 OK, good afternoon. So this is the second half of the course on algebraic topology. So by itself is a rather big topic in algebraic geometry in mathematics. But we will only kind of look a bit at the surface of it. So we had the first part about the fundamental group and covering spaces. And now I want to talk about. So the second part is about homology. So I've prepared some notes. They are somewhat preliminary. But you can look at them. And they are relatively close also to this book of Hetscher, Algebraic Topology. This would be chapter two, actually other parts of it. So now I want to start. So we'll have to get used to my handwriting. But I think it should be OK. So in topology, one starts somehow topological spaces. So if you want, in some sense, the shapes of some spaces. And algebraic topology is about somehow combining this with algebra. So what one wants is, so in algebraic topology, we assign some algebraic data, usually a group. So to a topological space x, a group f of x. And so then you can use somehow methods of algebra to find out something about x. If this group somehow encodes the information about x in a particular nice way. And so this somehow is supposed to help translate questions of topology into questions of algebra. And so you have seen one instance of this is the fundamental group, xx, x to the base point x. And this is some group which somehow tells us about loops in x, so about homotopy classes. And you have heard a few lectures about this. Now we want to assign different kinds of groups which are the homology groups. So these are just a whole set of groups, hn of x, correct, we do the hn of x with z coefficients, but we do not consider anything else, which are some a bn groups. So these are there for all positive integers. And they are a bn groups. Whereas the fundamental group, obviously, can be non-above. So now and what we actually will be studying is I will now want to introduce the topic of the study these homology groups. These are the singular homology groups. So singular. So now we will find out. I'm not sure whether we'll find out in this lecture, but it turns out to be that homology is actually in many ways simpler and easier to handle than homotopy groups. But in the beginning, it's considerably more complicated because it needs more technical kind of baggage. So the singular homology is simpler and easier to handle than homotopy groups. And the fundamental group is the first homotopy group. But we need more algebraic background. So we need, for instance, some homological algebra. And as you will see, the definition is also much less intuitive. For a fundamental group, you just have these loops and x and you can kind of deal with them for the homotopy for the homology groups. It's somehow a little bit more complicated. We have somehow more technical objects, simpler. So the question was just easier. Yeah, sorry about the handwriting. So singular homology is simpler and easier to handle than homotopy groups. So we start with some simple thing in this homological algebra. We talk about chain complexes. So what is that? So a chain complex, so C star D, is first a sequence Cn, N in Z of a billion groups, and some maps between them, which we all call D. And homomorphisms, D from Cn to Cn minus 1 for all n. So we have somehow, it looks like this, we have somehow here Cn plus 1, and we have D, and we have Cn, we have D again, which is in fact another D. We could call this Dn plus 1, this Dn, but we just write D, Cn minus 1, and so on. And with one property, namely that the composition of D twice is always a 0 map, such that D composed with D is equal to 0 at every stage. So if here we have another D, so if you first go from here to here, if you use D twice from here to here, this is just a 0 map. So the second D is 0 on the image of the first D, and the same is true if one goes on. This is a chain complex. And if we have a chain complex, we can define its homology, and the single homology we will deal with. Yes? What is the difference between? What? With the minus 1D, the down one is different. So the question is, what is the difference between the different Ds? Yes. And what I said is that I, by use of notation, always give the same letter D to all the different maps. We have the D from Cn to Cn minus 1, so we have a Bn group. And between any two successive one, we have a map which we call D. We could call it Dn. And then we would have here Dn plus 1, Dn, and we would here have here Dn plus 1, Dn. And this is then true for all n. But I want from now on always write D, because you should be able to see which D I mean just by seeing from where it starts. The one which I here call Dn is the one which calls it Cn. So if I write D from Cn to Cn minus 1, it's a different D than if I write D from Cn minus 1 to Cn minus 2. I mean, it happens all the time in mathematics that one uses the same notation for different things. In particular, as here, they are very closely related. It seems reasonable. OK. Oh, I wrote the other round. Yeah, indeed. This is precisely the reason why I do not write the index, because it only leads to confusion. OK. So if we have such a chain complex, we can define the homology. So the homology of the chain complex H star of Cn. Maybe I can just write Hn of C star, which is also which actually depends on C star and D. This is for all n and z. Is defined to be Hn is defined as Hn of C star D equal to the kernel of I write for the moment again D from Cn to Cn minus 1. And if you want, you can write the n. But this is, I don't. Divide it by the image of D from Cn minus n plus 1 to Cn. So why does this make sense? So note that we have that D composed with D is equal to 0. Thus the image. So if we apply D twice, we get 0. That means the image of the D we apply first, this one, lies in the kernel here. Thus the image of D from Cn plus 1 to Cn is contained, is a subgroup of the abiding group, the kernel of the next D. And if you have a subgroup of an abiding group, we can take the quotient group, which is again an abiding group. So Hn of C star D is the quotient group. So at least we can make this definition just for later usage. I will always write Cn of C star to be the kernel above. The kernel of D from Cn to Cn minus 1. And Bn of C star to be the image of the previous map. This would be called, elements here are called cycles, n cycles, and elements here are called n boundaries. So the entomology is the quotient of the n cycles, of the group of n cycles by the group of n boundaries. So this was for the moment enough of these generalities. Now we come back to defining single homology, the singular homology. And this will be done by making some chain complex of which it is the homology. And so in fact, so singular homology is the homology of a chain complex made from continuous maps, say sigma from delta n to x from simplices to the topological space x. So a simplex, so simplices are 0 simplex will be a point. An interval will be 1 simplex, a triangle is a 2 simplex, and whatever, 3 simplex is a tetrahedron, and so on. And obviously, if we say this, these are some sub spaces, some sub topological spaces of rn, and they carry the subspace topology, and then it makes sense to talk about continuous maps. And so now first, I have to say a little bit more about simplices before I can talk about more things, because I need to actually do some things with them. So I want to define them properly, and some maps between them. So we have a definition. So if we're given k plus 1 points, we can define, we can, so I call them maybe p0 to pk, I can define their convex hull. It's just what one thinks. It's the, if you write down the points, it's just everything between them. What is this formally? So this is p0 in these square brackets, p0 under pk. This is the set of all linear combinations of these points where the sum of the coefficients is 1. So all positive linear combinations. So sum of all sum i equals 0 to n, ti, pi. Now these are some vectors in i n, so I can form the linear combination. So these are some points in i n. Such that two things hold. First, the ti are all non-zero, and the sum of the ti is equal to 1. And if you think about it, if you do it for these three points, you would precisely find, you get all the points in the triangle spent by these. Or if you have two points like this, this precisely describes the interval. And the same way. So OK, this to k. So the question was whether the sum goes from 0 to k. And obviously, it goes to k, as you correctly said, because we only have k. Thank you. So if these points, or rather the difference with the first say, pk minus p0 until p1 minus p0 until pk minus p0, if these vectors are linearly independent, which precisely means that these points do not lie in a lower dimensional vector space, lower than dimension k, then p0 to pk is a k-simplex. So it's called a k-simplex. So as I said, we just have these. For instance, these three points, for them, the difference with p0. So p1 minus p0 and p2 minus p0 are linearly independent. And we find that this is indeed a two-simplex. Mostly we'll use the standard-simplex, which I call delta n, which is the convex hull of some special vectors, e0 until en. And so in this case, I mean, there are different conventions. I choose this one, e0 is just the origin. e1 is the vector where the first coordinate is 1. The other ones are 0. And then so on. So e2 would be 0, 1, 0, and until en. So in other words, and then delta n explicitly can also be written as you can easily check as the set of all s1 to sn in rn such that si are all non-negative. And the sum from i equals 1 to n si is smaller equal to 1. But I mean, this is because here the first one is 0. So it doesn't count what you take as coefficient for the first one. OK. So in this case, so just as an example, delta 0 will just be 0.0, delta 1 will be the interval from 0 to 1. 0 is e0, 1 is e1, delta 2 will be the triangle in the origin like this e0, e1, e2, and so on. So these are some very simple things. And so I also want one can also use these points to define a map from a standard simplex to a simplex or to the convex hull of some points. So for points p0 to say pk in i n, there's a continuous map, which have been denoted by the same letters with round brackets, p0, pk from delta k to the linear span of p0 to pk, which sends such a linear combination, sum i equals 0 to k ti times ei, any point in this thing? What? Yeah, which one? Yeah, the sum is equal to 1. And they're smaller equal to 1. Yes. Maybe it's an exercise to understand why. Because obviously, the difference is here that e0 is equal to 0. And therefore, equal to 1 becomes smaller equal to 1 there. OK. But here, note that here we are writing the standard simplex as a linear combination where the sum is equal to 1 of the ti. But the e0 is 0. And I sent this to the same linear combination of the pi. And you can easily check that this is basically a linear map. And so certainly, it's continuous. So we have a continuous map, in particular, from the standard k simplex to any k simplex. We have a continuous map. But the map is also continuous if these points are not in linear input. And we want to use this to define something which will later allow us, huh. Anyway, so there's this tiny problem that I introduced quite a lot of notation. And then I'm going to use it. But as I don't have so much blackboard, I will have to wipe it away. So you should consult also your notes. So in particular, we have the face map, which I, the face maps. So fi, which I actually usually don't call fi, but just by what they are, e0, ei hat to en. So if I write the hat over an index over such a coefficient, I mean this thing is not there. So this thing has only n components. The ith component is not there. So this is meant to be e0 to ei minus 1. Ei plus 1, this is a very standard notation. But once I explain it. So this is a map from delta n minus 1 to delta n. How does it go? Maybe I'll write down the map just explicitly, although it is written here. You would have to translate it. So if I have an element, so if I have an element, sum i equals 0 to n minus 1 ti, ei. So this is any element in this thing. I send it to the same linear combination of all until n with the i excluded. So this will be the sum i equals 0 to i. I can't call it i because I don't, because I have the i here. From k equals 0 to i minus 1, tk, ek. And then afterwards, I've taken this away. So I have to shift by 1. So this is plus sum k equals i to n minus 1, tk, ek plus 1. So that's what this notation probably means. And notice that the image is, as always, just this thing, the corresponding simplex, convex hull, all except for ei. But you should notice that this is indeed, in the standard sense, a phase of the simplex. So this is the phase of the simplex e0 to en, opposite to the point ei. So if you make a, I mean, I can just make some pictures to, I mean, if you have, for instance, if you have the interval here, this would be delta 1. Then f0 of delta 1 of delta 0 is, for instance, equal to, we take away the 0. So this is just the point e1. And if you look at it, this is one of the faces of this thing. So if we have delta 2 e0 e0 e1 e2, then what is f0 of delta 1? It's, what is f1 fc f0 of delta 1? Well, this is the linear combinations of e1 and e2. We talk because it's all e0. So this is just the interval e1 e2. So the span of e1 and e2. And you can really see this is, and if you try to make a picture of your mind of the next case of tetrahedron, you can also say that you always get the phase opposite to the vertex you have thrown away. So I think this is enough. Now you will, most likely at this point, have started asking yourself why I talk to you in such a complicated way about something as simple as simplicity. And that is certainly a quite justified question, because these things are rather trivial. And we make all these kind of slightly intricate definitions with them, but it doesn't change the fact that the things are still trivial and do not contain anything interesting. But now we want to use them to define our homology, and then these things become important. So as I mentioned at the beginning, the homology, so single homology of a topological space x, is defined using continuous maps from simple c's. So sigma from delta n to x. So we somehow, if you want, use the simplices as some kind of probe to find out how x looks like. So we have somehow our thing here, and here we have our x. And we have this map, sigma. And it somehow maps it in a very complicated way. It's just a continuous map. And we want to somehow understand something about how x looks like by studying these maps. That seems to be a crazy thing to do. And because as you can imagine, there's an incredible amount of such maps. If this is im, you have kind of uncountably many such maps and they can do just about anything. But if we do homology, we somehow finally get only the interesting information. So let's now look at the definition. Let x be a topological space. So a singular n-symplex, we usually forget the word singular, n-symplex in x is a continuous map. Sigma from delta n, so from the standard n-symplex to really just a simplex in our n-toe. And now, in order to get something useful, out of this we want to make it first considerably more complicated by looking not just at these maps, but at formal linear combinations of such maps. So technical word is we look at the free abelian group generated by them, but maybe that will not. So definition. So let x be a topological space again. So x is still a topological space. The singular chain complex, c star of x, so I hope you still remember that we talked about chain complexes, which were the sequence of abelian groups together with differentials with these maps d, which always d, if you want to call it dn, goes from cn to cn minus 1, and so on. And now we have to, in this particular case, we want to define such a chain complex. So we have to define some abelian groups and some differentials between them. So is defined as follows. So we call cn of x is the following thing that we might want to make sure that you understand what it is. So it's the set of all formal sums, sum i equals 1 to n, ai sigma i, where sigma i from delta n x is a singular simplex in x, so it's a continuous map. And the ai are some integers. And n is any positive or non-negative number. So I don't know whether you are familiar with such formal sums, maybe you could get some feedback. So in principle, this is just a shorthand for saying we have a map, so you can see that this is for all n bigger than 0. I will explain a bit more in a moment. And we put cn of x equal to 0 for n smaller than 0. So we have these abelian groups. Hopefully we'll see in a moment cn of x for all n. For n smaller than 0, they are defined to be 0. For n bigger than or equal to 0, they are somehow defined in this complicated way. So I can write this also differently. And then I explain what it means. So equivalently, we write an element of cn of x as sum, so all over all sigma r sigma times sigma for sigma r, the n simplices to sigma n simplices in x and a sigma in z, and only finite the many non-zero. So maybe this notation explains what makes it a bit easier to understand what the meaning of a formal sum is unless you know it already. Namely, it's just a shorthand for something else. So here, sum a sigma times sigma just stands is another notation for a map from the singular simplices in x, namely, sigma is sent to a sigma. And so the n chains, so another word for describing these n chains, so cn of x is also just the map, the set of all maps from singular simplices in x, similar n simplices in x to z. With the property that, if f is in cn of x, only finite the many, f of sigma are non-zero. So let me try to briefly put this together. I just want to make sure that you know what such a formal sum is. So in future, I will only use the notation of formal sums, either in this form or in this form. So an n chain is such a formal sum. But if you are not familiar with formal sums, you can view it as some kind of shorthand. Namely, the formal sum is another word for talking about a map, which associates to every simplex, every sigma, an integer with such a property that only finitely many of these numbers are non-zero. And the correspondence is by sending the map sigma goes to a sigma to the formal sum, sum a sigma times sigma. I mean, it's standard notation. Maybe you are familiar with it, but I wanted to make sure. And from now on, we use it. So I have defined to you these sets for the moment. Why are they groups? So cn of x is an obedient group. Namely, in the obvious way, you can add to such formal sum by summing the coefficients or in the language of functions of maps like this. You just sum the maps. Which one? You don't understand. No, I cannot. So this one. So you don't understand the notation in this. So I have defined so. So what's written here is sum a sigma times sigma. So a sigma is an integer. Sigma is an n-symplex in x. And what this is supposed to be a formal sum, where only finitely many of this a sigma are non-zero. And this notation can be viewed as a shorthand for the map from the set of all possible sigmas. So the singular n-symplex simplices to z which associates to sigma a sigma. So I just write down the same information as this map in terms of this formal sum by writing as a coefficient of sigma the thing it is mapped to. In other words, I just have to associate to every sigma a number of which almost all of these numbers are non-zero. It's supposed to be finite formal sum. And that's what this notation says. This is more or less clear. So I want to say that this is in a BN group. And if you understand the notation, that's fairly obvious. Namely, if I take the sum, I can just define sum sigma a sigma times sigma is map plus sum sigma b sigma times sigma, where again only finitely many are non-zero. Is defined by just adding the coefficients. So this is sum sigma is plus b sigma. And you can in the same way form the difference. And the zero element is the formal sum where all the coefficients are zero. And here you can also write it in this form with some finite sum where you only write the terms which are non-zero, where the coefficient is non-zero. And this is seeing the same as viewing this as maps. And then you sum the maps from singular and simplest to z just as maps. If you have one map to z, another map to z, you take the sum. That amounts to the same. You sum the image here. So this corresponds to, just as a remark, the sum corresponds to this plus, it corresponds to f plus g. So taking the sum of two maps, namely, f plus g of sigma is equal to f of sigma plus g of sigma. No? OK. So if you do not understand this notation, you will have to discuss it later because, obviously, from now on, these chains are the main thing that we are talking about. So if before the next lecture you have to understand this, otherwise you will not be able to follow anything at all. So now, we wanted to define these, we were in the process of defining c star of x comma d, the chain complex. So this consists of some ABN groups, cn of x for all n, which we have defined. And we have to define the d. At every n, you have to define a map. If you want dn from cn of x to cn minus 1 of x. So we want to define, which I always call d, we want to define this d. So we define d from cn of x to cn minus 1 of x for n. But in the following way, we use this boundary, these face maps. So if we first define it on a simplex. So an element here is a formal linear combination of simplices. So we can also just look at the simplex itself and define what the map is for this. So this would correspond to having, as a sum, just 1 times sigma and nothing else for 1 sigma. So we define it like this. So we define it on a simplex, sigma from delta n to x. So we define it on such a singular simplex as follows. We say d of sigma is equal to sum i equals 0 to n minus 1 to the i of sigma composed with e0, the i to en. So we take this face map, which gives us a map from delta n minus 1 to delta n. Then we apply sigma, which is a map from delta n. So we start here with delta n minus 1. We map it by this map to delta n. And then we apply sigma to get to x. And so this is a map from delta n minus 1 to x. So we see that d of sigma is a linear combination of a singular n minus 1 simplicity. So this is indeed an element in cn minus 1 of x, as it should be. And so, for instance, if we just look at the, so we have here, for instance, our simplex, e0, e1, e2, we map it with sigma to x. And then, so we take the, basically, we take the restriction of this to the different sides, except that we parametrize them with delta n minus 1. So we have here, again, e0, e1. And so we first map this in all possible different ways on one of the phases of this thing. And then we map it into x. And we take a linear combination of them with these signs. And the signs are obviously there so that we have a hope that we get a chain complex so that d composed with d is equal to 0. And we will now see why that is the case. So I haven't yet finished, though, because I have said that what it is for is a simplex. But I should tell you what it is for such formal linear combinations of simplices. And it's obvious how to do it. We just take the same formal linear combination of the d of the simplices. So we just define it in general by linearity. So d, cn of x, cn minus 1 of x is defined by linearity. So in other words, d of sum sigma a sigma times sigma is defined to be sum over all sigma a sigma times d of sigma. And that certainly makes some sense. I mean, here we have a finite linear combination. So only find the many of the a sigma are not 0. And this will still be here. And we get a few more if we apply all these. We get n terms for just 1 before. But if we're finitely many and out of each of them we make n, we still have finitely many. So it's a bit, as you see, as I kind of, if you want promised you, it is a bit complicated. You make things rather complicated before they can become easier again. Now we want to finish making this chain complex. So we have to prove that the composition of d with itself is 0. So lemma e composed with d, so by which I mean, obviously we have cn of x with d, cn minus 1 of x, again d to cn minus 2 of x. So this combination, this is the 0, a is equal to 0. So if we apply d twice, we are supposed to get 0. And this is supposed to, I mean, this somehow comes from choosing these signs. So let's just try to do it. First, there's one thing that is very easy to check, but maybe you should check it. If we take first, say, a0 to ej, so ej will leave out to en. And we compose this with e0 ei, so this was wrong, en minus 1, ei. So en like this. We take this composition. So we here first take the map to the j's face of this n minus 1 simplex. We compose it with, and now we take this by mapping this face as the i's face into this simplex. And the claim is that this is slightly more complicated than you might have thought. So it is indeed, here we leave out the j's. Here we leave out the i's. So I would think this is obtained. So this is obtained by leaving out the j's and the i's. So this by itself is a map from delta n minus 2 to delta n. So this is e0, ei, no, ej, ei, so en. So we leave out just the j's and the i's. So this, as you can see, is a map from the delta n minus 2 to delta n. So if it looks like this. So if j is smaller than i. But if it is the other way around, then we have here ei. Then ej plus 1 is left out. And if you think of it, it's obvious how the maps are defined. If you remember the formula I wrote down, you just write down, you see immediately that this is what happens. Because somehow doing this means you leave this out and shift everything by 1. And if you do it here, you shift also the one you have left out by 1. So this is if j is bigger equal to i. So you can check this. This is really simple. But given this, we can prove this thing. We first prove that d, composed with d, if you apply to a simplex is 0, then by linearity, it will always be 0. So proof. So by linearity, this was defined in this linear way, it is enough to show that d, composed with d, of sigma is equal to 0 for sigma and n simplex. I mean, for complete this, I say what I mean by this. By linearity, it's enough to show because then if I have d composed with d of sum sigma a sigma times sigma, d is supposed to be linear. That is, this is just defined by applying the d to this. So this is nothing else than sum sigma a sigma d composed with d, or say d of d of sigma. I can write like this. And this is 0. So the whole thing is 0. OK, now let us do the thing. So we just compute. So we just apply the definition. This is the first one, so it's d of sum i equals 0 to n minus 1 to the i sigma composed with e 0. We leave out e i, and we go until e n. This is the first application of d. And d is linear, so we can just apply d to this. And d applied to this just means that we, OK. So this is, by definition, this is sum i equals 0 to n minus 1 to the i sigma composed with, and now we are supposed to apply these other face maps. So e 0, e i hat to e n, composed with, so we are supposed to now apply the d to this, which means that we compose with the face maps of the n minus 1 simplex. So this is e 0, e j to e n minus 1. So this is a map from delta n minus 2 to delta n minus 1. And here we go to delta n, from delta n minus 1 to delta n. Then we apply sigma, and we make this linear combination. And I forgot something. Namely, there was also a sign. It's minus 1 to the j times this. So this becomes minus 1 to the i plus j. So this, I hope the step is clear. And now we have to remember what these are. To leave out, this one means leaving out the j. Here we leave out the i. And this, according to what I wrote here, corresponds to this in these two different cases. So we have to also distinguish these two different cases. So this is sum i equals 0 to the n minus 1 to the i, sum j smaller than i, which is corresponds to this case, minus 1 to the j times sigma, composed with e 0, e j is not there, the i is not there, and then we have the n. Plus, the other case here, sum i equals 0 to n, sum j bigger than or equal to i minus 1 to the j sigma composed with, what is it, e 0 to the i, not there, e j plus 1. Now we can exchange. So the last one, we can also write in a different way. This is the same as sum, so this part here, the same as sum i equals 0 to n, sum j bigger than i minus 1 to the j minus 1. We have to shift it by 1 so that it becomes j, sigma composed with e 0, e i, e j, e n. But now we see that if we exchange i and j, these two are precisely the same sums. So this sum is equal to this sum with one tiny difference. Here we have minus 1 to the j, here minus 1 to the j minus 1. So we have precisely twice the same sum, but with the opposite signs. So that's it follows that d composed with d of sigma is indeed equal to 0. So we find that d is indeed a chain map. So we have that c star of x comma d is a chain complex. OK, so this finishes, in some sense, a definition. So now we know how if we have a chain complex, we define its homology. So we have a chain complex, so we define its homology in the usual way. And this will be the singular homology that we will be talking about for the rest of the lectures. So we have the group of n cycles is z n of x, which is defined to be, again, the kernel of d, if you want dn, but it's always d from cn of x, so cn minus 1 of x. And so this is the cycle n cycles, singular n cycles in x. And pn of x is the image of d from cn plus 1 of x in cn, which are called the n boundaries in x. And we know as d composed with d is 0 that pn of x is a subgroup of zn of x. And so the nth homology group of x is hn of x equal to zn of x divided by bn of x. So we know that cn of x is in a bn group. pn of x is a subgroup, so the quotient is in a bn group. So thus we have the homology groups hn of x for all n in z. And by definition, the cycles in negative degree were 0. So we have hn of x is equal to 0 for n smaller than 0. And the other ones are the interesting ones. So now I think it does take a while to somehow digest this. Also, it seems fairly crazy to expect that you can get any useful information out of this. So we have these crazy maps from the singular simplices into x, which are just continuous. So there's an enormous amount of them. Now we take formal sums of them, so they're even more. And then we do this crazy thing that we define as differential, and we take the homology. So one would normally expect that this is completely impossible to handle. It's huge and so on. But we will find out that in many cases, actually I mean, in many cases, this will be a finitely generated bn group, because these two things are more or less of the same size. And that one can get some nice information from it. I mean, somehow the very vague idea, which you always has to do with the following. So if you have some, I mean, this is just some kind of vague information why this should be of any use. So if we have such a map from delta n to x, and x somehow looks like this and has a hole here, then obviously we cannot map delta, this thing, in such a way that it contains the hole. It always has to kind of be on one side. But now, assume we look just at the boundary of it. So the sides of this thing. So if delta n really looks like this, this is delta 2, we have a map of this thing like this. So there's obviously no map from this to this which contains this hole in the middle. There's no continuous map because we have this hole. But on the other hand, obviously, if you look at the sides of this thing, there's a map from this side to this. There's a map from this side to this. And there's a map from this side to this. So the faces of the triangle can be mapped to it, but not the triangle itself. And so this means, in some sense, we have a map from, so the faces, if you just look at this thing, this will be a one cycle in x. And it would be a boundary, namely the boundary of the map of the triangle into this thing if we could find a map from this into this which just fills this. But as there is a hole, we cannot fill it. So somehow, therefore, this homology will tell us something about the holes that our space has. But things are done in this complicated way to somehow get this out. And we cannot directly talk about the holes, but it's all in terms of these maps. But this is somehow the thing. With these maps, the difference between a boundary and a cycle is somehow that it's a, if we have a cycle, it's always some kind of thing that closes up in some way. There have been higher dimensions. I mean, that's intuition. I'm not saying it's true in the strict sense. So if we have a cycle, it somehow closes up outside. And it will be a boundary if we can fill it up inside x. If there's a hole, we can't. That's the intuition. But the definitions are easy to work with the definitions because intuition doesn't work so well for this. So now let's see whether we can. Let me see where I am. So the first thing we want to prove is something very, we want to prove that this homology is functorial. So that means if we have a continuous map between topological spaces, we get a map between the homology groups. And this will allow us to show that we can, for instance, see from the homology whether two spaces are homeomorphic. So I want to show you see homology is functorial. So that means if f from x to y is a continuous map, then we get a homomorphism, f star. We get homomorphisms, f star, from hn of x to hn of y of a being groups. So we want to define this. And we actually just define it by going back to this homological algebra, but saying in general what we do with chain complexes. Because we anyway will need that later. So for this, go back to chain complexes. So the definition, let c star d, d star d. So I go even a bit worse than what you complained about before. So we have a chain complex here, the differential I call d. We have another chain complex, the differential I still call d. So we have here infinitely many different maps which are called d, and another set of infinitely many different maps which are different from these which are still called d. But it's always just the boundary map in the chain complex. So that means, so the chain complexes. So a chain map, f star from c star d to d star d, and I will usually not try to d, is a sequence of homomorphisms. Maybe now I will actually give them a number fn from cn to dn. A chain complex is given by a sequence of maybe in groups, cn, and maps between them, such that all the maps with the d's commute. So it's commute with the d's, so such that. So if I look at cn, we have a map to dn, and we have also the boundary map d to cn minus 1, and here we have our map fn, and here we have the map fn minus 1. And this should always commute. So it doesn't matter whether I first apply the d and then the corresponding f, or whether I first apply the corresponding f at a different level and then the d. And so we have all these d's, and all these f's, and everything commutes. So if f star from c star to d star is a chain map, we can define a map on homology. It induces, which I call again by a map, I again call it f star, because there are not so many symbols I have. It's induced by f in the same way to f star from hn of x from hn of c star to hn of d star, in the following way. So if alpha is a cnn cycle in cn, so if alpha is in cn of c star, so this means after all that d alpha is equal to 0. So alpha lies in cn and d alpha is equal to 0. Then we can define its homology class. We call by denote alpha its homology class. It's homology class in hn of x. hn of x, after all, is cn of c star, which, as you remember, was cn of c star divided by bn of c star. So we just define the equivalence class of this element in this quotient. We put it by this bracket. And now we want to define a map from the homology here to the homology here. And well, we just say we take the class of alpha to the class of f star alpha. So by the same bracket, we define this also in, if we define d star, so we define this thing f star by f star of alpha, so this means alpha is a class in cn of c star. And we just send it to the class of f star alpha. Now, in order to be able to write down this, we have to know that this f star alpha actually is in cn, that d of it is 0. Otherwise, we are not even able to write the equivalence class. OK? So we have to prove that this is well-defined. So this is well-defined. The first thing is that f star alpha is in cn of d star. So that we can do this. And this just follows from the definition. Because we use this commutation thing. Namely, if we take d of f star alpha, then we want it to be a cycle. This should be 0. Well, we know that d commutes with f star. So this is f star d alpha. And d alpha is equal to 0. This is f star of 0. OK, so that's also difficult. And the second one is I want that this depends only on the class of alpha, not on alpha itself. Now, we say the image of the class of alpha is the class of f star alpha. So the class of f star alpha should only depend on the class of alpha. So if the class of alpha is equal to the class of alpha prime, so this means the difference is the boundary, where beta is in cn plus 1 of x. Then what? When if we take f star of alpha prime, what is it? Well, so we write down f star of alpha plus d beta. f star is a homomorphism. So this is f star of alpha plus f star of d beta. And again, f star commutes with d. So this is f star of alpha plus d of f star beta. But so it's d of something. So it's a boundary. This is an element in dn of d. So it follows that the difference is a boundary. So the homology class is the same, f star alpha. So we see the map is indeed verified. So time is essentially up. So let me see whether I can say something sensible. No? OK. So I think I should stop then. It doesn't make sense to now attempt to finish the proof because it takes another 10 minutes or so. But now we will kind of in the most obvious way say what the push forward is on. So if f from x to y is a continuous map, we know that a singular simplex, no sigma, is always a map in x, is a map from delta n to x. So how do we get from this and the singular change linear combinations of these? So how do we get the map from such things to corresponding things in y? Obviously we're composing this with f. So we apply the sigma will be, so f star will map sigma to sigma f composed to sigma. This is now a map from delta n to y. And then one, no, by linearity, one extends this to chains. And then this will give us a chain map on the singular chain complex. And therefore it will give us a map on homology. But we will do this next time. And then we will see the consequence. What is the difference between this application B and the differential form? Well, by itself, they are completely different things. Obviously because one is an analysis and one is here. But indeed, there is some relation. So it's a rather deep fact that the homology, actually the homology, but this is more or less the same, of a manifold can also be computed using differential forms. And this gives you some kind of, you get some kind of chain complex out of the differential forms. And the D that you have for differential forms will give you a D for this chain complex. And you find out that the homology that you get in this way is the same as the homology we have here. It's a very non-trivial fact. It's a big theorem. And you can also see there's some formal similarity. Because if you look at how the D for the differential forms is defined, the differential forms are always anticommuting when you commute them. And so when you write it down, you get all these signs, alternating signs, which precisely lead to the fact that D composed with D will be zero on differential forms. And this is somehow for the same reason as here. So somehow things are very parallel and very similar. But obviously you are in completely different worlds. But in the end, the results are closely related also.