 So we were talking about morphisms. So the last thing we had shown, that if we look at this map, so phi i from ui to an, which is the user map that sends a point a0 to an, so here we have a i, a n and a i is not 0 because when ui sends this to a0 divided by a i until this one we leave out, that this map is a morphism, is an isomorphism. This was, we know that morphisms to an are given by n tuples of regular functions. And so we proved it that way. Now we want to give some simple, some applications. So just some facts. So first, every variety is isomorphic to a quasi projective variety. So here, I mean sometimes I had said that a quasi projective variety is just anything that we can say that we sometimes say that even if you have something which is locally closed in an, it can be viewed as locally closed in pn and so on. But here we don't, here we really mean that a quasi projective variety is a locally closed subset of some projective space. And the other statement is that every variety has an open cover by fine varieties. Recall that in a fine variety is a variety which is isomorphic to a closed subset of some an. So as I said, this statement is quite useful for theoretical purposes because it means one can prove things that are facts which are local on a variety by just proving it on fine varieties. And this is quite useful also. There is the concept of an abstract variety, which we do not consider, which is somehow motivated by this. So an abstract variety will be a variety which will be some space which in a suitable sense has an open cover by a fine varieties. So the generalization goes this way. And if one goes to the general, the modern algebraic geometry, one has a notion of schemes which are more abstract and more complicated. And you have a notion of an affine scheme, which is a generalization of an affine variety. And then a scheme is something which has an open cover by a fine scheme. So it's again a generalization of this. OK. So this is how by here things are quite simple. So let X be a variety. So if X is a locally closed sub-variety in some projective space, then by definition it's quasi-projective. So there's nothing to show. So we can assume X is locally closed sub-variety in AM. So we can put, let's say, put Y to be so phi 0 to the minus 1 of X, which is now, so phi 0 is the map from u0 to n, which is Pn is locally closed. Because we know that phi 0 is an isomorphism, in particular homomorphism. And so we therefore know that X is, and it's also locally closed sub-variety of Pn, because it's irreducible, because X is irreducible, and phi is an isomorphism. And so it's a homomorphism, and being irreducible is a topological property that is not the union of two closed subsets, unless one of them is equal to it. So this is a locally closed sub-variety, and obviously is an isomorphism. OK, so this is basically a triviality. And so the second one is not much more difficult. So we have here a local statement, so every variety has an open cover by a fine variety. So if X is quasi-projective, we have that we can write X equal to the union i equals 0n of X intersected ui. And X intersected ui is locally closed sub-variety in, or at least isomorphic, in n. So if X is a variety, it's either quasi-projective, or it's a locally closed sub-variety in n, and it has an open cover by things, which are isomorphic to locally closed sub-variety in n. So as we have here a local statement, we have to just show it is enough to check it in case X is a locally closed sub-variety in n. So we can assume, OK. So what do we have to show? We have to show it has an open cover by a fine varieties, so it's equivalent to say that every point on X has an open neighborhood, which is in the fine variety. That's the same statement, only set in a different way. So equivalent for every point in X, there exists an open neighborhood in an X, which contains p such that u is a fine. Such that u is a fine. And the fine meets isomorphic to a closed subset in some an, in this case in an. So we have assumed that X is a locally closed sub-variety in n. So we can write maybe X equal to, say, Y without Z, where Y and Z are closed sub-varieties or closed subsets in n. And now we have to take any point and find the neighborhood. So let p point in X. So then there exists a polynomial in kx1 to xn, which does not vanish at p. Maybe a non-zero polynomial, a non-constant polynomial, such that with f of p is different from 0. And always find that polynomial which doesn't vanish at a given point. So then, let me see, however, I made a mistake. I mean, I was a bit too fast. I mean, that's also true, but that's not what I want. So what I want is that there exists an f in the ideal of Z with f of p different from 0. This is because p does not lie in Z. It lies in X, which is Y without Z. So then, Z is a closed subset. So the zero set of the ideal of Z is Z. So it's clear that this is the case. So now we put take Xf, and remember what Xf is. This is the set of all points p, maybe now they're called q in X. Subset f of q is different from 0, is an open subset in X. And we see, I can actually put Yf. So you see, this is an open subset in Y. And we know that Yf is a fine, because we proved that we have a fine algebra set. We take this thing, and it contains p. And we see also that Yf is contained in X, because we take away, so f is in the ideal of Z. So the zero set, so Z is contained in the set, which we take away when we make Yf. So thus, Yf is in a fine open neighborhood. OK, so that was simple. But as I said, it is sometimes quite useful. So now we want to describe morphisms between quasi-projective varieties, again coordinate-wise, in a very similar way as we did with fine varieties. So that it's something that your coordinates are locally given by polynomials or something like that. And so let's do this, because as I said, the definition of morphism is a bit abstract and useful for theoretical purposes, but usually not for studying examples, so theorem. So let X in M, Y in N be quasi-projective varieties, and let phi from X to Y be a map. So then the following are equivalent. First, phi is a morphism. Second, phi is given locally by regular functions. So by this, I mean the following. So for all points P in X, there exists an open neighborhood U in X, which contains P. And regular functions H0 until Hn, which are regular on the whole of U with no common 0s on U, no point in projective, what we see in a moment, such that phi of Q is equal to H0 of Q until Hn of Q for all Q in U. So I can also write phi is equal to H0 Hn on U. So this is the first statement. The other, and you can see that we need that we have no common 0s, because otherwise this would not be a well-defined point in projective space. And the other statement is with polynomials. So it's locally a polynomial function, a polynomial map. So by this, I mean a similar statement. So for all points in X, there exists an open neighborhood U in X and polynomials. So maybe I call them F0, so Fn, the variables on the space we start with, which are homogeneous of the same degree. And as before, with no common 0s on U, such that the map is given in the same way. So phi of Q is equal to F0 Q until Hn of Q in U. So I can again write phi is equal to F0 F. OK, so this is rather similar to the affine case. It's also rather simple. So we will prove it by reducing it to the statement in the affine case. Now we have proven, and for this we use the result we just proved, that we have this open subset U0 is isomorphic or Ui is isomorphic to An. OK, so I mean, so if phi is a morphism, then it's also morphism as a morphism to Pn. So I can just look at that case. So if phi from x to Pn is a morphism and P a point in x, well, then it will be true. You know that Pn is covered by this open subsets Ui. So the point P will map to one of the Uis, maybe to more than one, but will map to one. So there exists an i such that P of P is in Ui. And for simplicity of notation, I assume that it maps to U0. I mean, otherwise you can manage with indices. So assume P of P is in U0. So let U be a neighborhood, open neighborhood of P such that, so P in x such that phi of U is contained in U0. For instance, I could just take the inverse image of U0 on a phi. So then obviously if we take this map phi 0, you know phi 0 was this map from U0 to n, this isomorphism, and compose it with phi from U to n is a morphism. So by what we know is, we can write it as an n tuple of regular functions, regular functions regular on the whole of U. We have this result, how morphisms to a n go. So in order to get back phi, we have to compose with the inverse of phi 0, which after was the map, which took an n tuple like this and added one to it at the beginning. So then we have phi is equal to U0 composed is phi 0 and this is therefore equal to 1 H0 H1 for every point q, we do this. And so we see these are, one is obviously a regular function on the whole of U, it's regular everywhere. And these are regular functions, so the map is given by an n plus 1 tuple of regular functions. This proves this statement. Now, 2 to 3. So assume our map is given by an n tuple of regular functions. So assume phi is equal to H0 to Hn, I mean on U. So we have an open cover, so that for every point p, there's an open neighborhood such that and so on. And the HI are regular with no common zeros. Well, we can make, we are supposed to just find for every point p and x an open neighborhood. So we can, where something is true. So we can always make our open subset smaller. You can simplify it. And by making it smaller, we can assume that these regular functions are again the quotients of homogeneous polynomials of the same degree. So the smaller can assume that HI is equal to fi divided by gi with fi and gi in k x0 to xn homogeneous of the same degree. I'm not saying that for all i, the f i's and the g i's have the same degree. It just means if I fix i, then f i and g i have the same degree. And g i non-zero g i without zeros U. We've done this many times. Now I want to prove this statement. So I just have to clear the denominators. Write that for all i, let say li equal to what? I take, I think, fi times g0 times g1 until I leave out gi until gn. So then the fi and the ni are homogeneous, all of the same degree. It's the same as the degree of the product of all the gi, because the gi has the same degree as ever. And I claim they have no common zeros on U. Let me, you can check. And we see by definition, we have that fi on U. p is equal to h0 for hn, where we have this. And now we just multiply the whole thing with the product of the gi's, which then we get l0. Because after all, multiplying by, if we do this for any p, we multiply with the value of the product of the gi on the coordinates. This is just the same constant for all of them. And it's fine. And we know that the gi don't have zeros on U. So we can multiply this without the li getting a common zero. And so this is this statement. And now finally, we have to prove that if we are given a map by polynomials in this way locally, then it actually is a morphism. So this is in 3 to 1. So to be a morphism is a local thing. If I have a map, that it's a morphism is something you have to check just locally. If you have an open cover of the source and each open subset of the cover is a morphism, it is a morphism. So let phi, assume phi restricted to U can be written as, say, h0 for hn for some polynomials as before, homogeneous of the same degree, and with no common zeros. Then the only thing we have to show is that phi restricted to U is actually a morphism. And that's so we can, by making U possibly smaller, again, we have to only find an open cover, can assume that instead of these things having no common zero, one of them actually has no zero on U. Because the set of common zeros is the section. So we can just take away the zero set to get the smaller open subset. We can assume one of the hi. So I will just assume, say, h0 has no common, has no zero. Because obviously, U is the union of the sets where one of them is non-zero. That's the statement. OK. So then we can put hi. And so for i equals 1n, we put hi equal to h large hi by h0, which is a regular function on U. Because it's a quotient of homogeneous polynomials of the same degree. Well, so in some sense, we just retrace the steps. So we therefore have that phi. In other words, phi is of the form 1h1 to hn. But that means if we take phi0 composed with phi, this is done by dividing everything by that. This is the map h1 to hn like this. And this is a morphism because it's given by polynomials. But as phi0 is an isomorphism, if something is a morphism, after I composed with an isomorphism, it was morphism to begin with because I can compose with U0. OK. I mean, to begin, in principle, we are still on U. So phi is equal to U is a morphism. And therefore, phi is a morphism. So OK. So this is some. So we have this concrete description, I mean, which is at least you can somehow write down coordinate-wise what morphisms are. It's not so super practical because you only have this open-cover substat, so you can only do it locally. I mean, but still, we have a concrete way of saying it. Just maybe a couple of examples. We want this one. So I will give a couple of examples where actually the morphism is given just by an n-tuple of polynomials. We don't need to define it locally, but there are also many where you need. So the first one is very simple. The projective transformations. Just if we have a matrix, a 0, 0 to a 0, n, a n, 0 to a n, n. So this is n times 1, n plus 1 times n plus 1 matrix with coefficients in k. Then this defines us a way, a morphism of pn to itself in the obvious way. So let a from pn to pn. I associate to a vector, say, p0 to pn. I associate this matrix applied to this vector. So I mean, strictly speaking, as a column vector like that. So I just maybe write it out like this. I just apply this matrix to the vector. And well, this would now set it on the basis of vectors. But here, I associate to the thing in projective space the corresponding vector in projective space. So this is called a projective transformation. Sometimes also projective change of coordinates. And we can see this is a morphism. Why is that? Well, we can write it down in coordinates. Because what do we see? In coordinates, it is just given by the linear polynomials. We have, for instance, the first component of the image is a 0, 0, b 0 plus a 0, 1, b 1, and so on. And so this just means that we just have a 0, 0, x 0 plus a 0, 1, x 1. That's just what it means. This is the map, which is the polynomial, which associates to b 0 to bn, a 0, 0, b 0, and so on. So this gives us the first line. And then it goes on just applying in the same way the coordinates. And this is a n 0 x 0 plus. So this thing is given by an n-tuple, an n plus 1 tuple of polynomials of degree 1. So it's a morphism. And in fact, it's an isomorphism. Because obviously, if we do it with the inverse matrix, we get the inverse map. So if we take the inverse of this morphism, is the morphism given, ah, yeah. You see, I forgot to say it. But this is not necessarily a morphism, because if we have, if this has a kernel, then this will map to 0. So all the coordinates would be 0 in the image and would not be well defined. So we have to have an invertible matrix. If the map is not invertible, this will not give us a well-defined morphism. But once it is invertible, it obviously gives us an isomorphism, because we can take the inverse matrix to define the inverse map. OK. So obviously, I mean, this is not so exciting, but it is one very often, these are somehow, if you want the most trivial morphisms which don't do anything at all, you just in some sense, really change only the coordinates on Pn. It's just if you put another system of coordinates on Pn, you would, that's more or less what it amounts to. But what does actually affect, but is actually not so easy to prove, is that all isomorphisms of Pn to itself are of this form. And at any rate, very often, you want to classify things only, I mean, to understand things only up to isomorphism, or up to isomorphism of the ambient projective space. So somehow, we often just ignore projective transformations. What? Yeah, I think it's not so easy to prove, yeah. I mean, the question is how we don't really have a reasonable criterion, how to show that something is not the morphism, OK? And so I don't think I know how, with the things that we have discussed so far, I don't know how to prove it. I mean, one needs slightly more advanced techniques, I think. I mean, I haven't tried very hard, but I think it is a bit more. So another thing that one often considers are projections. So this is, say, when x, Pn is a sub-variety, and W, Pn is another sub-variety, no, actually a projective linear, so projective subspace of Pn. So by this, I mean that you have some linear subspace of Kn plus 1, and you just look at the points in projective space which correspond to that, the image in projective space of some dimension. Then we can project from it. Do you want to do K? So we assume it does not intersect. W. Then we can project from W. So what does it? So in particular, if we have such a linear subspace, it has dimension K. So it has co-dimension N minus K. So it can be taken as a zero set of N minus K linear forms of N minus K polynomials of degree 1. It's just a linear subspace of dimension of co-dimension N minus K intersection of N minus K hyperplanes. So there exist N minus K linear forms. So homogenous polynomials of degree 1, H0, so Hn minus K plus minus 1, such that W is a zero set of that. And now the projection from W is the map given by these linear forms. So the projection is, let's say, PW from X to P N minus K minus 1, which is just given as H0, Hn minus K minus 1. So for every point P, we just associate this. And this claim is a morphism. So it's given by polynomials homogenous of degree 1. So that's fine. So they should have no common zero on X. But that's also fine because the common zeros of these is your W. So the H i have no common zeros because W intersected X is N. So this is a well-defined morphism. Well, in some sense, it's not completely well-defined. I call it, here, the projection from W. But the way I've defined it, it doesn't just depend on W. It also depends on the choice of these. So PW depends on H0, Hn minus K minus 1, and not just on W. But if, say, we have another set of such forms, L0 to L N minus K minus 1, then you can see these are just two bases of the same vector space, namely of the vector space of the corresponding dual vector space. And so there will be a matrix which transforms one basis into the other. And so this means then there exists a linear projective transformation A from Pn minus K minus 1 to Pn minus K minus 1 such that it makes one into the other. So such that, say, H0 to Hn minus K minus 1 is equal to A composed with L0 and N minus K minus 1. So we see that the projection from a linear subspace is well-defined up to a projective transformation in the target. So if we somehow view them as something where nothing really happens, it's just a choice of bases, then they are well-defined. And I mean the most useful case is the projection from a point. So if P, so in particular we can, so in particular if P is a point in Pn which does not lie in x, we have pi P from x to Pn minus 1. So for instance, I mean again well-defined up to projective transformation of N minus 1. For instance, if P is the point x0, so say which one did I want? So 0, 0, 1 in Pn, then we see that we can just take the coordinate functions to define it. So we have pi P. One way to write pi P would be x0. And maybe I should say some words about projection. You can always view it like that also in this case, but I will not detail that. That you have this, you know, have x in this projective space. You somehow take a projective space of a complementary projective space which doesn't intersect this one and you make some projection to it. So by, in a linear way, so let me, so we can look at it in this case where it's a bit simpler. So for instance, if we identify Pn minus 1 with a hyperplane, just a 0 set of xn in Pn in the obvious way. So a0 to an minus 1 is identified with respect to a0 to an minus 1, 0. Then we find that if I define Pp like this, pi P of a point Pq will be equal to the intersection point of the line through the points P and q with this hyperplane. So you can somehow, so this is what you usually call projection. You know, you have a, so that the picture is really like a projection. So you have here a given point P. Here you have a point q. And here somehow is this Pn minus 1, which is actually just a 0 set of xn. And what you do is you just take the line through P and q and it will intersect this thing in one point and that's the image point. OK. So that's just one example. Later, you know, anyway, so now for the moment, that is as much as I wanted to say about this. Now we want to look at something which is in some sense simpler, but then turns out to be in the end, not so much simpler, which is products of varieties. So it's just the obvious thing. If x and y are two varieties, you want to consider the product x times y and you want this to be a variety and whatever. And why would one want to study such a thing? So one thing that one wants to use it to study is again to study morphisms because you can describe a morphism by its graph, which is a subset of the product. And this is sometimes useful, a useful way to study a morphism. You can say things about morphisms sometimes nicer by talking about something about the graph or you find it useful to prove things that way. And we will actually introduce two important notions. Using these products, which are separateness and completeness, which you know that with our risk etopology, spaces are not housed off. Therefore, they are not really compact. I mean, they were all. But we want somehow some replacement for the house of property and for the compactness. And so what we think that we will call separateness will replace the house of property. It will somehow say that morphisms from varieties behave as if they actually were morphisms between house of spaces. There are continuous maps between house of spaces, even though they are not. And one has then completeness, which replaces compactness. And so again, we'll prove some theorem that morphisms between projective varieties behave as if they were house of compact spaces and continuous maps between them. Although, again, this is not true. This is anyway what we will use it for. And this will require to look at the products in a somewhat careful way. So now I will start with the easy case, which is the product of the fine varieties. Because it's easy to see what the product of the fine variety should be. Then next time we will deal with the product of projective varieties. And the problem is that if you take x in Pn, y in Pm, then there is no projective space in which the product lives in a natural way. So you have to somehow deal with that. So let's start with products of the fine varieties. And that's actually simple. And you even had an exercise where you have solved this problem for us in some way, at least in part. So this is easy. So if x in Am and y in Am are, say, closed sub varieties, well, then as a product, we take the product. And x times y is the product of the two, which is just An plus M. And we will see that this is, again, a closed sub variety. So let's do it right properly. So let x, An, y in Am be closed sub varieties. The product of x and y is just the product. I mean, it's a bit crazy. So the set of all PQ in An times Am, which is the same as An plus M, such that P is in x and Q is in y. So that's rather obvious. And now the point is that this is also a closed sub variety. First, we can easily see it's a closed subset. So if x is a zero set of some polynomials, f1 to fn, maybe I should say as usual. So I denote so that x0, x1 to xn be the coordinates on An. And the first factor, y1 to ym, the coordinates on the second factor, Am. And then we take x1 to xn and y1 to Am as the coordinates on product. And so if x is closed, it's given we can write equal to the zero set of, say, f1 to fk for some finite number, polynomials f1 to fk. For fi are polynomials in x1 to xn and y as a zero set of some other polynomials, g1 to gl in the other variables. And then x times y is just a common zero set of these in An plus M, so x times y. This is clear from the definition because what does this mean? So the fi only have the first coordinates. So if I take a point in An plus M as a pair of pq, where p is in An and q is in Am, then the first conditions say that the point p is in x. And the second condition says that the q point q is in y. And so this is the set of all points in An plus M. The one component lies in x, the other one in y. So we then can see, so this is a closed subset. It's a bit more difficult to see. We want to see also that the product of varieties is a variety, so irreducible. So that is a bit more tricky. And we have to take some effort. We will prove it slightly more general way so that we later can use it also for projective varieties. So we prove the following lemma, which is general statement about topological spaces. So let x and y be irreducible topological spaces. Now, we want to make one assumption about the topology being somehow just a little bit reasonable. So you see, it's not quite clear what kind of topology we'll have on the product. So I mean, here we will have the risky topology on the product. But now here I have a general statement. So assume we have a topology on the product is such that the following maps are continuous. So if I take, say, Jp from y to x times y, which sends any point q in y to the pair pq. So just you have the product x times y. And you just fix the point p here. And you just map every point in y to corresponding to 5 in that direction. And the other round also, so this is for all p. So it's continuous for all p in x. And also the other round, that iq from x to x times y, which sends p to pq, is also continuous for all q in y. So if I just embed this as the final direction, it should be continuous, which one would assume that most topologies one can imagine on the product will have this property. Then the product is irreducible. So if both factors are irreducible and we have this property that these embeddings are continuous, then the product will be irreducible. So let's see how one proves this. It's a bit trick. So we have an indirect proof. So we assume that it is reducible. So assume x times y is equal to S1 union S2, where the Si are closed. In other words, we then show that would then give that x is reducible. So then we have to see that, and I think we also assume that y is irreducible. And we will want to show that x is also reducible, not irreducible. So we define 1, 2, not so many, but take Ti to be, so intersection of all q in y, eq to the minus 1 of Si. So where eq was this map. So in other words, this is the set of all points p and x, such that if I take pq, this is an Si for all q in y. So if somehow this would be S1, then we see that T1 would be somehow the set of all things where all the fiber lies in it. So we know that the map Jp is continuous, now that we had assumed for all p. And y is irreducible, thus it follows that the image of Jp is also irreducible. So Jp of y, which is just p times y, is irreducible. I mean, with the topology that we have on x times y, because we know if you have the image on the continuous map of something irreducible, it's irreducible. So it follows that this thing must either be contained in S1 or in S2, because otherwise we could write it as a union of two closed subsets in the intersection. So it follows that p times y is contained in S1 or p times 1 is contained in S2. But what does it mean? It means for every point p, we have that p times y is either contained in S1. So I could also, to rewrite this, I can also it's a set of all p and x, particular distribution, such that p times y is contained in Si. This is equivalent. So this means, and this is true, for all p. For all p and x, it's either in S1 and S2. So it means that x is the union of T1 and T2. And on the other hand, I didn't say it, the Ti. So this, the IQ is continuous. Therefore, the inverse image of Si by IQ is closed. And Ti is an intersection of closed subset. So thus, it follows that Ti is closed. So it follows that x is a union of T1 and T2. And these are closed subsets. And obviously, neither T1 nor T2 is equal to x, because otherwise, the corresponding Si would be equal to x times y. Thus, it follows that x is reducible. And so we have proven our statement. So we can apply this to our, this general statement out, our concrete product of a fine, closed subset of fine space. So we assume still that, so this is the corollary. We assume that, again, x in a n, y in a m are closed sub-varieties, but actually would also work otherwise. Then it follows that x times y is also closed sub-variety. So we want to apply the lemma. And so x and y are reducible. So we have to check these two, these statements, that these obvious maps of inclusion of the factors into the product are continuous. But that's very easy. So if I take a point, so only to check, let's say, for q and y, we have that iq from x to x times y is continuous. Obviously, we also have to prove it for JP, but we can imagine that the proof will be the same. So OK, so let's do it. What is it? So I mean, note that. So we write q equal to, say, b1 to em, that's a point in y. Then iq is just a map, x1 to xn, b1 to em. It's just a map that's taking an n tuple of points in a n and sending it to this n plus m tuple, where these are just fixed numbers. And this certainly is continuous because, coordinate wise, it's given by regular functions. These are constant, they are certainly regular, and these are just polynomials, they are also regular. So it's a morphism. It's a morphism, and therefore it's continuous. So it's very simple. So this proves, and then we apply the lemma. I mean, obviously the same you would have to do, as I said, for JP, but that is the same statement. I mean, the same proof. So finally, I wanted to show you the universal property of the product, the proposition, universal property. So it's some kind of property about maps to the product which defines, somehow determines the product up to isomorphism. And it's somehow, very often, when one uses the product for something, one will actually only use this universal property. So let's see what it is. The first statement is the projections, p1, see. So we assume again, just to know that x is in an, and y in am are closed sub varieties. And we have again x1 to xn, the coordinates on an, and y1 to ym, the coordinates on am. So p1, which I can just write x1 to xn from x times y to x, isomorphism, and the same for the second projection. So it's just, and as you can see, this is obvious because I've written it down. I mean, I have given you the coordinates of the map. They are obviously polynomials. And so this is a morphism, so this is OK. And the second statement is the actual universal property. So let's add any variety. So the morphisms from z to x times y are precisely the following, the maps, which I could write fg z to x times y, which sends any point p and z to f of p g of p for f from z to x and g from z to y morphisms. So I'm saying that I have told you in some way what all the morphisms to the product are. They are given to give a morphism to the product of x and y is the same as giving a product a morphism to x and a morphism to y. And the morphisms to the product are just given by using the first morphism as a map in the first direction, the other morphism that's mapped at the other direction, OK? So I mean one can also, yeah, you can also say it differently if you want. You can say a map, so equivalently, p from z to x times y is a morphism. So this is even only if p1 composed with c and p2 composed with c are morphisms. It's the same statement, because p1 composed with c is f and p2 composed with c is g, OK? So let's prove it. So one was obvious, now two. It's also not very difficult. So in some sense. So if h from z to x times y is a morphism, so maybe then if I come and we know that p1 and p2 are morphisms, then it follows that the composition with p1 and 2 are morphisms, why do I call it h? Then we have that f, which I defined to be p1 composed with p and g, which I defined p2 composed with p are morphisms, as compositions of morphisms. And we see that p is equal to fg, because it's just a map, which sends the point to the fact that if I compose it with p1, I get f, and with p2, I get p. Just says that the map is like this, OK? So conversely, assume I have f and g morphisms. Then recall that x and y are a fine varieties, actually sub varieties of a fine space. So such morphism will be given by tuple of regular functions. So then there exists, say, f1 to fn in whole z of z. So regular function is on the whole of z. And g1 to gm in oz of z such that f is just f1 to fn, and g is just g1 to gm. Well, then it's not very difficult to see how we can write our f, g in terms of regular functions. So this is now given by an n plus m tuple of regular functions. Therefore, it's a morphism. OK, so that was maybe all I wanted to say for today. So next, so this, as you see in the case of the fine varieties, the product is a rather straightforward thing. You just take the product, and the office things work. Now we will see that for sub varieties of projective spaces, it's a bit more complicated. Because, as I said in the beginning, that if you have, say, x in pn and y in pm, you have to wonder in what projective space the product might be. And there's some trick, which is maybe a little bit of a cheat. Namely, you have an embedding. So a map from pn times pm into some much bigger projective space. And if you have x in pn and y in pm, then you define as the product the image of x times y in this bigger projective space. And then we just call that the product. And we will then see that we also have some universe property and that one can do some things with it. Anyway, we see this next time. Thank you.