, it is briefly explained. But maybe it's not relevant now to do this. I want to briefly tell you why such a thing could be true. So the main step is anyway this. And we will use this theorem again for another theorem, which will be more important for us. So therefore, I want to explain about this. So we know that Cn of A plus B is contained in Cn of x. But they are usually not equal. I mean, unless some extreme cases. So normally, they are not equal. So the difference is given by, in some sense, simplices which are too large. So by some A sigma times sigma, where for some sigma, sigma of delta n is not contained in A and not contained in B. But you can imagine, here you have A and you have B. So if the image is somehow not contained in one of them, you could try to subdivide it into smaller ones so that each of them is contained in one of them. No? So I mean, if you think of it, maybe this was not such a. So assume this is our picture. So say this is A and this is B. And we have the image of our simplex looks like this. So it's sticking out a little bit here, a little bit here. So if I, for instance, would instead replace this map by a map where, by kind of two maps, where each of them, one maps to this half and one maps to this half, then it's OK. And so in order to prove the result, we have to somehow be allowed to subdivide our simplices. So the problem is the image of the simplices, sigma of delta N, can be too large, but we can subdivide. So delta N is equal to the union i equals 1 to k delta N i, where this is the image of another n simplex but so that such that the image is contained in A or in B. And how do you achieve this? Well, if we have such a simplex, we have the so-called barycentric subdivision. So we just divide, so we take a point in the middle, we divide all the sides by half. Hope I get it right now. And so we have, instead of one, two simplex, we have one, two, three, four, five, six, OK? And they are somehow smaller. And each of them I can view as the image of a two simplex mapped into this thing. And then we want to replace the delta N by, so the composition of all these maps, the sum of the compositions of all these maps from delta N into this thing, identifying it with this, and summing them all up. And so there are obviously two main problems. There are a number of problems with this. So the first is that you have to show that you can make such a subdivision if you make it often enough. You will get that the image of these smaller simplices indeed is always contained either A or B. And the second thing, which is maybe not obvious at all, is that if instead of using this big simplex, you use what you get by subdividing it, so just the sum of the simplices which you obtain by subdividing it, that this gives you the same on homology. This you also have to show. And it turns out that you can show all this, but it's quite complicated. So show that you get the same on homology. You have to prove some chain homotopy and so on. And you just have to struggle quite a lot. But anyway, the basic idea is that it can be that your simplices are too large. And if they are too large, then you subdivide them. And then you have to work very hard to turn this into an argument. And I mean, I do not claim that I gave you a serious hint of how this is done. OK. Anyway, this is the. And then it's very easy from this to prove the excision theorem. So let me now try to see whether I can give you some 10 minutes. So that's not very much. So I did not think this. Oh, yes. So there are two things. There's one kind of thing that I wanted to state in addition. We had this, if you already did it, you had this exercise, which you were supposed to finish before the lecture of Thursday. But actually, there was supposed to be a lecture of Thursday, but I have to go to the dentist, so there's no lecture on Thursday. But you can give it to me on Thursday in my mailbox or whatever. And the last one was about reduced homology. So if you recall, at the topological space, we had this map. Again, cn was a singular n chains. And we have the differential, d, from cn of x to cn minus 1 of x. And in the exercise about reduced homology, I defined another differential from c0. So we defined d, whatever, d sharp from c0 of x to z, which was just that if I have a 0 chain, so some a sigma times sigma, and in some sense, this is just a point, a 0. It's just a formal combination of points. This was mapped to the sum of the coefficients. And then, otherwise, so we always use the usual d. And then, so we define h0, so hn, sharp of x, the reduced homology is defined to be hn of x, the usual homology, if n is greater than 0. And it's called, it's the kernel of this d sharp divided by the image of d from c1 of x to c0 of x if n is equal to 0. So somehow, I changed the 0 homology. The others stay the same. And the advantage of this is that it somehow simplifies some arguments because one of the things that one finds that you use for this is that if I take this d sharp composed with d, this is the 0 map. This was part of the exercise. And the result that one proves, or that you prove, is that if x is path connected, then h0 sharp of x is 0. So somehow, the fact that you make h0 smaller by taking the kernel here, instead of being z, if it's path connected, it's 0. And that makes a number of arguments simpler. And I say this here because of following results, which I state without proof, but which are simple modifications of what we had before, namely, so the first is a kind of corollary. So first, so the long exact homology sequence also holds for reduced homology, I mean, where you, OK? So we have a we can, in the long exact homology sequence, we can change what we have at level 0 by this. And it still is exact. So it ends delta h0 sharp of x of a goes to, I think, actually can have h0. But anyway, this is not so important because if x and a, for instance, are connected, it means it ends with 0 here. So if x and a are path connected, we get already h sharp 0 of a is equal to 0. So somehow, the homology sequence is simpler. And also the excision theorem, oh no, that doesn't make sense. OK, so this is something I wanted to mention. And this, what? Well, yeah, I think, I mean, in order for it to make sense, I should take the reduced homology, which would be the quotient. But I mean, I will only apply this when, say, x and a are connected, in which case it's anyway 0 here, and they don't care about what's there. OK, so now time is almost up, but I also start a bit late, so I can say a couple of more things. So now I want to give an application. I don't think I will finish it, but I can first set things up a bit more. So I want to give one application of this excision theorem to prove some theorem. So first, I define something which is anyway useful. So let x be a topological space, and a in x a subspace. Well, in the exercise, we were told what the retraction is, but I can maybe just repeat it. So a map R from x to a is a retraction, so a continuous map is a retraction if its restriction to a is the identity. And we had seen that this means that the induced map on homology by the inclusion is injected, I think, was an exercise. And now we want something else. So a is called a strong deformation retract of x if you have such a retraction, but you have more, namely, if there is a homotopy, f from x times 0, 1 to x. So such that f of x 0 is equal to x for all x and x. f of x 1 is an element in a for all x in x. And f of a t is equal to a for all t in 0, 1, and all a in a. So what you can see is that this, so in other words, you have that f in particular is a homotopy from, so we have the identity on x. So let me see. So you can see, so if you put r, this is a map from x to a, and we also have the inclusion of a, then this says that f is a homotopy from, say, so how does it go? From the identity, so we get here at 0, we get the identity. The identity of x to i composed with r. And so in particular, it means, in particular, we find that a and x are a homotopy equivalent. Because you have a homotopy equivalent from 1 to i composed with r, and if you take the other composition r composed with i, this is the identity on a. Because the map r, so here if you look at it, f of a t is a for all a in a. So we have a, and in this case, if you have a strong deformation retract of x, we have that a and x are homotopy. So I should maybe have explained it slightly differently. So we start with a being this retraction, which we here take as the map at 1. And then we have a homotopy between this and the identity. So the statement, OK. So this is a particularly kind of typical example of a deformation retract. So one example would be, for instance, if you have, say, one thing is so, if we take the 0.0 in i n, then this is a deformation retract. You can just, if you have any point, you can just make the homotopy by, so what would you say? You take f of x, t should be t times x. Then you can see that now I don't know whether I got it the correct way. It's precisely incorrect way round. So it is more natural in this way, obviously. But then I would have to change it here. So if you, therefore, I could say 1 minus t times x, which is less natural. But you can see when t goes to 1, then we go to 0. The map is always the identity at the point 0. But precisely it has its property. And so this, after all, shows that i n is contractable, which you already know. OK. So next time we will use this to make an application. Maybe I can just say what the application is, so that you, which is kind of funny, something which you might consider obvious, but which is not so easy to prove. So it's called often invariance of dimension. So let u subset i n v subset i m. The non-empty open subsets assume u is homeomorphic to v. Then it follows that n is equal to m. So an open subset of some i n cannot be homeomorphic to an open subset of another i m. So somehow the dimension, you would think that an open subset of i n has dimension n in some sense. The dimension is something which is invariant under homeomorphism, which is a power r not client. We will see how one can prove this using the excision theorem. OK.