 OK, so I will start. So we had started to talk about products. So I had told you that if we have products, so I first was talking about products of the fine varieties, close varieties of A n, so if x in A n and y in A m are close sub-varieties, then we have that their product, x times y, lies in A n plus m, and is a close sub-variety too. So it's easy to talk about products of close sub-varieties of A n. And in fact, I'm going to need this later, is that this also works if they are locally close. So this is an easy exercise. If x in A n and y in A m are locally close sub-varieties, then it follows that x times y is locally close sub-variety. I didn't prove it, but it's easy to see how you would modify the proof. And then one thing that we had was the universal property. And that said, basically, that if I have such product, p1 from x times y to x and p2 from x times y to y are morphisms. So x subset A n and y subset A m are close sub-varieties, but one can again see that one can easily generalize the same proof. We prove it for locally close sub-varieties. And these are morphisms. And secondly, we have that a map phi from any, so if z is any variety, then a map phi from z to x times y is a morphism, if and only if the compositions with the two projections are morphisms. This is one reformulation of what we stated the other, what we proved the other time. So that's as much as we had done for close or locally close sub-varieties of a fine space to study the products. And now we want to look at the case of quasi-projective varieties. And the problem, as I had said, is that if x is in pn and y in pm are sub-varieties, it is not by itself true that x times y will lie in some projective space. So as the most general things we are studying are quasi-projective varieties, we want the things actually to lie in some projective space and not in the product of projective spaces, although we could also do that. But anyway, that's not how we do things. We somehow have to make this thing lie in some projective space. And we do this, so thus we find an embedding. So this moment just means an injective map from pn times pm to some larger projective space. And we will identify x times y with sigma of x times y. So we study, so instead of actually taking the product, we look at the image here. It's the image under an injective map. And so then this thing has a structure of a quasi-projective variety when x and y are quasi-projective varieties. We will see that. OK, so first I want to define this embedding, which is called the, or this map here, which is called the Sega embedding. Yes? Why is this project this space y? So I mean, so I don't precisely understand the question, but I think you are asking whether, why it is not true that the product of projective space is a projective space? Well, it just isn't. If you look at the definition, it's a different thing. No, it's not the same. No. The product of two projective spaces is not a projective space, no. If you look at the definition, it's really different. You have pairs of n plus 1 tuple and then n plus 1 tuple, where on each of them it's up to multiplying by a constant. And that's really a different thing than taking some other tuple up to multiplying by one constant. It's really a different space. And if you, for instance, if one does algebraic topology, so if you do it over the complex numbers, you can look at the homology of projective space. And you will find that this has z in all even degrees until dimension n for pn. And for pn times pm, it looks quite different. So it has also some homology groups which are larger than z. So it's really not the same. So it's, anyway. So I mean just to connect to that. So you can even see, well, whatever. I will tell you what n is. For some n, which I will now specify. In fact, I think it's the first thing I write down. So now we come to the definition. And as a first statement, I will answer your question. So we put n to be n plus 1 times n plus 1 minus 1. OK, so it's very precise. n. And so we want to take pn like this when we start with these. So I want to define this Seger embedding. So definition Seger embedding after some Italian mathematician, Benjamino Seger, who defined this. So we take x0 to xn, the homogeneous coordinates on pn. We take y0 to ym, the coordinates on pm. And we also want to somehow take the coordinates on this large pn. So these I call zij, which depends on two indices, where i goes from 0 to n and j goes from 0 to m. Are they coordinates on pn? You can see that the dimensions match. Here we have n plus 1 times n plus 1. And then the projective space has dimension 1 less. So you have a homogeneous coordinates. But somehow I want to not numerate them from 0 to n, but I want to enumerate them by pairs of such numbers, which somehow correspond to the coordinates of pn or pn. OK, now we define our map. So we define a map, sigma. Or if you want also sigma nm, but I will usually forget nm from pn times pm to pn. So this is just a set theoretic map that I'm defining. I'm not defining morphism, no, because we don't know. This is, from our point of view, not a projective variety. So we don't know what a morphism is supposed to be. So we just define a map. And how we do it? So if we have two points, a0 to an, a point in pn, and b0 to bm, a point in pm, we send this to all the products. So the ij's coordinate here will be the product of ai times bj. So this is a point here. OK, and sigma is called the Sagan value. So first we want to see that sigma is well defined. I mean, that's kind of trivial. So if you, I maybe just say it, if you multiply all the coordinates here by lambda, you get an equivalent point here. If you multiply all the coordinates here by mu, you get an equivalent thing. And if you do this, for both of them, you multiply all the coordinates by lambda times mu. And so you see that the class depends only on the class. So this is well defined. OK. And I write sigma and m, or sometimes just sigma, if I remember who n and m is, the image. So this would be some subset in pn. So now we want to somehow, so on pn, we have these charts, ui. So these open subsets ui, which cover it, where one of the coordinates is non-zero. We have them also here, and we have them also here. And we want to somehow say something about how they are connected. So and introduce notation. So for i equals 1, 0 to n, we put ui equal to, I mean, the corresponding ui on pn. So the set of all a0 to an in pn such that ai is different from 0. And for j from 0 to m, we put uj, the corresponding thing on pm. So this is nn, p0 to pm in pm, such that pj is non-zero. And you see that, obviously, this is a certain abuse of notation because sometimes i is both lower than n and m. And it defines two different things. But the idea is just that to make life easier for me, whenever I have an i, it means I'm in pn. And whenever I have a j, I'm in pm. So you just, so that one doesn't have to write so much. And obviously, we also have, for the same i and j's, we have uij, which is the set of all points in pn, plargen, whatever aij in plargen. I cannot have the same index, such that aij is not 0. These are the corresponding charts on the plargen, the corresponding open subsets. And we know that these open sets are isomorphic to the corresponding affine spaces. So we have a ui from, which way does it go around? Yeah, an to, or maybe I'll write it like this. We have isomorphisms. So we have an isomorphic via ui to ui. And the inverse map is fi. And the same way we have am isomorphic via uj to large uj. And the inverse map is phj. And finally, a largen is isomorphic via uij to uij. And the inverse map is phij. So this just is to introduce the notation. So what else do we need in order to see whether i? And so we see that, for instance, pn obviously is just the union over the ij of the uijs in the same way as before. These are these open subsets, so one coordinate is non-zero. This is an open cover. And so sigma, which is the image of this map in pn, is equal to the union over all ij sigma intersected uij. And these things, this is also an open cover of sigma. So it's a cover of sigma by intersection with sigma to open sets, such that I will also write this one as sigma ij. So we have a lot of notation. And finally, we have one more piece of notation then we can actually start. So I write sigma ij like this to be the map from an plus m to uij. So how does it go? So I can take a point of an plus m is a point of an. So a pair of points, p in an, q in am. So we take a pair of pq and we send it. So first we go to the corresponding open subset in the projective space. So this is ui of p and uj of q. And then we apply the map sigma. So here we are in, these are some points in pn times, this is a point in pn times pm, we can apply sigma. And this will be a point in uij. In fact, it will be a point in uij intersected sigma. And so in fact, by definition, we have that the image of this map is precisely sigma ij. We see that if you, so one direction is clear, we know that this maps it into this. But it's clear that for the ij coordinate here to be nonzero, we need that the i's coordinate here and the j's coordinate here is nonzero, which means precisely that it comes from here. OK, and now we are in a position to state the result about this. So now I have to wipe it out, but you have to know all, you have all the notations in your mind. So theorem, sigma is injective, and the image is a closed subset. A closed subset, yeah. And sigma is closed in pn. Later see, it's also closed sub-variety, but that's not immediate. And in fact, we can see how it's given. So sigma can be written as a 0 set of the following set of equations of degree, of polynomials of degree 2. So I take zij times zkl minus zilzkj. And this is for all possible ij's and so on. So i comma k run from 0 to n, and j and l run from 0 to m. Because first is always the coordinate, which, OK. So we have a certain, so by itself, this is n plus 1 times m plus 1 equations, but some of them are maybe trivial. But anyway, so this is what we have. So I claim that this set, the image, is the common 0 set of these equations, of these polynomials. Well, yeah, it would, let me see. I expect it's, you can write in matrix of which it is a determinant. But this is just, these are just four terms. So it's not, by itself, a determinant. So I expect you can write down some matrix so that it is all the 2 by 2 minus. But somehow that's not particularly useful, I think. So I mean, by itself, you just have this equation. The point is here is that you see here, you have ijkl. And then you exchange. So here i goes with change, that goes with l. So you exchange how they are matched. But I expect you can somehow most like find some matrix so that this will be the determinant of some minus. But anyway, OK. So I think I will, if I now state the whole theorem and then prove it, the obvious disadvantage is that I state the theorem, then I have to wipe it out and prove it. So maybe I will now prove the first part, then I state the second part and prove it and so on. I think that's easier to follow. OK, so let me do that. So we start with proof. So first, the injectivity is kind of trivial. Let's just, in order to get started slowly, we can still do it with sigma from a0 to an b0 to em. If that is equal to the corresponding thing for some other tuple, say, a0 prime to an prime b0 prime to em prime, well, then what? What does it mean? It means there exists some lambda in k, which is not 0, such that I can get the corresponding point here. I mean the sigma of it, which we know what it is, by multiplying by lambda. So such that lambda times ai bj, maybe with a prime, I do it like this, is equal to ai bj for all i and j. So there's one lambda which does it for all of them. After all, this here is the vector which has these products as entries, and this is the vector which has those as entries. OK, so exists one such. Now these are non-zero vectors, so we certainly find that one coordinate, there certainly is one coordinate which is not 0. So choose, say, i0 and j0 such that ai0 times bj0 is non-zero. So then we can just, you know, this holds for all i and j. So then we can do the following. So we have, so then we have for all i, we have that if we put, I mean we can write out, ai times bj0 is equal to lambda times ai prime bj0 prime. And so we can divide here, ai is equal to, what is it, lambda bj0 prime divided by bj0 times ai prime, times ai prime. So what is it, and this holds for all i. So that means, so this is some non-zero constant, OK. So we have some non-zero constant so that if I multiply the vectors of the i primes with this non-zero constant, I get the other vector. So that means that a0 to an is equal to a0 prime an prime, OK. And obviously the same proof with the role of i and j exchanged will show in the same way we have that b0 to bm is equal to b0 prime until bm prime. OK, so this shows the injectivity. Now we have this more interesting fact that the image is closed, actually given by some explicit polynomials. So if I, so let w be the zero set here on the right-hand side. So on the right-hand side of the equation, on the left-hand side of the blackboard. So clearly we have that sigma is contained in w. Why is that clearly? So if you're in the image of sigma, then the ij's coordinate is ai times bj, where the a is a vector in pn, and b is a vector in pm. So if I have here ai times bj times ak times bl, obviously it's equal to ai times bl times ak times bj. Multiplication happens to be commutative. So that is clear. But now we have to see the other direction. So now let aij, again index indices go from 0 to n and 0 to m, ij, be an element of the point in w. We have to show it in sigma. So we choose a coordinate where this thing is not zero. So we choose k and l such that akl is non-zero. Obviously this is a point in projective space, so it certainly has a non-zero coordinate. And then we just write it down. So we have our vector. This is a reasonable vector. We can certainly multiply it by a number, by a non-zero number. We take this one. This is the same vector. We just have multiplied that. But now we can use this equation. So we can move the indices from one side to the other. So this is, so I can take akj, but using this equation, times a, what is it, i, l. What's this in l? It's still i and j are running. So here something has really changed, but this is OK by the equation. But what is that? I'm saying here we just have the ij coordinate of this thing. Is the product something where we have j here? And the product of something where we have an i here? So that means that this is actually equal to sigma of the vector. So here we have the one that runs here, the i. So a 0l until a nl, ak0 until akf. This is how the map sigma was defined. And so we see it's in the image of sigma. OK, so this proves the first part. Now I should state the second. So I had this map sigma ij, which I'm sure you will remember. So sigma ij from a n plus m. So this is now the second statement of the theorem. From a n plus m to this ij, I say is an isomorphism. So I find that this map sigma ij, so this sigma ij, after all remember this was sigma intersected u ij, we find that this is actually isomorphic to a n plus m. As it should maybe be, this somehow is supposed to be what we get for p n times p m. And so locally, p n is equal to the open subset u i of a n is isomorphic to a n. The open subset uj of p m is isomorphic to a m. So we might hope that. And remember that the map was we take a map pq and it's mapped to sigma of ui of p to j of k. OK. So let's see. So it is correct here. I only canceled it. OK. So obviously we can assume, so we do the proof. So we can assume that i is equal to j is equal to 0. Obviously this makes no difference. So if I look at the map p0,0 composed with sigma 0,0, so what does it do? This will do what it takes an n tuple a1 to a n. And another one b1 to bm and sends it to what? So maybe I can first, maybe I can do it slowly. So if I apply this to this, what will we get? So first we apply to sigma 0,0, we map this into protective space. So this is p0,0 of sigma of the thing starting with 1 a1 to n 1 b1 to bm. And then the sigma will consist in taking all possible products of these elements. So this is equal to the vector c ij. ij goes from i from 0 to 1, j from 0 to m such that c0,0 is equal to 1. c i0 is equal to ai, c0,j is equal to bj. This is for i bigger equal to 1, j equal to 1. And c ij is equal to ai bj for i and j bigger equal to 1. That's just the definition. And then we apply p0,0 to it. So this is the same thing. So this will be c ij, ij where i,j is different from 0,0. So we just leave out this one. So maybe I write it like this. So if you write it in coordinates, it means, you know, so I have here just the coordinates are just either ai or bj or ai times bj. So it's always a polynomial in the coordinates on a n. So it follows that this is a morphism because it's given by regular functions on a n times a n. So phi,0,0 composed with sigma,0,0 is a morphism. And thus sigma,0,0 is a morphism because we just composed with the inverse of an isomorphism. And we know now that by definition, this map sigma,0,0 is surjective onto sigma,0,0, large sigma,0,0. Because that's the large sigma,0,0 was defined as the image. So sigma,0,0 and a n plus m is irreducible. So we know that if we have a surjective morphism from something irreducible, I mean a surjective map. So this is by itself is a morphism. If you have a surjective morphism, then the image is irreducible. Because this is a continuous map. And if this was a union of two closed subsets, then the inverse image would be the union of two closed subsets and so on. So it follows that sigma,0,0 is irreducible. So thus it is a quasi-protective right. So we can define the inverse map. So sigma,0,0 is an isomorphism because the inverse map. For instance, if I just define it in coordinates, I claim it's just z n,0 divided by z,0,0 until z 1,0 until z n,0 divided by z,0,0 and then z,0,1. Now if you start with a1 to an and b1 to bm, and we do this, so the first coordinate will then become 1. And we see that if we apply this to it, we get back precisely what we had here and also the other way around. So this is the inverse. And this is a morphism, no, because it's given by regular functions on the open subset u,0,0. So this was the second part. Now we come to the third part of the theorem, which I should find a way to answer here. So the third one is something very simple, but anyway. So for if I take a point q is a point in pm, then the map iq, actually I shouldn't call it, maybe I call it iq bar, from pn to p large n, which sends p to sigma of pq is a morphism. In fact, we'll see it's given by polynomials of degree 1, and the image will be a projective subspace of pn, so just a linear subspace. Anyway, so this is quite simple. And you just write down the definitions. So this is kind of obvious. So let q be given in so as equal as p0 to pm. And then what is iq? Well, it's just a point is just mapped to all the point in p large n, which is given by all the products of these coordinates with these coordinates. So in other words, iq as a map is just I take bj xi ij. So the ij coordinate is bj times xi. And so you see this is a map which is given by polynomials of degree 1, so it certainly is a morphism. And we can see that the image is just, in some sense, also projective space. So the image is the set of all ej ai, where a0 ij such that a0 to an is n. This is the projectivization of some linear subspace of an plus 1, of tan plus 1 with a large n. So it's a projective linear subspace. OK, so this was the third one. Now, we proved this just because we want to use it. I mean, this is not really a big deal. But we want to use it to prove that the product is irreducible. Because if you remember, we had this lemma for what we used for affine varieties, that if you have the topology on the product such that maps like this are continuous, then and both factors are irreducible, then the product is irreducible. And so therefore, this will follow. So let's see what the statement is for x in pn, y in pm, quasi-projective varieties. We have that sigma of x times y is a quasi-projective variety. And if x and y are both projective, then sigma of x times y is projective. I think that is all. OK, so this is a statement. So I maybe will, for simplicity, only prove the statement with a projective. So we assume x and y are projective. And then we prove that sigma of sigma of x times y is a projective variety. It's easy. You can easily adapt the proof for the quasi-projective case. If it's quasi-projective, it is something closed minus something other closed. And then you have to write a little bit more. But it's the same thing. So first, I want to show that x and y be projective algebraic sets. So we first want to show that sigma of x times y is a projective algebraic set. And it's similar in the quasi-projective case. So sigma of x times y, I can write, if I want, this union over all ij of sigma of x times y intersected uij. That's not very exciting. But I know that this thing lands in uij, if and only if we are in the image of the map phi ij. So I can also write it like this. Maybe you can see whether you believe me. I take sigma ij of phi i of x phi j of y times phi j of y. So this would mean I take x intersected ui, then I map it to an, and then I apply sigma ij. And the same like this. So this is the same thing. So now, in order to show that something is a closed subset, it's enough to prove that the intersection with any open subset of an open cover is closed in that open subset. So let's see. So phi i of x times phi j of y is a closed subset of a n times a n plus m. Because x is closed in, say, in a n, then x is closed in pn. If I intersect it with ui, it's closed in ui, and then phi is an isomorphism. And so this thing will be closed in phi of x is closed in a n, phi j of y is closed in a n, and the product of two closed subset like this is closed in a n plus m. And sigma ij is an isomorphism. So sigma ij from pn times pm to pn to sigma ij subset pn is an isomorphism. Thus, it follows that sigma ij of phi i of x times phi j of y is closed in sigma ij. And so we have an open cover of sigma by this open subset sigma ij. And the intersection of the image is closed in the open subset for each open subset in the cover. So it follows that sigma of x times y is closed in sigma. And sigma is closed in n, such that sigma of x times y is closed in pn. It's closed in closed. So this proves that. So it is closed if we have two projective algebraic sets, then the image of the product under the sigma embedding is also a projective algebraic set in p large n. And now we have to show it's irreducible. And we want to use a lemma that we had last time. This said that if we have a topology on the product, x times y, such that if x and y are both irreducible and we have a topology on the product, such that this map iq, p is sent to the pair pq and the map jp, q is sent to pq, is continuous for all pq, then the product will be irreducible. So we apply this here. So we have sigma from x times y to sigma of x times y is a bijection. So we could just say that we, so thus we only need that iq, which is the same as sigma composed with iq, iq is a map p maps to pq, that this map is continuous and also jp continuous for all pq. So we only need that. Then it will follow that as x and y are irreducible, it will follow that this is irreducible. Well, but we know that it's continuous because we have seen it's a morphism. So therefore, I mean, if we wanted to directly apply the previous lemma, you would have to argue on x times y. So then you would have to say we take this bijection and we use it to put topology here. That comes to make this a homeomorphism. And then it would follow directly here. But then it's the same statement here if we do it composed with sigma. OK, so this proves the, so this was the last part of the theorem. So we have, what have we managed to do? So we have this map, sigma, the Seguin betting, from pn times pm to pn, p large n. And we see that it's injective. The image is a closed sub-variety. And in fact, if we apply it to a product of two sub-varieties of pn and pm, the image will be a sub-variety of p large n. And so therefore, we have somehow, if instead of looking at x times y, we look at the image, which is bijective to the thing we had before, then we get a variety, a projective variety in the sense that we had before. And so we just want to do that. We want to identify the product x times y with its image in p large n under this map. So the alternative, you could, so that, otherwise you would say this is not really an amplification, but we define as what we call the product of x and y, the image of the product. So that it's a variety. And then, by abuse of notation, this image, we just call x times y. So let me write this down or something. No swear m. So we have again. So if x and y, x subset pn, y subset pm, projective varieties, we identify x times y with the image pn. And in particular, we identify sigma of pn times pm with the image sigma. And so these are then, this is a quasi-projective variety. This would be a projective variety. So if we do it like this, we find that the product of varieties is, again, a variety. And let's see. So now we want to see that we can work with this in a nice way. So first, so in principle, obviously one could always, I could do without this identification, it's just, but if you always have to remember the sigma and so on, it makes it a little bit more complicated. So it's easier to think that we're actually working on pn times pm. It's just that what are the close subsets are just the inverse images of the close subsets in pn, in p large n under this map, and so on. And a function on this thing would be regular if and only if it's the pullback of a regular function on the corresponding set in the image. So a remark, if you still remember part 2 of the theorem, this said that, what was it? It said that this map sigma ij from an plus m to sigma ij is an isomorphism. So if we now do this, translate this with this identification that we instead of saying we are in sigma ij, we are in the corresponding subset ui times uj of pn times pm, then this just says, so this part just says that if I take ui times uj in pn times pm is open, namely this would be the intersection of sigma with uij. And if I take the map phi i times phi j from, and I'll take the inverse of this map anyway, it doesn't matter, ui times uj to an plus m is also an isomorphism. Because this is just the sigma ij is just the image of ui times uj. And this map, the inverse of this map is just phi i times phi j, if you look at how it is defined. So phi i times phi j is just a map which sends map pq to phi i of p, phi i pj of q. And this is, this is identification, the inverse map of sigma ij. OK. So we also have our charts, which are the obvious charts. So the things that we have proven in this theorem translate into some very simple statements. I mean, if we do it in these coordinates, so that somehow just tells us that we can work with the products in the kind of obvious way. You know, everybody would think that the product of two open sets is a product. And if you take the product of two such charts, then this will again be a chart on the product. And this is true if we make this identification. This is what we have proven. OK. We can maybe look at the simplest example, which is p1 times p1, because in particular we, so if I take, now I take sigma of p1 times p1, which I have also, if I want just 4p1 times p1, this lies in p large n, where n is 1 plus 1 times 1 plus 1 minus 1. So this is 3. So OK. And this can be written. I claim this can be written as the 0 set of just one equation, say, 0, 0, 0, 0, 1, 1, minus 0, 1, 0, 1, 0. So a power i, if you look at it, there will be more equations. I mean, the equations that I wrote down, it's for all, you know, there should be, what is it, for equations or something. But if you write them down, most of them are just 0 is equal to 0. And this is the only one that remains. There's one equation. OK. So you see that this thing is really the 0 set of polynomial of degree 2 in p3 gives us p1 times p1. And as I told you, for instance, we have here kind of some p1's in here. We have the point p times p1, and for a point p here, and the points q times p1, the point q here. And so sigma of p1 times p1 contains two families of lines, namely, so sigma of p times p1. This will actually be a line in p3, and for all p in p1, and sigma of p1 times q for all q in p1. So through every point in p1, there will be two lines of this kind. And I have some kind of picture in the notes. I mean, it's difficult to imagine how this should go. But anyway, this thing is covered by these two families of lines. OK, so finally, for the moment finally, I want to come to the universal property. So in the case of a fine varieties, we had the product. We had the universal property of the product. We want to show that the same universal property also holds here is as if, as our definition of the product, we take sigma of the product. And as I kind of tried to mention the other time, the thing which makes something into the product and behave like the product is the universal property. So therefore, if sigma of x times y fulfills the universal property of the product, it is for all means and purposes, the product. OK, so we have the universal property. It's the same as before. And it's also not much more difficult. We will reduce it to the fine case. So let x and y be quasi-projective varieties. You can assume that x is in PM, y is in PM. So then, first, the two projections, p1 from x times, yeah. So I now write it in this kind of abuse of notation. So I write p1 from x times y to x and p2 from x times y to y are morphisms. So this rather means you take sigma to the minus 1 composed with p1. Anyway, these are morphisms. And the second statement is if z is any variety, the morphisms from z to x times y, by which I really mean sigma of x times y, are precisely given by pairs of morphisms 1 to x and 1 to y. They're fg from z for x times y given by sending a point p to the pair f of p, g of p. And if one wants to not use the abuse of notation that the morphisms would be z goes to sigma of x times y, and this would be fg composed with sigma. You just take sigma of this. So it's the same thing. So to be a morphism, it's enough that its restriction to any open subset on a chart on an open cover is a morphism. So it's enough to see that p1 is restricted to x times y intersected ui times uj, so that this is a morphism. But we could use what I just said. We have this map phi i times phi j from ui times uj to an plus m. And then we have here the map p1 to an. And this certainly is a morphism. No phi i times phi j was a morphism, so it's a morphism. And the p1 given here is just a restriction of this map. So this is not here, so this phi i times phi j. And the same, obviously, for p2. So this shows that p1 and p2 are morphisms, because in the charts, they are morphisms. And now for the second part. So we have two statements. So the morphisms are precisely those. So if we are given a morphism, we have to show it's of this form. And if we are given something of this form, we have to show it's a morphism. So first, let h from that to x times y be a morphism. Now we know that p1 and p2, the two projections, are morphisms. So then f, which is p1 composed with h and g, which is equal to p2 composed with h, are morphisms. And by definition, h is fg. It's the map which sends the first component. We have f of this and the second of this. We have n, h is equal to fg. OK, so this is the trivial direction. So now we want to go the other route. So let f from z to x and g from z to y be morphisms. We have to show that this fg is a morphism. So we put zij to be, what is it, the inverse image by f of the open subset. So ui, so x lies in pn, y lies in pm, and we have the ui and the uj there. So the inverse image of ui intersects it. As f and g are morphisms, this is an open subset of z and this is an open subset. So this is open in z. And clearly, z is covered by these zijs, if I go over all i and j. And so thus, fg is a morphism, if and only if. It's restriction to each of the zijs is a morphism. Because we know that something is a morphism if it's restriction to all open subsets of an open cover as a morphism. So if and only if fg restricted to zij is a morphism for all ij. OK, now we just look what it is. So I take this map zij and I apply to it the map f comma g. So point p is sent to f of p comma p of p. Well, this will send us to x times y intersected ui times uj. OK, we want to show that this map is a morphism. We compose with the phi i times phi j. This phi i and phi j, we know, are isomorphisms from ui times uj to an times n. So this is phi i of x times phi j of y. So this map is an isomorphism. So in order to see that the original map fg was a morphism, it's enough to show that the whole composition is a morphism. But notice now we are in the affine case. So we are just looking. So this map is just phi i composed with f phi j composed with g, which is the map zij to an plus m. And so now we are looking at a morphism from zij to an, and the morphisms are from zij to am. And this is the product morphism according to the universal property for affine varieties. This is a morphism. And actually, OK, this is not necessarily closed. So for quasi-affine varieties, like I said in the beginning of the lecture, so we can reduce the statement to the affine case by using this cover. So that means we have this universal property. So again, this says just that it's very easy to see what the morphisms from anything to the product are. They are just given by pairs of morphisms to the factors. So I can give a couple of trivial examples I mean to just what morphisms are. So for instance, if f from x to y and g from z to w are morphisms, then I can take the product of these morphisms, f times g from x times c to y times w is a morphism. And here, obviously, what I mean by this is that the map pq is mapped to the pair f of pg of q. And this is essentially trivial because what is f times g? So as a map from x times z to this thing, I'm saying that f times g is just equal. So we start here. We take f composed with p1. So we first project to x, and then we apply f. It just means we take f of p, comma g composed with p2. And so therefore, it is a morphism. And I mean, this is not very exciting. So in particular, if x is isomorphic to y and z is isomorphic to w, then x times c is isomorphic to y times w. That's kind of clear. So if f, so assume you have f from x to y is an isomorphism, g from z to w isomorphism, then f times g from x times z to y times w is a morphism. But it is even an isomorphism because the inverses, obviously the product of the inverses. So this is all very simple. Now, my time is essentially up. So in the next lecture, we will talk about two important properties that are already kind of hinted at before. So one is separateness. So if you have, so actually the statement is that if you have the diagonal in x times x, so if x is a variety, if you look at its diagonal, so this is a close sub-variety. That is actually a non-trivial statement. So and this is the, this is called separateness. And it has some nice properties. And you can somehow imagine that it is related to things behaving a little bit like, it's a little bit something like a Hausdorff property, that the diagonal is closed. It is, for instance, if the topology on the product of x with itself was the product topology, then it would be equivalent to then Hausdorff would be equivalent to the diagonal being closed. But the risk topology on the product is not the product of topology for the risk topology. So that's not quite true. But somehow still the fact that the diagonal is closed will give us some properties which are similar to Hausdorff if we study maps. For instance, we find that the graph of morphisms is closed like you used to, you know, and similar things. And the second statement will be completeness. And that is what should replace the compactness properties. So it is, so completeness is a bit more difficult to describe. But one consequence is that, you know, if you have a variety which is complete, then the image of any morphism starting from that variety will be closed. And this is, in some sense, you can view as a new way to make projective varieties. You just take the image of any morphism. And it is not true that all varieties are complete. In fact, what is true is that a variety is complete if and only if it is projective. So it's really a special property of being projective. And that's also compatible with what I have tried to say, that somehow projective varieties are what we want to replace compactness. So OK, and we will prove that. And this actually is slightly tricky proof. And then with that, we should be more or less done. And afterwards, we come to rational maps. Anyway, so the next time will already be on Friday. Because I think I have to skip some lectures, so I have to catch up.