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Let $C$ be a category. For the purposes of this question, I would like to avoid cases where the answer might be "no" simply because $C$ is "too large", and so I will ask that $C$ has a set of generators, i.e. a set of objects $\{X_\alpha\}_{\alpha \in A}$ such that for any two morphisms $f,g : Y \to Z$ that are not equal, there exists an $\alpha\in A$ such that $\hom(X_\alpha,f) \neq \hom(X_\alpha,g)$ as maps $\hom(X_\alpha,Y) \to \hom(X_\alpha,Z)$.

(I am most interested in the specific cases where $C$ is either small or presentable. More generally, $C$ could be a "site" or even a "colimit sketch", and this question would still be apt.)

I would like to advocate for the following terminology: a cosheaf on $C$ is a covariant functor $F : C \to \mathrm{Set}$ that takes colimits to colimits. (Similarly, a sheaf is contravariant functor $C \to \mathrm{Set}$ that takes colimits to limits.)

Given two nonequal morphisms $f,g: Y \to Z$, does there necessarily exist a cosheaf $F$ such that $F(f) \neq F(g)$ as maps $F(Y) \to F(Z)$?

Note that the usual version of the Yoneda lemma doesn't suffice: the corepresentable functor $\hom(X,-)$ preserves limits, not colimits, and the representable cofunctor $\hom(-,X)$ is a sheaf, not a cosheaf. Using representable sheaves, one immediately has the answer "yes" if $\mathrm{Set}$ were replaced by $\mathrm{Set}^{\mathrm{op}}$. Perhaps there is a faithful cocontinuous functor $\mathrm{Set}^{\mathrm{op}} \to \mathrm{Set}$ that I am not aware of?

Some applications of cosheaves that I care about are listed at; the proposition there also provides one reason I care about this question.

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Ok so it's weird to me that corepresentable functors are not cosheaves... Is there a "cosheafification"? Does the cosheafification of $\text{hom}(Y, -)$ suffice? – Dylan Wilson Jul 1 '13 at 5:47
@Dylan: If $C$ is small, the category of cosheaves on $C$ is presentable. Since colimits are computed pointwise and "colimits commute", the inclusion of cosheaves into all functors is cocontinuous. It follows from the special adjoint functor theorem for presentable categories that this inclusion has a right adjoint, which you could call "cosheafification". So in that sense the answer to your question is "yes". It is a little funny: usually, "forget" is the right adjoint, and "free" is the left adjoint. So "cosheafification" is "forgetting" data from a functor to a cosheaf. (Namely, ... – Theo Johnson-Freyd Jul 1 '13 at 20:39
@Dylan ...cosheaves are determined by their values on any generating set, so somehow the data being "forgotten" by cosheafification is the values of the functor on all the other objects.) – Theo Johnson-Freyd Jul 1 '13 at 20:41
up vote 7 down vote accepted

No. Let $C$ be any category with a zero object. If $F : C \to \text{Set}$ is a cosheaf, then in particular it sends the zero object to the empty set. But since the zero object is also the terminal object, every object in $c$ is equipped with a morphism $c \to 0$, from which it follows that $F(c)$ is equipped with a morphism to the empty set. Hence $F(c)$ is itself empty, so $F$ is the trivial cosheaf. Now take $C$ to be, say, the category of finite-dimensional vector spaces.

Edit: In fact the free example suffices. Let $C$ be the free category on a pair of parallel morphisms $f, g : c \to d$. Then the coproduct $d \sqcup d$ exists and both inclusions $d \to d \sqcup d$ are isomorphisms. It follows that if $F$ is a cosheaf then $F(d)$ is the empty set, from which it follows that $F(c)$ must also be the empty set and $F(f) = F(g)$ must be the unique morphism $\emptyset \to \emptyset$, so $F$ is again the trivial cosheaf.

Edit #2: Here's another example. Let $C = \text{CRing}$. If $F : C \to \text{Set}$ is a cosheaf, then in particular it sends tensor products to disjoint unions, sends $\mathbb{Z}$ to $\emptyset$, and preserves epimorphisms. Hence it sends $\mathbb{F}_p$ for all primes $p$ to $\emptyset$. Hence it sends $\mathbb{F}_p \otimes \mathbb{F}_q \cong 0$ (where $p \neq q$) to $\emptyset$. And now it follows as above that $F$ is the trivial cosheaf.

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Awesome. I should have thought of the $\mathrm{Vect}$ case. In Edit 2, maybe I'm being dumb, but why must a cosheaf preserve epimorphisms? Not every epimorphism is split, e.g. $\mathbb Z \to \mathbb Q$ is an epi in $\mathrm{CRing}$. – Theo Johnson-Freyd Jul 1 '13 at 17:55
I do like that your examples don't need* the cosheaf to be cocontinuous, but just to distribute over finite coproducts. (*Except for the question in my previous comment?) – Theo Johnson-Freyd Jul 1 '13 at 17:58
Then again, you only use "preserves epimorphisms" to show that $\mathbb F_p \mapsto \emptyset$, and this can be seen because the two inclusions $\mathbb F_p \to \mathbb F_p \otimes \mathbb F_p$ are isomorphisms. – Theo Johnson-Freyd Jul 1 '13 at 17:59
@Theo: a map is an epimorphism if and only if its cokernel pair ( is trivial, which is a type of pushout. But yes, I also noticed later that you don't need that fact to show that $\mathbb{F}_p \to \emptyset$. – Qiaochu Yuan Jul 1 '13 at 18:40
Ah, great. I probably knew that at some point. – Theo Johnson-Freyd Jul 1 '13 at 20:42

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