Is the dual notion of a presheaf useful? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T21:37:41Z http://mathoverflow.net/feeds/question/32071 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32071/is-the-dual-notion-of-a-presheaf-useful Is the dual notion of a presheaf useful? Dylan Wilson 2010-07-15T21:45:20Z 2010-07-18T13:19:59Z <p>It seems that there is a common theme in mathematics where, if we want to find out about a category C, then we look at $\hat{C}$ (the category of contravariant functors from $C$ to $Set$). There are all sorts of good reasons for this (Yoneda's lemma being a big one, and the fact that this is a topos). There are other versions, sometimes you look at contravariant functors into some other category, like groups (e.g. for algebraic groups). But I have never seen the dual notion: covariant functors from some other category INTO $C$. </p> <p>Is this notion as useful as the notion of a presheaf? I guess it's like looking at a category "over" $C$ as opposed to a category "under" it. I guess, hidden in this question is the question: Can one learn anything about a category $C$ by looking at presheaves of $C$? For example, does the difference between presheaves of sets and presheaves of ableian groups tell us anything about the differences between the category of Groups and the category of Sets? </p> <p>EDIT: There seems to be a bit of confusion of what I mean by "dual of a presheaf." I don't mean "copresheaves" (which I didn't know existed before I asked this), I mean what you get by reversing the arrow of the functor not taking the opposite category. So I'm looking at functors into the category of interest, as opposed to out of them. I can see how this is confusing because usually "dual" doesn't turn around functors, just arrows inside categories. So I guess I mean presheaves in $C^{op}$... (but covariant instead of contravariant?)... who knows.</p> http://mathoverflow.net/questions/32071/is-the-dual-notion-of-a-presheaf-useful/32079#32079 Answer by JosÃ© Figueroa-O'Farrill for Is the dual notion of a presheaf useful? JosÃ© Figueroa-O'Farrill 2010-07-15T22:43:27Z 2010-07-18T07:35:18Z <p><strong>Edit</strong></p> <p>I seem to have misunderstood the nature of the duality in the question. This answer is not relevant. I'll keep it in case it has any archaeological interest.</p> <hr> <p>In <a href="http://en.wikipedia.org/wiki/Local_quantum_field_theory" rel="nofollow">local quantum field theory</a>, one does use the notion of a co-presheaf (or is it a pre-cosheaf?) of operator algebras. You can read about it (perhaps not with those words) in Streater and Wightman's <em>PCT, spin-statistics, and all that</em> and perhaps also in the links from the above wikipedia page. This is perhaps not surprising as it is the states of the system which form a presheaf.</p> http://mathoverflow.net/questions/32071/is-the-dual-notion-of-a-presheaf-useful/32100#32100 Answer by Gregory Arone for Is the dual notion of a presheaf useful? Gregory Arone 2010-07-16T01:39:11Z 2010-07-16T01:57:17Z <p>Covariant functors from the category of pointed sets to the category of pointed topological spaces are sometimes called $\Gamma$-spaces, and they have been important in algebraic topology. One reason is that $\Gamma$-spaces model infinite loop spaces (and therefore connective spectra) and are very helpful for understanding stable homotopy theory.</p> <p>$\Gamma$-spaces also serve as a model for particularly well-behaved covariant functors from the category of pointed topological spaces to itself. Of course, these functors play an important role in topology as well. I like to think of Goodwillie's Calculus of Homotopy Functors (and also of Michael Weiss's Orthogonal Calculus) as a kind of "sheaf theory for covariant functors". In these theories, covariant functors are analogous to presheaves and linear functors are analogous to sheaves (The definition of a linear functor is essentially a homotopy-invariant version of the definition of a sheaf). The process of approximating a general functor by a linear one is analogous to sheafification, and so forth. These theories provide methods for studying certain types of functors, but of course they also tell you something about the category of spaces itself. </p> http://mathoverflow.net/questions/32071/is-the-dual-notion-of-a-presheaf-useful/32162#32162 Answer by Donu Arapura for Is the dual notion of a presheaf useful? Donu Arapura 2010-07-16T12:41:12Z 2010-07-16T12:41:12Z <p>With no disrespect to my fellows answerers, the examples so far are a bit exotic. If $X$ is a space/manifold, the collection of continuous/$C^\infty$ functions on open subsets with compact support forms a copresheaf (the maps are extensions by $0$). For an even simpler example, start with a presheaf of modules $M$, then the dual $U\mapsto M(U)^*$ will be a copresheaf. You can replace <code>$()^*$</code> with your favourite contravariant functor.</p> <p>Also, I seem to recall that Bredon thought the notion was useful enough to include a discussion of them in his book on sheaf theory.</p> <p>That said, I have no very good explanation for why (pre)sheaves are more common, and seemingly more useful, than the dual notion. I suppose one explanation, is that in some cases by dualizing you can get rid of the "co"-ness without loss of information. I've used this trick myself.</p> http://mathoverflow.net/questions/32071/is-the-dual-notion-of-a-presheaf-useful/32222#32222 Answer by Tilman for Is the dual notion of a presheaf useful? Tilman 2010-07-16T20:21:25Z 2010-07-16T20:21:25Z <p>You can define a category whose objects are all functors $X\colon I \to \mathcal{C}$, where $I$ is any small category, and whose morphisms are given in the same way as for ind-categories: $$Hom((I,X),(J,Y)) = lim_{i \in I} colim_{J \in J} Hom(X(i),Y(j)).$$ (If you restrict attention to filtered categories $I$, this is precisely the category $Ind-\mathcal{C}$.)</p> <p>There is a fully faithful functor $F$ from this category, let's call it $Cat-\mathcal{C}$, to $\hat{\mathcal{C}}$: $$(I,X) \mapsto (c \mapsto colim_{i \in I} Hom(c,X(i))).$$ This functor is an equivalence of categories if one gets some set-theoretic problems out of the way. For example, if $\mathcal{C}$ is small, then any contravariant functor from $\mathcal{C}$ to sets is a colimit of representable functors, which shows that $F$ is essentially surjective.</p> <p>So, basically, this is the same construction as the Yoneda embedding! </p> http://mathoverflow.net/questions/32071/is-the-dual-notion-of-a-presheaf-useful/32342#32342 Answer by Dylan Wilson for Is the dual notion of a presheaf useful? Dylan Wilson 2010-07-18T06:33:43Z 2010-07-18T08:02:21Z <p>Sort of an answer to my own question that I just stumbled across: A "stack over a category $S$" is a functor from some other category into S that satisfies various properties. <strike>If these properties are similar or dual to the ones that a presheaf must satisfy in order to be a sheaf... then I would say that the notion I was talking about in my question was that of a "pre-stack," and that the answer is "Yes, they are useful... they give rise to the notion of stacks!"</strike> EDIT: Considering my original question simply asks if the notion of functors into a category being useful, it doesn't really matter if the property of a stack that separates it from any old functor into $S$ is related to the property of a sheaf (that separates it from any old functor out of $S$)... What matters, I think, is that stacks are a type of functor INTO $S$ as opposed to out of it, and they are, indeed, useful. But it would be nice if someone could clean this answer by erasing this sentence and replacing it with some "big-picture" description of what (if anything) a stack over $S$ tells us about $S$... in relation to, say, a sheaf on $S$. Am I still incomprehensible? It is rather late...</p> <p>Of course, I may be spouting nonsense... so I'm making this answer a community wiki in the hopes that someone who knows something about stacks can verify or falsify what i've written, and hopefully expand. </p> <p>(Meta-note: I put this as an answer because I wasn't really sure what to do... I don't think it properly belongs as an edit to the original question... it's basically a different question "are stacks basically what I was thinking of?" but it was sufficiently related and similar to my original question that I don't think it merits a whole NEW question... so I put it here as community wiki. If this is not the correct thing to do, then I trust a moderator or someone much wiser than me will take down this post!)</p> http://mathoverflow.net/questions/32071/is-the-dual-notion-of-a-presheaf-useful/32358#32358 Answer by G. Rodrigues for Is the dual notion of a presheaf useful? G. Rodrigues 2010-07-18T13:19:59Z 2010-07-18T13:19:59Z <p>Let me try and take a stab at this question. I will give not so much as an answer to your question, but more of a rambling collection of remarks. The post turned out to be <em>much longer</em> than I wanted, because as B. Pascal once complained, I did not have the time to write a shorter one. Expect some inconsistencies, a lot of hand-waving, etc.</p> <p>To put some order in my thoughts, I will try to argue the following four points. Fix a category $\mathcal{A}$, which is the category "of interest". Then:</p> <p>(1) Functors into $\mathcal{A}$ are interesting.</p> <p>(2) Functors into $\mathcal{A}$ tell you everything about $\mathcal{A}$.</p> <p>(3) Part of the perceived asymmetry follows from the fact that in many cases the interesting category $\mathcal{A}$ is equivalent to a presheaf category (or some localization thereof).</p> <p>(4) There is not really an asymmetry between functors out of and functors into $\mathcal{A}$, but more of a duality.</p> <p>The first point is easy to settle, as Qiaochu Yuan already gave a class of interesting examples: for a category of interest $\mathcal{A}$ and any category $\mathcal{I}$, a functor $\mathcal{I}\to \mathcal{A}$ is a diagram of shape $\mathcal{I}$ in $\mathcal{A}$, so hell yeah, functors into $\mathcal{A}$ are important. You may think this a rather pedestrian example, but below I give more examples.</p> <p>To explain (2), instead of working with categories, that is, in the $2$-category of categories, let us go one level down to the category of sets. A set $X$ is determined completely by its elements, that is, maps $\ast\to X$ where $\ast$ is a singleton set (any one will do: for the sake of determinacy, take the singleton set comprised of the element $\emptyset$). Now in a general category, we cannot speak of elements, but we can (and do) speak of <em>generalized elements</em>.</p> <p><b>Definition:</b> Let $\mathcal{A}$ be a category and $a$ an object of $\mathcal{A}$. A <em>generalized element</em> of $a$ is a map $x\colon d\to a$. This is also written symbolically as $x\in_{d} a$.</p> <p>We do not need generalized elements to develop category theory, but personally, I found them useful to build upon the intuition gained from working with more "concrete" categories. A very nice discussion of generalized elements is in Awodey's book on category theory. Now the kicker: Yoneda's lemma tells us that if we know all the generalized elements of an object $a$ then we know everything about $a$, including of course, the arrows <em>out of</em> $a$ (note: and by duality, if we know all the arrows out of $a$ we will know everything about $a$). Two possible objections may be raised:</p> <ol> <li><p>The original example involves a $2$-category: does not make much of a difference, as my reply to Kevin's comment (see above) still applies. And besides, there are $2$-categorial versions of Yoneda lemma.</p></li> <li><p>Yoneda's lemma works ok, but you need the knowledge of all generalized elements and these range over a potentially proper class of objects: that is true, but in virtually every interesting category one can trim down this proper class to a (small) set, even a singleton set. I will not spell out the proper definitions; they are intimately tied with "smallness" conditions and the adjoint functor theorems (and yes, the category of (small) categories satisfies them).</p></li> </ol> <p>For (3), let $\mathcal{C}^{\mathcal{B}}$ be a functor category. For size reasons, $\mathcal{B}$ has to be small (note: if you have no scientifick problems with creation ex nihilo, you can always spawn a larger universe by invoking the axiom of universes and sidestep this particular size issue). Since functor categories inherit most of the good properties of the codomain category, you will want $\mathcal{C}$ to be as good behaved as possible (e.g. complete, cocomplete, abelian, symmetric monoidal closed, etc.). It is not true that the structure of $\mathcal{B}$ is irrelevant for the structure of $\mathcal{C}^{\mathcal{B}}$, but it is true that in general, $\mathcal{B}$ only needs a bare minimum of structure. This asymmetry between the domain and the codomain categories is reinforced by the fact that many of the interesting categories $\mathcal{A}$ are equivalent to functor categories, even presheaf categories (or some localization of them). Here are two examples.</p> <ol> <li><p>Let us consider the category of groups $\operatorname{Grp}$, undoubtebly a category "of interest". There is a category $\operatorname{Th}(\operatorname{Grp})$ that has all finite products such that $\operatorname{Grp}$ is equivalent to the category of product-preserving functors $\operatorname{Th}(\operatorname{Grp})\to \operatorname{Set}$ where $\operatorname{Set}$ is the category of sets. This equivalence can be generalized to a <em>very large class</em> of categories of "algebraic flavor" and even some that at first sight do not bear the least resemblance to "algebraic categories". A few extra remarks about this example. First, the category $\operatorname{Th}(\operatorname{Grp})$ is a category constructed to make the equivalence work (it is the free category with products on a group object), in other words, it does not exactly fall within the class of categories "of interest". It's a similar to the example of diagram categories in (1), where the domain, a free category on a graph, is just a categorial construction to make the identification of diagrams with functors, not an interesting category by itself. Second, by replacing $\operatorname{Set}$ by another category $\mathcal{B}$ (with at least finite products), you can now speak of groups in $\mathcal{B}$. This gives another class of examples where functors into a category are interesting.</p></li> <li><p>If $(\mathcal{C}, \mathcal{J})$ is a site (a category with a Grothendieck topology), by first taking the category of presheaves and then a suitable localization, one obtains the category of sheaves. This produces a host of geometric categories, like manifolds and schemes.</p></li> </ol> <p>Much like in example 1, the interest is not so much on $(\mathcal{C}, \mathcal{J})$ and even less in the codomain, which is usually the category of sets (other categories for codomain also work, but the categorial requirements for everything to work smoothly are fairly strong), but in the (pre)sheaf category.</p> <p>For my last point (4), a lot could be said, but I will just point you to two articles by F. W. Lawvere in the TAC Reprints, "Metric Spaces, Generalized Logic and Closed Categories" and "Taking categories seriously" (google for them, they are available online). In them, Lawvere makes several remarks about the duality between spaces and algebras of functions which are directly relevant to your question. To show that there is not so much an asymmetry but a duality between functors into and out of, let me give you two examples.</p> <ol> <li><p>In your (that is, the OP) last post, you speak about stacks. A stack on a category $\mathcal{A}$ can be defined as a functor with values in $\mathcal{A}$ satisfying some conditions -- this is the fibered category approach to stacks. But a stack can also be defined as weak $2$-functor with values in the $2$-category of categories (satisfying some extra conditions). For reasons that I will not explain, the first approach is better, but nevertheless the point should be clear. There are actually many examples of this "duality", that identifies some category of functors into $\mathcal{A}$ with some category of functors out of $\mathcal{A}$.</p></li> <li><p>Let me end up with an example from physics that further illustrates this duality. Quantum field theories are notoriously hard objects to define (let alone study). Several years ago, V. Turaev defined the notion of a Homotopy Quantum Field Theory, HQFT for short (check his papers in the arxiv if you are interested) which is a very simple, "toy" example of a QFT. If $X$ is a topological space, we can define a category that has for objects manifolds $M$ equipped with a homotopy class of maps $g\colon M\to X$ and a morphism $(M, g)\to (N, h)$ is a cobordism $W\colon M\to N$ with a homotppy class of maps into $X$ extending $g$ and $h$. An HQFT is a monoidal functor from this category into another monoidal category (usually, the category of finite-dimensional complex linear spaces). I am omitting lots of details, but the gist is that an HQFT gives us invariants of manifolds $M$ by mapping $M$ into some fixed background space $X$. But we can turn things around, for the category of $X$-HQFT's is a (functorial) invariant of the homotopy type of $X$, an invariant cooked up by mapping manifolds into $X$.</p></li> </ol> <p>Hope it helps, regards, G. Rodrigues</p>