 Morphisms, I have a space of morphisms here and I should associate a morphism in something with a discrete topology. If all the morphism spaces are locally path connected, then a continuous map from a space into a discrete thing is just the same as a map of sets from pi zero of that set into the other set. So this is true if all Cx, y are locally path connected just because taking pi zero in that case is left adjoint to giving the discrete topology. Okay, so morally this thing is like left adjoint to that functor except not actually true. So that's that construction. Now if I take H of CDB, that's an ordinary category. That's an obvious functor to the category I talked about yesterday. An object here is an embedded sub-manifold. Just forget that it's embedded and take the induced smooth structure. That's an abstract manifold with the smooth structure. If I have a morphism here, I can take because of this condition, it has kind of canonically parametrized colors near the ends. And then I could just take the diffeomorphism class as a corbotism with this induced CN and C out and so on. And that diffeomorphism class does not change if I have a path in this space. So path in this space then becomes an embedded bundle over an interval and then the two endpoints are gonna be diffeomorphic. So if I write pi zero here, this would be the kind of isotopic class or the class in pi zero and the diffeomorphism classes then well-defined. And this preserves composition so this is a functor. So that's how these are related. And I claim that if the dimension of V is big compared to D, then this is an equivalence of categories. And in particular that's gonna be true. If you take that co-limit over bigger and bigger V, how am I doing on time? Good. Claim HCDV to cop D as an equivalent. Let me just claim that it's weakly contractable. Yeah, that's maybe a better claim to make. And this means weak equivalent. If there's any doubt, it's better to just say weakly contractable because that's all you ever need instead of worrying about if it's one or the other. So let's do that. Okay, why is this an equivalence of categories? Or maybe as you had also said, regarding this question of identities, not that I wanna dwell on that. A non-unitial category having identities, the way you usually write it, it looks like extra structure, but really it's a property because of this easy exercise that identities are unique if they exist. And if I take HCDV then that does have identities like composing with a cylinder. If I take Pizzerio of the morphism spaces, I can like deform back to where I started in some way that I'll not write down. So this is unital, just a comment. Why is this an equivalence of categories? Well, why is it essentially subjective? Oh, this is if the dimension of V is big compared to D. Essentially, subjective means anything is isomorphic to something from there. So given some abstract closed D minus one manifold, why is it the underlying smooth manifold of some sub-manifold of some Euclidean space? Well, that's precisely what witness embedding theorem tells you. If the dimension of V is at least twice this number, then the excess of embedding into V, then just take the image of that. It's gonna be defimorphic. Choose M, I can embed it into our twice this, 2D minus two. And then dimension of V just means it needs to be big enough that there's a linear embedding of this into V. So it's essentially subjective if dimension of V is bigger than or equal to 2D minus two by Whitney. Beading. Then we have to check fully faithful. Why is it full? That means surjective on the home sets. That's very similar. If I have two embedded things and I have an abstract comportism between them, why is that the underlying thing of an embedded comportism? Well, you need to embed that abstract thing into V cross an interval. And sort of extending the given embedding of the endpoints and a version of the Whitney embedding theorem also tells you that. So full is the same argument. Maybe you need 2D because you're trying to, or maybe even 2D plus one. You're trying to embed manifold of dimension D. Faithful has to do with the embedding spaces being connected. So if you have two abstract, you have two embedded things that are defimorphic real endpoints. Why can I realize that defimorphism by an isotopy? And a fancy version of the Whitney embedding theorem also tells you that the embedding spaces connected provided the dimension of V is high enough. Something like 2D plus two or something. Embedding of an abstract comportism between two embedded things. This is unique opt isotopy. If you pick two embeddings, there's a path of embeddings between them. And that means if you take pi zero of the morphism spaces there, you'll precisely get the morphism sets there. Okay, that's the relationship. And that's the sense in which I've sort of lifted this thing. So under this construction, it goes to a category that's equivalent to cop D, provided dimension of V is big enough, or take the code limit. Okay, that was a long discussion about that. Now, if, okay, and these spaces are locally path connected. So there is a, so if I take HC and then take IOTA. So I have the same objects. And then a locally path connected space has a continuous map to its own pi zero. Just send a point to its path component. This is given the discrete topology. So I can regard that as a functor from there to there. Now this thing here is, of course, surjective. And if I regard it as a continuous map, it is one connected. So one connected means bijection on pi zero, surjection on pi one. But pi one, that's trivial. So that's, of course, true. And it's set up to be a bijection on pi zero. So that means, okay, where am I going? It's a little bit out of order. Let me get back to that for a moment. I want to take classifying space of a topologically enriched category. And I just do it the same way. Take the nerve. Now take the space of, I mean, take the same underlying set as before. Np of c means composable p-tobbles of morphisms. And now I've topologized morphisms, so just give this the product topology. Then the nerve is a simplicial space. And I just define the classifying space as before, as the geometric realization of the nerve, which means the disjoint union of Np of c cross delta p. Well, I just use the topology here and give this the product topology, this the quotient topology. So it's simplicial if this is a category, if it's only non-unital, then this is a semisimplicial space. So that's what I should have said. Semi-simplicial means I only have, I don't have degeneracies. And then I should not quotient out by the thing that involves degeneracies. So let's do that. Sometimes called the thick realization. Good. So, okay, where I'm going is, I have this canonical map here. I want to take b of that and ask how connected is it? Turns out it's always two connected, no matter what topological inverse category you start with. And that's what I was just starting to explain there. And zero c is just the set of objects, maps to n, zero, h, c, which is also the set of objects. So this is a bijection. And one c is the total space of morphisms. So in other words, the distro and union xy, c, x, y, n, one, h, c, given the discrete topology, is just pi zero of this space. But mapping a space to its own pi zero is always one connected. Same argument applies to n two. This is also one connected and therefore zero connected. So in general, for all p, the induced map of n p is two minus p connected. And then there's a general fact about semisimplacial topological spaces, that if you have a map of such things, where the map of p simplices is, well, k minus p connected, then the map of realization is k connected. P of c to np of h, c. This is always two minus p connected for all p. And this implies that c to n, h, c is two connected. Well, that's great because at the end of last lecture, we were looking for a space with a two connected map into b of cop d. And then we can just use this embedded compotism category. So if dimension of b is big, c d v to a h, c d v, rather than this was equivalent to a d, induces two connected map from b, c d v to b. So this thing I call x at the beginning of the lecture. And last time I can use this one. So it induces an equivalence from the fundamental group of it. So for classifying field theories, we can use this space instead. So maybe the first time you see this, it looks like that was a step in the wrong direction. I had just an ordinary plane category and now I have to talk about topologically enriched and so on. But that's until you know better. What turns out for this space, you can just write a formula for its homotype type in terms of something that looks simple. Where I don't know a good formula for the homotype type of this thing. So this is a version of a theorem. I proved with Ip Madsen would re-determine twice. That's a formula for not only the fundamental group of it, but the whole homotype type. So, okay, before I can write the answer, I have to give, I'll tell you one or two things. So that's the grass manion of d planes and v. R, what should I call that? No, x something, the d dimensional linear subspaces. So the usual grass manion of d planes and v. The usual topology. Over that there's a vector bundle called the universal vector bundle and it's orthogonal complement. It's just the space of pairs of x comma v in the grass manion cross v itself. So that little v is in the orthogonal complement of the space. So this is a vector bundle whose fiber dimension is dimension of v minus d. And then td comma v is a tom space, this bundle. So that means the one point compactification of the total space. After homomorphism you can also take the disk bundle and take the quotient by the sphere bundle. So this is a space, finite dimensional CW complex. Then you take that for r plus v. The r has to do with the morphism directions. And then you take the v fold loop space. So this means the space of pointed maps from SV td comma r plus v where SV means the one point compactification of v. So if v is rn then this is just SN and this is the n fold loop space. V is rn, r plus v becomes rn plus one and this becomes the n fold loop space. So that's what the theorem is about the homotype. In terms of things that are kind of familiar in homotopy theory, in the original paper we proved this in the direct limit where this co-limit over bigger and bigger v but it's also true for finite dimensional v. No all finite dimensional v, yeah. Even like if the dimension is equal to d. Let's see, is that true? Yes. So you always have the m to z here. So if d is bigger than v, both sides are a point. And yeah, no conditions. And there's also a version with orientation and any tangential structure you like. Let me not write that down, but. In the last few minutes, let me just explain what that says about pi one. Okay, so in a special case, okay, just to repeat what I said before, there's, as far as I know, no good formula for the homotype aside of this thing. But, so this is maybe evidence that this topologically enriched thing is a more natural object. Let me spend two or three minutes on the case d equals one and v equals r where I drew this picture before. Let me make the picture a little bit more complicated. Here's a morphism from five element set to a three element set. In that case, Grassman one of r plus v is Grassman one of r two. The space of one, unoriented one planes in r two, also known as r p one. I mean, also known as a circle, but let me write it this way. This vector bundle over a circle, if you think about it a little bit, this will be the Mobius band, the non-trivial one dimensional vector bundle over a circle. If your one point can pacify the Mobius band, you get r p two. Then you take loops of r p two and you take the fundamental group point. Well, the fundamental group point has to do with pi zero and pi one. So pi zero of loops of r p two. That's pi one of r p two, which is Z mod two. Whereas pi one, the fundamental group, loops of r p two, some base point. Pi two of r p two is pi two of s two, which is Z. So that tells you that the fundamental group point of loops of r p two, up to equivalence, it has just two objects. I mean, it has two isomorphism classes, but if I pick one of each, the isomorphism group is gonna be Z. So up to equivalence, this is gonna be, I think people sometimes call it, okay, I'm gonna write Z disjoint union Z, for this means the one object group point with isomorphism group Z disjoint union itself. So that's determining the universal function from this group point, this group category into a group point and it will be this group point. Okay, let me end there.