 pseudo-holomorphic curves and ended with a little bit more abstraction, today I'm going to go in the opposite direction. I'll begin with some more abstract discussion and then eventually we'll find our ways back to pseudo-holomorphic curves. So where did we leave off? We were talking about a category C and we were talking about the category of functors, the category of sets. And the object of this functor category is called a pre-sheaf. The functor from C off the set is called a pre-sheaf. So this is the category of pre-sheafs on C. And one of these pre-sheafs is called a representable when it's isomorphic to one of these pre-sheafs Hx, which is just hom in C to an object X. And furthermore, I argued that a pair X in C and U of X representing F, unique, up to unique isomorphism. So if I want to specify for you an object of C, I might as well specify the associated pre-sheaf. Great. So we had some examples such as the space of continuous maps, the compact open to pathology. This represents this functor. And this CW complex BO represents the map to isomorphism classes of vector bundles. And this is the first one I guess is on the top watch on this category of topological spaces. The second one is on the homotopy category of CW complexes. So that's a quick review of where we left off last time. So it turns out there's not only can you talk about these representable pre-sheaves as individual objects, there's an entire functor from C to pre-sheaves on C. I'm going to, I'm now going to call it pre-sheaves on C. So that's this category sending X, Hx. And there's a fundamental fact about this, this functor, which is that natural bijection forms a pre-sheaves from H of X to any F and the value of that pre-sheaf on the object. I'm going to pose an exercise. It's completely trivial, it's like two lines or something. But yet, it's a little bit tricky to figure it out. So it looks, it's tricky to figure out where the content actually is and whether you've proven it or whether you've just written down the tautology. Okay. So in particular, if I take F to be itselfly a native functor, either hom from Hx to Hy is equal to hom from X to Y. In other words, this functor Y, fully faithful, fully faithful meaning it's a bijection on morphisms. Okay. So what that means is that we can view C as a full subcategory of pre-sheaves of C. This is a, and this is a much stronger assertion than this assertion here about representing a functor, the object representing a functor being unique up to unique isomorphism. So you can think of P of C as consisting of sort of generalized objects of C in the sort of very vague sense of, but the sense which is similar to the way in which distributions are generalized functions. A distribution is something which, well, you can't evaluate it at a point, but you can integrate it against any smooth function. You can sort of pair it with smooth functions. And a pre-sheave is exactly the same sort of thing. It's not an object of C, but you can at least make sense of this mapping an object of C to it. Okay. So just, Mark, there's a precise version of this statement that C is the recode completion of C. Yes? I was afraid. Okay. So I think we're done with that. Questions? Yeah. So the question is why is this implication true? Yeah. So I'm just going to take f equal to h y. So f is h y and h y on x on from x to y. Now, I didn't tell you what the map was. And so you might complain that this equality might not be the map induced by the functor. Okay. So there's a more geometric subcategory of the category of all sheaves. And that is something which doesn't make sense on all categories. It only makes sense on some categories with the topology. And I'm going to stick to the case of topological spaces or smooth metaphones here. So then, basically, I need a notion of open covering, which has both of these categories have. f, the sheaf, then draw open covers, open sets, the following, the bijection goes to limit. Let me not write it like that. Let me not write it like that at all. The set of tuples, alpha i in f of ui, such that when you take their image in f, any overlap, they coincide. So this, what it says intuitively is that the specifying element amounts to local data examples. Yes. Yeah. Okay. So let's, yeah, that was many questions. So let me just go over the definition again, clarifying all of that. Okay. So we're only considering these two categories. Okay. For right now. So x is an object of one of these categories. And every open subset of x is therefore also an open subset of smooth metaphones. So I can apply f to that open set. And if I start with an element of f of x, so there's a tautological inclusion map from ui to x, just including that open set. And so that gives me a map, a restriction map from f of x to f of ui. So if I have an element of f of x, I can get a bunch of elements of f of ui, one for every i. Now, if I have two open sets, f of x maps to i and to uj because they're open subsets, and then both of these also map to ui intersect uj. Okay. And the fact that f is a functor means that this diagram commutes, this is the same map, just the map induced by pulling back under the inclusion from ui intersect uj into x. So if I start with an element of f of x and I look at the tuple of elements of f of ui I get for all i, then I can't just get any tuple. The only tuples I can possibly get are the ones that satisfy this condition that when I take alpha i and further restrict it to the intersection, that's the same as taking alpha j and further restricting it to the intersection. And then I made this sort of imprecise statement that the assertion that this map is an isomorphism should be thought of as saying that, well, to specify an element of f of x amounts to local data on x. I mean, would it even by local? Well, I mean, you can specify on some open sets and if you have agreements on overlaps, then it extends to your whole space. But I think maybe at this point an example would be clarified. Yes, question? Yeah, well, yes. Okay, so repeat the question. The question was the definition of a sheaf that you often first see is a sheaf on a fixed topological space. And if you took this definition and replaced C with a category of open subsets of a fixed topological space, then you would recover that definition. So this is a, anyway, it's a very similar flavor. It's a different thing. You know, it's the one you want to use if you're talking about representing functors. Okay, so let me write down an example. An example is that these unated functors are sheaves. Why is that? Well, if I suppose I have some smooth manifold m and I want to map x to m, well, specifying a smooth map from x to m is the same as specifying a smooth map on each UI to m and with the requirement that they agree on overlaps. There's not much content in that. But there are other things which are sheaves. For example, a and b are smooth manifolds. Then z maps to smooth maps, z cross a to b. That's also a sheaf. And it's basically the same reason because now the sheaf property is telling me that to specify a smooth map from z cross a to b, I can just specify on UI cross a and it should agree on the overlaps. Okay, so that one is also a sheaf. Usually the statement, usually the assertion that something is a sheaf is sort of a trivial statement. It's either, if it's true, the proof doesn't amount to much. But we remarked last time, this one is not representable. They're more sheaves than representables. What is something that's not a sheaf, not sheaves? U maps to the set of embeddings of U into Rn. So to specify a map from U to Rn, that is a local data. But the condition that that map be an embedding is not a local condition. So this map would be an injection, but not a surjection. U maps to set of continuous functions on U cross U. That's not a sheaf because it's not local data. If I have a point P comma Q and P is different from Q, then that's, if this is U and this is U and this is the diagonal of U and I try to make an open cover of U and look at continuous functions on UI cross UI, that's only going to give me functions over some of these boxes. It's never going to tell me what it does. There's something which is also not a sheaf U maps to isomorphism classes of vector bundles. So that's a little bit, maybe surprising at first glance that that's not a sheaf because I had this intuitive model that a sheaf is something which is local. And to specify a vector bundle, that's something pretty local. You can glue vector bundles together if I give you an open cover and vector bundles on each open set and you specify isomorphism between them and co-cycle condition. You can glue them together to get a global vector bundle. Okay, so the problem is here isomorphism classes. Okay, so why is this not a sheaf? Well, I mean here's an explicit counter example, right? Take a circle and cover it with this open set. And this open set, let's look at vector bundles of rank one. So the intersection is this union, that there's only one vector bundle on this piece, only one vector bundle on this piece, and only one vector bundle on their intersection. And yet the vector bundle, there are two vector bundles on the circle, rank one vector bundles. It depends on how I glue them together. And that's not something that this definition captures. Okay, I'm going a little bit out of order from my notes, but I think it makes sense. So to fix this last example, we need the notion of a stack and that following. So first, let's define a groupoid. This is a category. All morphisms are isomorphisms. So if you take any category and just throw away all the morphisms, which aren't isomorphisms, you get a groupoid. So take your favorite category. It gives you an example of a groupoid by just throwing away all the non-invertible morphisms, groups and isomorphisms of groups. So for example, if G is any group, we can look at this BG. This is a groupoid. It's just a single object and the end of set of all the morphisms of that object is G. That's an example of a groupoid. Another example is X is a topological space. Then vector bundles of X and isomorphisms of vector bundles over X is a groupoid. So functor F, not values and sets, but instead valued in the category of group oids of the stack when for all open covers, the following is an isomorphism. So sorry, X equals, yes, thank you. So again, I'm only taking C here to be as smooth manifolds are sets, or smooth manifolds are topological spaces. So here, I'm just going to write down an analog of the of the sheaf condition, but it will be a little bit different. So again, I choose alpha i in F of ui for all i. Okay, and then there's this agreement on the overlaps condition. But we're in a category, a category of vector spaces. It's usually a bad idea to talk about equality or isomorphism of objects. It's better to specify an isomorphism. Yes. Yes. The question is, what is group oids? Group oids is a two category. I was, I was, I was hoping not to have to say this. So this is a two category. It's not a category. And so anyway, in a, in a category, the collection of morphisms from X to Y is a set. And in a two category, the collection of morphisms from X to Y is a group void. And that means you have to be a little bit more careful about what you mean by associativity of composition. Maybe I'll leave it at that. Okay, the question is to repeat. So in, in, in a category, um, x, y is a set. In a two category, um, x, y is a group void, one of these things. And replacing it with its set of isomorphism classes is usually not what you want to do. When it's a group void, you have to be a little bit careful about what you mean by associativity of composition because um, x, y times hom y, z times hom z, w, to hom x, w, that is a functor between group voids. And so you can't, you, it is, it is the wrong condition to say that the two functors you obtain by composing in either order are isomorphic. You need to specify an isomorphism and require a co-psycho condition for, for, for quadruple morphisms. Anyway, just despite the extra complication, most of the things you want to ever consider with group void, with, with, with categories, you can just do with two categories. It works mostly the same. There are a few more diagrams to, to chase, but that's it. Okay, so maybe I'll continue writing the definition. Um, you take alpha i in f of ui. You can restrict alpha i and alpha j to ui intersect uj. And, um, we no longer want to say that they're isomorphic. They're the same since, since they're objects of a category, we should specify an isomorphism between them. So, so that's part of the data here, beta ij. That's, uh, from alpha i restricted to ui intersect uj to alpha j restricted to ui intersect uj. And, um, well how do we glue together vector bundles? If I have an open cover, x equals union i ui, and I have vector bundles on each of the ui, and on each double overlap, I have an isomorphism between the restrictions. Um, if I want to get a vector bundle on a total space, I'm missing a condition, namely the co-cycle condition. So I better have the, uh, beta ij times beta jk. I'm going to swap the order of these, so my composition is in the correct order. I think alpha j, alpha i. Uh, so then I can write this co-cycle condition in an order which makes sense, beta ij, beta jk is equal to beta ik. So this thing on the right here is a groupoid. Let me not tell you the, the definition of, well let me not write the definition of an isomorphism between two of these things, but there's natural way to talk about things here, being isomorphic. Basically, through, uh, the example that, uh, x maps to vector bundles on x, so I don't think I need to say more about that. Um, but basically anything you could want to parameterize is going to give you a stack. So, um, you know, x maps to, say x is, is a smooth manifold. Let me do things in the smooth category. Then I can ask to parameterize remount surfaces by smooth manifold. So what does that mean? Um, c goes to x. Um, well it says, I guess it's a maybe it's a proper submersion, um, with a fiber-wise, almost complex structure. Proper submersion, um, with fibers, with two-dimensional fibers, a fiber-wise, almost complex structure. So that, this is the, you know, the, this, this you would call the modular stack of remount surfaces. Okay, so the question, you, you, you should be the first question. Um, you ask is, you know, is it representable? Right? If, if it were representable, then that object representing it, I would call it the modular space of, of remount surfaces. Oh, so what's the answer to this question? Yes, Paul. So if the answer were yes, then it would probably be pretty hard to construct it. So, okay. Um, so everything representable is of the form, ham to something. And ham to something is always a set. And, and this is not a set because it has automorphisms. That's the, it could be equivalent to, it could be equivalent to, it's not equivalent to a set, is why? Uh, no, as long as you, as soon as you have any automorphism, it's, it's not equivalent to a set. It's not equivalent to a set indeed. It's not, yes, okay. Yes, I, I should, I should, I should write it, I should say it like that. Because it's, say it's value on a point, it's not equivalent to a set. Where the morphisms of the groupoid on the left hand side, the left hand side of what? F of x? Or this, this, maybe this groupoid here? So, okay, I'll, sorry. Okay, so, so, yeah, so, so, so, yeah, I think it must be about this. Okay, so, so morphism, morphism is a diagram. It's an isomorphism over x. So, so c to x is some family, and c prime to x is another family, and I, I give this map. And this map should be an isomorphism, not just any map. Well, it's isomorphism of families over x. It's, ah, yes. Um, so equivalent means, what do I mean by equivalent to a set? Um, right, so when you have, um, so, if you have two categories, you can ask for functor, um, in both directions, and for both compositions to be isomorphic to the identity. That's an equivalence of categories. It's called equivalence, not isomorphism, um, because, um, because you don't want to ask that both compositions are equal to the identity functor, just that they're, um, equivalent to it, or isomorphic to it. So a set, you can regard it as a groupoid by saying, well, the objects are that set, and they're no morphisms other than the identities of every, of every element. Yes, I, ah, why is it not equivalent to a set? Yes, so I'll, I'll, I'll finish my sentence. It's not equivalent to a set since, um, there exists, ah, there's just Riemann surface, whose automorphism group is not one. And as soon as you have an object of your groupoid, automorphism, well, it's even, even only, right? A groupoid is equivalent to a set, even only if the automorphism group of every object is trivial. Indeed, yes, the stack of vector bundles is not representable either. Okay. So, so despite the fact that these are not representable, um, well, how can I say, they, they, they are still nicer than arbitrary stacks. Oh, they're also nicer than arbitrary stacks, but they're even nicer than arbitrary pre-sheets, I guess. No, so let me, let me refer back to my notes rather than just, ah, making it up as I go. But in, in, in particular, um, for example, modular stack of orbital, of modular stack of, of Riemann surfaces, and here, um, just so I don't say it's technically wrong statement, let's say the genius is at least two, um, is an orbital. So that's probably the, the tamest you can, the, the, the tamest you can be without being actually representable by smooth manifold. So it's a, it's a quotient of a smooth manifold by, I just locally a quotient of a smooth manifold by finite group action. Okay. So, so, so let me say what, what, uh, of that, what that means. Okay. So if you have a stack of topological spaces, um, x and y are stacks on topological spaces, y is called representable when in top, uh, fiber product x cross y z representable. So we already have a notion of what an object is representable. I'm defining what it means for, for morphism to be representable. And what you should think of this as saying is basically that the fibers are representable, the fibers of the morphism are representable, whatever, whatever that means. Yes, an orbital is not a manifold, right? So, so Homstone orbital fold is, um, the question is, how can the modified stack be an orbital? Um, well, I haven't really defined for you an orbital. So, so, uh, the orbital do not form a category. They form a two category. They're one of these. And, and what that means is that the maps, um, yeah, an orbital doesn't have a set of point that has a, has a group point of points. Anytime you have a, a local model, you know, Rn mod G and you have a point, um, in Rn, uh, you should regard the, you know, the point in the quotient as having automorphism group that the stabilizer at that point. Y for, for every, yes. So in, indeed, I messed this up, um, every, in top and every map, e to y. Yes. Yes. A map from z to y is, is a map from the United Functor of, of z to, to y. Equivalently, it's an object of the group point y of z. Okay. So, so, so more, more, more generally, um, if he, any property of morphisms served under pullback. So, property of morphisms served under pullback is a property of morphisms such that, um, in any diagram like this, if x to y has p, then, then so is the pullback. So there are lots of properties like this, ones which an open map of topological spaces, an embedding of topological spaces, um, a closed embedding of topological spaces, um, admitting local sections, um, a submersion of, of, of smooth manifolds. There, there are lots of these. So then, you, you, you can, um, you can say what it means for a map of stacks to have this property, meaning all pullbacks have them. Maybe I'll just make a definition, orbifold categories with manifolds, um, back x such that one, um, exists a surjective etal u to x to the diagonal map. So in the presence of two, one is equivalent to one prime, which says that x is locally modulo gamma for m, with manifolds and gamma finite group. So in some sense, this, this stack of, of vector bundles is the unit of point modulo dLnr. So this is a quotient in, in this category of stacks. Yeah, um, what do I mean by locally? Um, so there are lots of, right. So I mean, um, there are a bunch of open embeddings from m mod gamma to x and it's an open covering and open embedding and open covering. Those are properties of morphisms in category of topological spaces or smooth manifolds. And so they make sense for morphisms of stacks also. So I can talk about a map of stacks being an open embedding or a bunch of maps, a bunch of a collection of open embeddings being an open cover. Um, there exists, what was the meaning of one was, so there exists a surjective etal map from, from some, from some u and from some smooth manifold u. But etal means local isomorphism, local homeomorphism or local, the few morphism. Okay. So there were quite a few more things I wanted to say, but um, one is easier because the question was that one prime seems easier than one. Um, because showing something is locally quotient, uh, requires actually, well, okay, I, I, I'm just going to assert the opposite, which is that one, one, one is easier than one prime because, um, right, if x is union of mi modulo, gamma i, then mi to mi modulo gamma i is etal. And this is a bunch of open embeddings, in particular it's etal, so this is etal. Oh yeah, etal, etal, etal does not mean proper, certainly not. Yeah. Um, going from one to one prime is a little bit harder, but so actually let me stay a theorem along on the line. So this is a, in the last 10 years, a theorem of Zung, which says that if x smooth manifolds and it has proper diagonal, there exists a surjective submersion, your representant to x, then x locally, um, mod g, or smooth manifolds in g, compact v group, drop condition 2. Yeah, so condition, what would happen if we dropped condition 2? So condition 2 is analogous to, uh, being how to start. So, um, you, you can get lots of, um, very, very bad things, like, um, you could take the, like the real line with a doubled origin, and then maybe you could, uh, try to quotient by the, by the Z2 action exchanging the two origins or something like that. So if you have, if you, if you drop two, then one in one prime are no longer equivalent. One is weaker. Basically, you take any, um, any, any, any smooth manifolds and take open sets and identify them. However you like. Uh, yeah, so, so we have this, yeah, so I think, I think maybe I didn't say this. So C, um, embeds into, um, to C, the embeds fully faithful into the category of stacks. So that means there's really no need to distinguish between, um, element of C and it's, it's your native functor here. So this is just, if we're thinking about stacks, then we're working in this category. And if I ever mentioned an object of C, I really mean its image in here. Not that those things are, were really any different in the first place, but, um, so, so if I, I mean a map from, from sort of f, this stack, then, um, f means a map from Hx. Um, and by Uneda, that's, that's the same thing as L object. So I apologize that this lecture was all about stacks. I thought I was going to get to holomorphic curves. I promise I will do only holomorphic curves next time. I didn't, I didn't really hear the last part, something about the image of, image of a, okay. So, so, so the question was to compare the definition of a sheaf with the definition of a stack. Yeah. So, um, when is a stack a sheaf? That was, that was the question. Yes. Okay. So, um, okay, so sets you should think of as inside group voids. And there's, there's this thing you can, you can serve, there's a, a, a reflection, maybe call reflection, just, um, take isomorphism classes go from group voids to sets. Um, so this inclusion is, is very nice, uh, sets into group voids. And so any sheaf is a stack. Now you can take, take a stack and, and apply pi zero to it to get a pre-sheaf, and usually it's not a sheaf. Um, I don't know if I could, I mean, if, if it were originally sheaf, it came from a sheaf, but, but if it's not a sheaf, then probably, if it's not, yeah, if it, yeah, if you have a stack, which is not a sheaf, and you try to apply isomorphism classes, usually you're not going to get a sheaf. Yes. I mean, and, and, and, and yes, yes, vector bundles, we won't surface any of the examples. If it were a really trivial stack, I guess you could probably, I don't know. Yeah. So the question was, um, at the beginning of this time, in the end of last time, I talked about isomorphism classes of vector bundles that is a functor on the homotopy category of, of CW complex. And I said that's represented by BO. And then today I, later I said, vector bundles, that's isomorphism classes. Um, that's not representable as a functor on top. Um, yeah. So, so, so there's no inconsistency since these are different categories. And you know, you, you can try to, you know, take the functor from say nice topological spaces to the homotopy category of CW complex and try to prove one is representable based on noting, noting the other is you realize you can't prove it. And by doing so, maybe you'll see why. Yeah. So, so what do I mean by proper and surjective submersion? So these are again, uh, properties which are preserved under pullback. And so I can, um, I can, I can define them for a map of stacks by, by saying, well, every pullback satisfies them. Ah, yes. What do I mean by POC is the free code completion of C. So let's see if I get this right. Um, so, so let's say D is a co, co-complete category. And that means it has all co limits. Then I can look at co-continuous bunkers from P of C, D, co-continuous means preserves all co limits. And I can restrict it to C. And this is an equivalence. That's what it means. The P of C is the free co-completion.