Categories, Sites et Champs - lecture 2 of 3

0:00 Thank you very much for being here tonight, and I would like to note that, as I told you, F-1 is not always exact. Perhaps some of you are wondering why it is so exact, and there is no reason to be exact. This remark is addressed to people who know vessels in topological spaces, because in topological spaces, S-1 is always exact, as you might have guessed, except on a number, S-1 is exact. In the case where S-1Y is in the case of a topological space, then it is exact. Otherwise, you have to do a certain hypothesis that is not really in the details. So, we will now look at a particular case where things happen. We will look at the particular case of an object U of Ci. So you will remember that I have after Ju the morphing of Ci, Cu is Jut with a notation and has an object, an object I put in U, I associate U, E, sorry, I associate E. There is another part of the site, but we will see later, which is more particular, which I call I-U, which goes from U to I, I-U, and therefore from C-U to C-U, and to V, I associate the fibrous product, or cross-V, as you can see, in the case of topological spaces, the intersection. But that's less interesting, because I made the hypothesis that there are products.

2:30 I mean, I did the beginning, but a little later we will see the sites without these metaphors. I'm listening to this morpheus, I.U. or the Seraphim. So for those who have learned the crystals via the topological spaces, in classical books, they all know I.U. They have never seen J.U., but nevertheless it is J.U. that is an interesting morpheus. So from now on we are not listening to it for a long time. So, if we apply the general theory that F is equal to minus 1 to the GU, we will see that GU minus 1, GU is equal to, that becomes definite. So, the question is, is GU minus 1 exact in this case? So, first of all, you would have to calculate what I did. The GU of Agnes, in the pre-messo, before you had GU. We are going to do the Jules d'Ague. We are going to do the Jules d'Ague. We are going to do the Jules d'Ague. We are going to do the Jules d'Ague. We are going to do the Jules d'Ague. We are going to do the Jules d'Ague. We are going to do the Jules d'Ague.

5:00 We are going to do the Jules d'Ague. I have to say how much it's worth. First, the definition, the inductive limit. The curve, which is seen in Y, which is the nationality of W in U. It's a bit complicated because an open cell is an arrow W in U. An object of the cell, which is called an open cell. So, if I apply my formula, I find this here, and this belongs to CQ, the inductive derivative of GW in U. G is defined on U, so it is defined by the expression of W in U. Well, if you get lost, it's not the same thing. So, if we look at what that means, what does that mean? What is U? It's W in U. It's W. So that's the inductive limit for the arrows that go from W to W, for the arrows that go from U to G, from W to U. And that is a final object. No, that's not our final B, sorry, it's a bit complicated, that's why the demonstration is completely complete. Family is not an object, there is a co-final part, which is the demotivation of V in U. If I take W equals V, it's co-final in this category. So I say that, but I don't demonstrate it, man in U is co-final. The category is here. It's a bit complicated. Since it's simple, it means that there is an inductive limit according to this category.

7:30 But this category is discrete. So the inductive limit according to the discrete category is a co-producer. Don't worry if you don't understand this task. It's a bit difficult. Thank you for your attention. In the notes, I say that the answer J-U-1 is Hegelian. But it's wrong in the notes because I don't have another hypothesis. I'm not giving the best hypothesis, but it's quite simple. I assume that the equation A is additive. I didn't say Abelian. In this case, I say that J-U-1 is Hegelian. First, I said that J-U-Z is Hegelian. So why is J-U-Z Hegelian? I repeat the formula J-U lag calculated on V is a co-produced function for S in 1, so it is a co-produced function.

10:00 It's an exact counter. So why? So you know the hypothesis I made tonight, if you don't know it, but I assumed that the inductive limits of pi 30 were exact. But I didn't say that the co-products are exact. The inductive limits of pi 30 are not the co-products at all. The co-products are not pi 30. But a co-product, it's not very difficult. Let's say that OTI is a big one, that is to say that OTI is extreme. A co-product is also an adductive limit according to the finite parts of the equation. So it's a small negative. If you take a rare category, suppose that the finite co-products are exact and the finite adductive limits are exact. So, before this lecture, why does it end here, the first one backwards? Because this is the mistake in the polycopy. In the polycopy, I said that the inductive limits of France are exact, so the co-products are exact, and it is necessary that the co-products end up being exact. There are two types of products, the finished product and the co-produced product, and the finished product and the co-produced product are the same.

12:30 But if you assume that chemistry is additive, the finished product and the finished co-product are the same thing. Since the co-produced product is the exact one on the left and the co-produced product is the exact one on the right and the co-produced product is the exact one on the left, the liquid category and the finished product are the same. So the finished co-product is the exact liquid category. So, if we assume that A is the liquid, the final cost-products are exact. On the other hand, I made a hypothesis a long time ago, it's written in the notes, that the inductive limits are exact. So, these two things are correct. So, why do we know that any cost-products are exact? Well, because any cost-product, as you can see, is the same thing as the inductive limit. I'll do it a little faster. The co-products for J in J are the components of the other three terms. And the family of the finite parts of I is different. It's an inductive I. When the ordered set J in I with J is not an ordered set, it's a different category. So you see, there are a lot of things. That's it. So, that's it. That's it. And that's it in logical spaces. So, we take the germs, we take the germs, and we follow them from word to word, it's important. So, I have all this, and I simply remember one thing. So, I demonstrated the following letter, which comes from another chapter, if you want. It's that if A is additive, then the counter J, U, H is exact.

15:00 So now, I'll say the proposition, if you want. Suppose that A is additive. The answer to this question is exactly the same. You see, from time to time, unfortunately, we need to know what we are talking about. It rarely happens, but from time to time. For example, in Geo-Visage, we needed to go back to the source and say what it was. You can't always use theories to come back to the base formula from time to time. So, now that we've got the answer, what is Ju-1? First of all, I want to show that it is exact. But J-1, J-3 are quite far away, so we already know that it is exactly to the right. It is exactly to the right, like any other theorem, J-1. So why is it exactly to the left? So what is J-1? How is it calculated? J-1 is associated with J-1. Before you take a zygote, you first have to have a vessel. I look at it as a pre-vessel, then I apply a zygote, and then I take a vessel associated with it. So this is exactly to the left, as we have already seen. This we just saw is exactly. And this is exactly. So it's exactly to the left. So we just have to say that minus one is exactly.

17:30 So now there is another interesting phenomenon. So now I look at the general case of x in y, I look at the particular case of x in y. So already, I did not say that ju-1 is always exact, the hypothesis has to be true. And there is another interesting phenomenon, proposition, is that this time the function ju-zac-zac is interesting, while before it was false. What are they? Tf-1, Tf-dac, proposed with... or associated. So we can ask ourselves, why don't we do the same thing with F-dac-dac? Indeed, by the way, I've never studied it, but... This one has an interest because it's at least equal to the Tf-1. If I do nothing to prevent it, it's going to be F-dac-dac, and then the component... Well, to say it like that would mean... Nothing prevents me from doing what you see here, which is composed with S, Dac-Dac. And that's a right-handed add-on, that's a left-handed add-on, so it's composed, it's a right-handed add-on. And I imagine, I don't know if it's the right thing to do, but I imagine that what we could do, we could look at this counter. I don't know what the name is, but apparently it shows that it's not interesting. Maybe it's a joke. In general, it does not seem to be used. However, in this particular case, it is interesting for the following reason. J.U. Daktang, who is a professor of G, sends the precepts in the precepts. But the proposition is that it sends the beams into the beams.

20:00 He remains adjoined. J.U. Étoile too. He sends the beams into the beams without changing anything. So, J.U. Étoile, J.U. Dagdag, who is a father of adjoining professors, is a professor of beams in beams. So, I'm reflecting. I'm going to show you later, but I'm going to write the theorem for you. So, we have the pixels. Earlier, we saw Ju-1. But we also have Ju-1. You can see it here. And we have the pairs of functions. Ju-1, Ju-1, Ju-1. So, demonstration. The only truth is that J.U. Zagdag, who, a priori, sends the pre-vessels over U into the pre-vessels over X, sends the pre-vessels into the pre-vessels. So here, you have to remember, to go backwards, you have to go backwards.

22:30 So how does J.U. Zagdag look? So, you take g, a very small number, over a number of a, you take v, which is in ti, and you can say, look, I had a wave of g, which is in v, and I apply the formula, sorry, of the equations, and the projection limit for u, u, t, you can write it down here, you can write it down here, Since the sum of W and U is sent in P, it belongs to C, U, V of G sent in U. I have simply applied the definition formula of the reciprocal image of the equation. This simplifies it. What does it mean? It's the projection limit for W sent in U and U sent in V. There is a projection limit on your category that has the same final object, which is the value on the final object.

25:00 We have the formula. So j, u, z, g, calculated in g, u, 3. When I say u, 3, b, this is what they look like, the hybrid, they look like u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g, u, 3, g For example, if we take a curve of V, we get S, a curve of V. Well, I say that the number J, V, J, U, sorry, Z, Z of G multiplied by S, well, that's worth G of U, 3, S. And if G is a vessel, well, that's a curve of U, U, 3, U, S, So, what I want to say is that all of this is equal to g of v, and that is to say g of u times v of 3 times u times v of u.

27:30 That's the question. But the answer to the question is yes, because u times 3 times s times v of u is equal to v of v of u. Well, here we used the hypothesis that there were hybrids, but it's true even if there are no hybrids. As it will be done later, we will do things differently, more intuitively. Now I'm going to come back. Are you convinced? Just in case, you wrote on the left, in the top. The history of the universe after H-U is based on C-U, and you said it was a peso on X. No, no, no. G-dag-dag is a peso on X. G-G on U. Do you have the numbers? Yes, I do. It's just that... It's just that I don't remember the name of the question. Because what you saw at the beginning was certainly the objects. U corresponds to the variety CXU. The objects and the arrow from B to V. But then it's on the other side that we have an arrow from B to V. You said it's G of the fibrous product. No, I said J, U, y, y, G. It must be a peso. It must be a peso over X. Yes, well, in fact, I would have said it's more of a peso over U.

30:00 Well, a big piece of clothing. Wait, wait, wait. You were right. I said the wrong things. Thank you for your attention. Thank you for your attention and see you in the next lecture. If we try to convert this morphism into a sphere category, we want it to be a localization. If we want to convert a sphere category into a sphere category, we want it to be a sphere category. So we would have to invert all the arrows and that would be the third adjoint, the component-connected function. So we don't have to define it like that, we just have to define it as a domain function. The J-1, we want it to be that, and so the G-1 is the thing that we want. No, but you want other notation, that's what you want. No, but it's not the notation, it's just that there, suddenly, I mean, in fact, it works, well, the three vectors are the same definition, it's just that the nautism in these crystal categories does not go in the same direction as the law, and that's true. Yeah, it was written there, J-U, I-V-I, J-U. No, of course not. Even compared to what we defined visually as the morphism of localization, your morphism is going to be good. Well, it's not there. In Hungary, they work. They are defined in some way. What you are trying to tell me is that there are no quotations that...

32:30 Well, I... There are no quotations. Everyone has their own quotations. There are no quotations that are made by everyone. For me, when you have an amortization of y, well, f of y, at the bottom, it sends the pixels over x to the pixels over y. Yes, that's good. So once we have that, once we have that. Just to be clear, I told him that if A is not additive, this thing is not necessarily exact. Okay, yes, so it's just for that, actually, because, well, I had looked, like morphism in the categories of Faisal, I had looked at the one where the inverse image is exact, because, well, that's it, so it's good, actually, it's just that we don't have the same morphism. In any case, this one is not always exact, and Zagzag, we will see, it is not always exact, it is not exactly the same, it is slightly different from Zagzag. So, well, I take it back. So we were there. So, that's Zagzag's formula. So I say that it commutes to couplers. Why? Because if I take a curve of V, then U3 is a curve of M, and if I take S, a curve of V, then U3S0U is a curve of U3B0U. And so, it proves that J, U, M, Y, G... So, J-U-dac-dac-g-e-b. So, J-U-dac-dac is an object. Finally, we see objects in objects. So, now, to make things a little easier for you, to make the object simpler, in any case, for those who know objects in topological spaces, we will compare with the other function I-U. So, I-U is the type of U in X, I remind you, so here is X in T-U and here is U-U-V.

35:00 So, I say, I already said, I don't want to repeat myself, I say that, so, proposition, I say that J-U is the same thing as I-U-Z and that J-U-Zag is the same thing as U-U-Z-O-A. It may seem a bit weird to have to explain it like that, but the three J-U-Z-O-A meters exist even without the hypothesis that there are polyfibers. So, the most simple example is J-U-Z-A-G-G-B. G, V3, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, U, The first formula is the formula of the right and the second formula is the formula of the left. Because I.U. Ettoile, when we work in prefectures,

37:30 we work in prefectures, and we are going to demonstrate that in prefectures, I.U. Ettoile, he admitted an adjunct to the left and I.U. Dag. And J.U. Dag Dag, he admitted an adjunct to the left and J.U. Ettoile. On the contrary, the prefectures are the same. And since Iu star sends me So, often, yes, we add a notation. We write I U exclamation equals J U minus 1. We have I U exclamation, I U minus 1. I U minus 1 is J U minus 1, J U equals 1. That's the same thing.

40:00 So you have u star equals i u star equals u u star equals u u star equals u u star and then u minus 1 equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u u star equals u So, it goes from C to U, then from C to X, and it's a counter that we call, in the Abélian case, or in the case of topological spaces, predominantly by zero. This is what we can demonstrate, but I'm not going to do it. Take a case where X is a topological space, and then take A equals 2K. You can find 0 if it's not in U, and Fi if it's not in U. So it doesn't change. It's a piece that doesn't change G on the green, but it's presented by the green. Well, I'm not going to go too much into detail. So we have a few new factors that are very useful, but... If you have F, it's a piece on the right now.

42:30 So, we have a term that consists of couplings by u. So, SU is, I don't know what you prefer in mathematics, I don't know if you prefer i or j. So, that means you have your threshold on x. You restrict it to u and you extend it by 0 in addition to u. And then you have other terms that you have, well, at least in fact. U-S, J, U, tag, tag. These are the supporting sections in the theory of equations. You can see that the factors cut by U and gamma U. This is what I wanted to show you. This is the theory of equations that separates you from the theory of equations. You are used to doing interesting things, right? So I don't consider that... Well, if we do the theory of equations, all this is very important because these are the formulas we use all the time. But the goal is not quite the same, so it is a twistor theory. There is one more word to understand all this. In the case of topological spaces, there is a special phenomenon that we usually don't see. It is called U-X-U. That is, I-U, that is, J-U.

45:00 So, in the case of topology, U-3-I-V is U inter V. So if I go to the composition of these two factors, what does that mean? That means that I go from Cu to Ci by JuC and then from Cu to JuC. How would I go? I go in U, then I go in JuC, then I go in JuC. And then, U.S.S.I., product hybrid, V, U over X, V cross U, over U. That's the general thing. And if I'm on topological spaces, that means I have an open V in U, I also have the open V considered as open X, and then I take its intersection with U, which is itself. So in the topological case, this thing is identity. In the topological case, it's not true otherwise. I'm not saying it's the only case where it's true. It's not always true. So, when I say state, for example, it's not true. So, the component is identity. So, it proves that ju and 3 are composed with iu and 3. For example, it's identity. But that's iu-1. For example, iu-1 is composed with iu and 3. It's identity. So, we deduce it. EIU star, or if you prefer JU-1, is totally true. So when you have joint factors such as the component and the identity, one of the two, I don't know which one, is totally true.

47:30 So in the topological case, JU-1... Well, I'm not sure it's JU-1, it's not my fault. So let's stay with the star. The problem with identity is that iu-1 is composed with iu-1, which is identity, and so it's iu-1 composed with iu-excavation equals identity. So I find that iu-3 and iu-excavation are completely the same, and it's in the 4th plus. Well, there are other things that are in the books that I don't know about. I don't know if we can start right away with the chatty level. Yes, I'm still going. It's a new paragraph. Well, so we'll forget all that. However, now I'm going to talk about things that are more related to the spirit of the course, on the fields. It's the locally constant beams and then the recalibration of the beams. So a new paragraph is the socially constant beams. So, we are always on a cycle. X is a cycle. You can think of it as x or 2. So, first of all, you know what it is if you have an object M in A,

50:00 you know what the object Ni is. It is the associated object with M. So this is the object Ni. So what is So we say that f is a constant beam if f is an isomorph to an emitter. I didn't say equal, it's not the same. So I'll give you an example of a constant beam. You can take the equation here. I'm going to take a beam f, which is a subbeam of an infinite function. So we continue. Let's continue with the real. I'll take the next beam, which is the constant times the exponential of a coordinate, t. The exponential is the beam of the functions which are locally the multiples of the exponential function. You see, I have a constant beam on the rift, and I have the continuous function on the rift, the complete value, and then I have the beam of the beam, which is there, to a beam, I don't know what it's called, a constant beam on the rift, what's it called? S? S? S is not very... no. What did I forget? No, not a beam, a section.

52:30 So, to the function a of t, which is the function of a constant, I associate the function a of t with the whole t. You can see that phi is a monomorphism of vessels, and that the constant vessel is mixed and isomorphic to its image by phi as subvessel of C0. As we said, the vessel that is here is obviously isomorphic. So this is an example of a constant beam, but which is not the beam of locally constant functions. An isomorph beam is a quantum quantum beam. So you see, I gave you as an example of a constant beam, the beam of locally constant functions, but there are isomorph beams as well. These are not locally constant function spheres. For example, the spheres that are also the multiples of the exponential function. We can take another function, but it's not the same because it's never the same. Well, so we have the definition of a constant sphere. So, another definition, we say that the S is locally constant, The ratio of f to d is constant. It's a constant ratio. So a ratio which is constant is a ratio which is constant.

55:00 One more minute. So an example. This is the example I gave yesterday. Consider the equation of x in y, where x is c to the zero and y is also c to the zero. The equation is z. You have seen that this beam was not a constant beam over x, it was locally, in the same way, y squared, but not exactly, you remember? It's linear. So it was a locally constant beam, two times two, which was not constant. So it's quite subtle. We can stop for a minute, and then we can continue. Before, I would like to write about the progression of constant functions. Also, you have the functions of localization, constant, this is in the form of I, and then state, or localization and constant. So, I'll give you another example, a level of S3, the question of localization and constant. This is an example of a level of S3, but we can also say that it is an example of a state perverse.

57:30 So, the following example, x is equal to c, and then you see the partial difference between x, z, z to the z, minus alpha, which is alpha, which is not equal to z, so it's a bit complicated. So you have to look at B as an amortization of T0, and then we're going to apply alpha to the nucleus. So, first of all, we're going to put ourselves outside of zero. First, we're going to put ourselves outside of zero. So, outside of zero. So, if I put myself outside of zero, I say that... First, I'm going to put myself... I'm going to use this differential equation. First, on an open disc, outside of zero. So, I said that when a disc is open outside of zero... So, I'm going to put myself on the disc... I have the operator D on L, and I say that these morphings of these two are parallel to each other by a vertical morphing. So a morphing of these two, it will be a multiplication. So what is the function? I have A of L, a primitive of the function alpha over L. So this primitive does not exist on C-0, as you know, but it exists on D. So here I multiply by exponential of z, and here I multiply by 1 over z, exponential of z.

1:00:00 You can have fun because I have done it, but I don't have the pressure to calculate it, but I have done it. You can see that it's the same as this one. That is to say, you have a function, you start with the z, the z is there, you apply p. And if you use it with an excellent or negative answer, it's the same thing as if you start by applying an excellent or negative answer, which is a derivative. So it's the same thing. And of course, the most applicable are isomorphisms. If you use it with an excellent or negative answer, it's the opposite. If you use it with an excellent or negative answer, it's also the opposite. So it's isomorphisms. So it shows us that if we take the nucleus of T on DL, we know it, it's a matrix of time, the nucleus is isomorphic, but be careful because that's on D, not on any region. So T on DL, as in the case of the vessel, is an isomorphic, so it proves that this is an isomorphic. So we have a complex vessel of vessels that are isomorphic on D. So it proves... It proves that K-alpha, K-alpha is a vector that defines everything. It is a vector on X, but equally, equally, it is a vector equally constant. Now, globally, what do we have to do?

1:02:30 So here, Y is equal to C minus the origin. So if there is Y-alpha, if we hear the words C-alpha, C minus the origin, it means that Z is on Z. All of these terms are equal to zero, because F will be a multiple of F with no alpha, i.e. F with no alpha with two L, and so on. So, this formula is reduced to zero. This shows the course of the difference between the two. And, similarly to Céline, but not globally. So, we think about a Conex universe, which is very specific. So, here's another example, which is basically... If you take alpha equal to 1.5, it's basically the same example as yesterday. So, here's another example, because it's always so consistent. So, you see that... So, we could do later, but it's always so consistent. We'll come up with a method. I explained it to you earlier. But you see, it's very easy to be at the opposite, at the opposite side of things, and so on. There is a theoretical theory that says there is a category equivalence between the 40% of the pieces in the list and the objects. But you can't do anything about it. There is another theoretical theory that says there is a category equivalence between the 40% and the objects. There is a difference between the local and non-local questions of the terms of the lecture and the presentation of the questions.

1:05:00 We don't have a look, but in fact, we can see the results. So, before going further in this direction, I would like to make a parenthesis that interests people who have studied mathematics. This is why we call it the mode of K-I, so it is the sphere of K-module. But very often, we need a sphere with a moving halo. It is itself a sphere. So what is an algebraic K-sphere? What is an algebraic K-sphere? The halo sphere is not the same. In the case of Algebras, we have a K-algebra, we call it nano, which is generally a K-algebra. So a K-algebra beam, as you can imagine, is a beam on a x-axis. That is to say that for every u, R of u is a K-algebra, and then the operations of restriction are commuted by multiplication. There are also addictions and duplications, so we will ask you if there are addictions and duplications in local morphisms. We will not ask you if it is a K-algebrite system. An example of a K-algebrite system is the K-algebrite system. Another example, if we want a real or complete variety, it can be Sainte-Louis or Montserrat-le-Mort. And there are other examples, which are a complete analytical variety.

1:07:30 There is the name of the differential apparatus to the north, DI. It is an important example, the Z, which is not commutative. I'm going to take it for granted, but that's to say that it is not asked for R because it is commutative. If you want to learn more about this topic, you can visit our website at www.scienna.com or at www.scienna.com. Now I have defined the container where I discussed what is a morphism. So pay attention, it's just a little process. If you have M and M of 2R, a morphism of P of S of N, that's the data for all U's, of P of U, of M of U,

1:10:00 which fit the diagrams that we have here, then it's a morphism of P. So it's a morphism of P. So, whatever is U, R of U, it is R. So, what is annoying is that we have already defined the category of modes of R. So, if the mode of R is an algebra, it goes into the mode of Kx. But, obviously, this function is not so ideal. It is the same thing as the mode of R. On an anode, the A modules are sent into the M modules. Now, they are identical, but not exactly identical. They are not the same. The equation is there, it is injective. You have two R modules. You have Ohm over R, 2M divided by N. It is sent into Ohm over Cali, 2M divided by N. This line is injective, but not subjective. This counter is faithful, but not... So, what I mean by that is that everything I said about beams does not imply anything on the category of modules R. I think that if the theory works the same, the theory of R modules, of beams of R modules, is not a particular case of the theory of beams. So, maybe we are right, we would have a more general theory of modules. That's why there is an exchange between the stacks. It allows us to understand what we are talking about. So, I'm not going to... it's just an example, I don't want to go too far. So, let's see what these examples are. So, a definition.

1:12:30 What is a titanium? Well, it's the result of a conflict between an anode and an atom. At the moment, it's not very competitive. We don't have the number of titanium, the number is not competitive. There are many examples for those who do not know algebra, but there are plenty of them, so I don't know how to explain them. For people who do not know the difference between squares, there is a variation of the difference between squares, but the function is the same. So as an example, for example, the points of the B, the points of the difference between squares, is a menu. This is an example of the topology, you can see it here. But sometimes, you have to be careful, the Overarm complex, an experiment by the way, let's say that you have a algebraic system in the right half of the anemones, there, it is clearly found, it can pass next to the modules because there is the Overarm complex, but without introducing the modules. This is not a complex, it is a complex of C modules, it is not a complex of OX modules, because the morphemes are not OX linear. This is the complex of Rarm and not the complex of Rombaudi. So, let's stay in parallel.

1:15:00 So, earlier, we saw that the texts... We are going to refer to another eucalyptus. To go back, we saw that it was a locally constant text. And we are going to refer to the locally constant text of Rombaudi. Locally constant text. Locally constant text of Rombaudi. So, that means that it is locally, that is, locally... and, of course, a person, LX or M. So, what's interesting, let's say, we're going to take, maybe, let's say, a new definition. Let's suppose that the heart is a body and, through a system of cards, it's the vessels, which are locally constant, that bind. That means that L is locally isolated from KX. So, we can do the same thing with OX. So, a vessel, now it's X, OX, A space M is locally free. Locally, M is an isomorphism that is defined only locally. In other words, there is a return to M. Whatever D is, there is an isomorphism, M is an isomorphism that is defined only locally.

1:17:30 The definition of an isomorphism can be found everywhere, and if it is a local isomorphism, then it is a global isomorphism. If you have a local isomorphism, then it is a global isomorphism. But here, it is not a global isomorphism. So, a thing like this is also called a vector fibrin. In the case of the rich, it is called a vector fibrin. So, just now I gave you an example of a local system of rank 1, which is not very common, so here is an example that is a little older, but if you want to find a talent among us, this is what I was saying, it is totally unrecognizable. So, for example, take the P1 and take Mr. Tours, and you will have 20 points of the fundamental element. So, if you take a map, the map, which is an open menu of complementary points, and so if I have a point C, you can say a coordinate, Z is the coordinate of the map, and you have a denominator, so it's going to be Ux, which is going to be U, and similarly, it's going to be Ux, which is going to be U, by S, 9, S, and Z. So, what is the shape of the circle? In the terms of C, it's the shape of S.

1:20:00 Thank you for watching this video. And so on and so forth. And so on and so forth. And so on and so forth. And so on and so forth. And so on and so forth. And so on and so forth. And so on and so forth. And so on and so forth. And so on and so forth. And so on and so forth. And so on and so forth. And so on and so forth. And so on and so forth. And so, it's an example of a right fibrous which is not trivial, that is to say that the vector is not... It's very analogous, you see, so there is a parallel line. So, we're going to move on to the superficials, and we're going to talk about the reclamation of the pieces, we're going to say that the two sections, the piece of the constant diameter and the space-time measure, so in fact we're going to do stacks, and we're going to say, it's clear that the data, as I don't know if it's too much, it's completely...

1:22:30 So, we're going to give you a way of doing this, especially in your field, or in the field of liberal thermodynamics, and we're going to give you a theoretical idea of how to do this. So, a field has the property that if you have locally defined sections, or if you have two-to-two intersections, well, the sections are globally defined. So, now we're going to do the same thing in a rain field. So, we're really... it's really a field. Is there a recurrence of X and Y? And above all... So I changed the notation, in fact, on the questions, surely, of this class. It's not too many questions. It's more simple. Is I'm going to write about a family operating in a classroom? Most people are going to do the difference. So the family is indicated by the letter Y and then its image. So you have a line of Y in the object. So, if this application is injective, it must be identified as I or U, I call it I, he calls it I, he calls it S, I call it I, he calls it U. So I can identify I and S if the application is injective. If it is not injective, the two notions are a little different. In general, the coordinates are the same.

1:25:00 I prefer to consider that this application is injective, because it is indexed by Hilbert space. But rather than indexing the objects of S as they are, for purely notationary reasons, I prefer to inject them by me. In particular, Ui times Uj times Ui is a notation of our Uij. And the same goes for O. The application of the objective is marked in the same way as the application of the objective, but it's not the same thing, it doesn't work the same way. So, when I did SOS, we come back to the general discussion, the two theories work in the same way. This is an interesting thing, maybe I forgot to say something, something like that, what does it mean, what does it mean, it's not a notation, it's the same thing, it's a reflection of the existence of Hilbert and Witten, I don't know if I should have said it or not, I hope it's clear to you, so I'm going to look at Tessou's family, on Hilbert and Witten, and then... On UiJ, there is an isomorphism, Keta Ji, Fi is an isomorphism to Fj. And then it's Etat J.

1:27:30 It's not equal. On Keta Ji, Fi is an isomorphism between Fi and UiJ, and J is an UiJ. It makes no sense to say that there is equality. On the other hand, if I take an intersection of 3, 8, k, zeta, if I make an isomorphism of zeta, k, z, which I compose with zeta, y, y, I would have an equality with zeta, k, y. It's not an isomorphism because you don't have time. Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, Lx is equal to Lx, So there is still the question of the third interception, composed with fj, ui, ui, k, then in sk, ui, ui, k, there I have theta, there I have theta, k, j, and we see that in form of si, zeta, ui, ui, k, sk, zeta, ui, ui, k, and there I have theta, ui, ui, k. So the idea is that in the symbol,

1:30:00 All of these are related to the equation of the equation of the equation of the equation of the equation of the equation of the equation So you can see that if we start with a TSO-S, and when it comes to natural isomorphisms, these isomorphisms deserve this condition, which we call the impossibility condition. So the project is the reciprocal. Now we're not going to start with a TSO-S, we're going to start with a TSO-SI family, and we'll come back to it. I'm going to read a few of them.