Do cocontinuous SET-valued functors separate points?

Question

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 http://ncatlab.org/nlab/show/cosheaf; the proposition there also provides one reason I care about this question.

Answer 1

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.