Lawvere, axiomatic cohesion

0:00 There is a differential variety, with these properties, which are united by a metric tensor. There is a relational approach, which is difficult to conceive, but which has already reached the head, which is to say, what we call the metric, that is, the metric tensor. You take B, I take C. I'm not going to give you theoretical names. I'm just going to look at what this sequence of things is for certain values of B. So I start with B equals zero, B equals the category. So why do we keep them together? So wait a minute, because I've made some terrible notation mixtures. I'm going to explain what's going on here first, and then I'll make the connection with this B equals zero. I'm going to examine this first, otherwise we'll see that here... You have the category of the sets and the category of a single object. The category is reduced to an object and only the identity on this object. So you have a terminal function here that I note like this. This function is of course a natural one. A natural one on the right and a natural one on the left. We note 1 and 0 because the category 1 is in fact the set that we are going to note like this. The element of 1 is going to be the star. So the star, I send it on the set to an element, by 1, and by 0 I send the star to the empty set, and these are well joints, we will check for people who do not know what I mean by joints, these are well joints marked on the left because that means that if I take a set E here, if I go down, well it comes on star, now the adjunction that is here means that the star goes back to the empty set.

2:30 The fact of the junction that we want to show is to say that there are as many mortals from here to there as there are mortals from here to there. Well, there, in fact, there is only one, and then there too there is only one. So there is the junction, the property of the junction, it has a good place here, which sends E to the star and which sends the star to V. That's the junction on this side. And then the junction on the other side, well, this time it's again with two stars and two angles like that. This time you send one of that to one. One is the set of one element. As a consequence, a star-to-star morphism is as much as a E-to-A morphism. So, what we are simply saying is that in the category of the whole, the empty object is the initial and the object is the terminal. Which means that we have an adjunction like this. And the center of the beam is not the same, it is not the same? Which beam? There. The one that is empty. You have to say it like that, yes. It's an identity then. The fact that the morphisms from A to A are the same as the morphisms from A to A. But there is not a morphism from E to B. No, but there is on the other side. That's why here the right joint and the left joint are the same. The right joint is 1 and the left joint is 0. Let's take an example, because then we'll see what the probability is for the same thing. Here you can see that there are arrows in different directions to say which way to go, right or left, that's what it is in the T.U. Well, I'm trying to get the same drawing as here. This drawing here is a piece of that, if you want. Yeah, I see. There's one that goes up, you see. No, no, it's the other one. The other one is the same.

5:00 The one here is the Y. It's the Y. So, the general situation, you can see what's going on, but there are even fewer after, which are not very interesting, and it would be interesting if it was more complicated. I'm going to give them to you right away in the case of any b, or b equal to z, sorry, when b is equal to z, what happens? I'm forced to ... Have you finished looking there? Subtitles by the Amara.org community So, I give you the sequence, we will call it the sequence of Robert Wood, the sequence that interests us, you leave, so like here, you leave, instead of putting 1 here, you put 7 here, so you will have 7, and here you will have the genesis of 7, which therefore goes from 7 to the power of 7x. In other words, these are the countervariant factors on 7, the factors of 7 in 7 but which are countervariant. You have this factor there. Then you have here, so I entered the notation earlier, you have here the power of 11. So that, indeed, the genesis of 7, I entered the notation of 1 here. So here, the 1 here, I can take the power of that, I can take 11, I'm going to go around it to make it visible, and it makes me a factor from A to A.

7:30 So, after that, 1 is the denominator that I have defined here. 7 is the same thing as 7 to the power of 1. Ah, I put 1, it's 7. Yes, in my case, I wrote 7, I put 1, it's 7 from time to time, knowing. Then there is another adjoinment which is going to be here, and which is going to be e to the power of 7, the exclamation which is equal to u. So from this adjunction, we have made a ball here. Then you have e to the power of 7 to the power of 0. There are ups and downs. And to finish, since this is a power of something, it is an adjoinment of the kind that exists on the left. and what is called zero-existence in the orientation of the object. These are the things, but this is what happens in fact. If you start from the category B equals 7, you have the Yoneda plunge. This plunge has a left adjoint, which has a left adjoint, which has a left adjoint, which has a left adjoint, and so on. The interesting thing is that you have to work to demonstrate that the category that would be like that from its Yoneda is in fact equivalent. This is to show you that the existing ones on the right and on the left become more and more restrictive, of course, to the point of forcing the category to be the category of all the points of the sets. If you start from any category and you have all that, you have to understand, in the case of Wood and Rosebuth, how, step by step, you force the category to be simply these sets, objects where there is no structure.

10:00 Any presence of structures under your category must disappear, you must be sold, crushed, crushed by the fact of its successive existence. And are we showing that... You are destroying coherence. Do the U, V, Z, E, V, and X in a certain way look like that without understanding that they are positive things? Yes, laterally, yes, but you have to prove it. That's right. A priori, you're talking about a... There are not several sequences like that, I mean. No, the adjoint is unique. You're talking about the rayonetta here. The rayonetta of B is here. Now, if you have an adjoint here, it's necessarily that. It's necessarily laterally that. But you have to show it. Then, etc. Yes, the sequence is the same. The same is by replacing... By replacing 7 by B? Yes, that's it. If there is a sequence that is like that, the theorem indicates that in fact B is 7 and obviously that suddenly the unit of B is the unit of 7. The x is that. Then there is a flat elliptic, it can only be that. A flat elliptic, it can only be that. Of course. Exactly. Yes, the equations are indicating that it is not too good. These are the two points I wanted to start with so that you understand that successive clarifications are restrictive, even if you haven't seen the demonstration yet. So now, a much simpler example, I'll come back to that, so that you understand what the request is. I look at this simple question, I want to put it like this. On this simple drawing, Lovire made very interesting variations in the colloquium on categories at Tours, which I had organized at the time with Pierre Danfus, to show how, from abstract data, i.e. an UIO of this type, we could calculate a simple equation, which was replaced by the 7-power vector equation. And in this case, they had a configuration of this giant from which they showed very nicely how to do the differential calculation from there.

12:30 By the fact that, starting from this data and from these two factors A1 and A1, if we impose that there is an isomorphism here, what does that mean? What does that mean? Obviously, this is not the case here. Zero is not equal to 1. And now take, for example, the category of Hegelian groups. What I did here, for example, when I take an object E here, an object star, as I have the identity here, the identity gives me something between 0 and 1, and then as I have the identity here, by the junction on the other side, it gives me something from 1 to 1. So I have an arrow here from 0 to 1, which is composed of two. And what interests Lovir is the nature of this arrow, whether it is an objective, sub-objective, and so on. It is the constraints on this arrow that will express more or less cohesion. Here it is not cohesion at all, because zero is not equal, it is the distinction. The fact that zero and one are so distinct means that there is not much cohesion. If, on the other hand, you took the category of Abelian groups, you would do the same thing, there would be an element. Here, you also have 0 and 1, but as you know, in this case, 0 is equal to 1. The terminal object is the initial object, it is the same, it is the group. The group, the null group, is the one that has only one element. So, the difference between this and this is precisely that here, the gap from 0 to 1 is not... I don't know what they are asking us to say in their lecture, whether it is epimorphic or something like that. I was waiting for someone to have notes of the last lecture to be exposed. Did he ask something about... About the article. About the article, maybe. If I look in my notes, if I find something in these notes... The important point is that, it's the fact...

15:00 He even puts equalities, often, to talk about qualities, types of qualities. So, for example... Epic. That's what he asks, I think. That it be epic in the first place. At a certain point in the lecture, he asks for it to be epic. So, it means, in all cases, that it must mean zero and one. Well, now the abstract scheme that he envisages is going to be the scheme that looks like this, a preamble that I made for you to show the interest of this idea of the constraint of Aparana-Juan. In tradition, we don't do that. In tradition, we observe what happens. For example, when you have the category of beams and pre-beams and the inclusion of pre-beams, The computer has a left-handed adjunct, but it still has properties of exactitude, for example, it commutes to the product. So in this case, we have an adjunction that exists, but there is not the second one, but there are properties of exactitude of the left-handed adjunct. That's why this pre-pesto question is not at all exemplary of what is there, and even less of what we will discuss later on the IAO. This is an example of the IAO.