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Let $C$ and $D$ be categories and let $F : C \rightarrow D$, $G : C \rightarrow D$ be two functors such that they are either both covariant or both contravariant. Under what most general hypotheses is the following statement always true:

A natural transformation $\eta : F \rightarrow G$ is a monomorphism (epimorphism) iff each component of the the natural transformation is a monomorphism (epimorphism).

What I would like to know are the weakest possible restrictions on $C$ and $D$ for which the forward implication holds.

Also, can someone direct me to a source that discusses this?

I should perhaps mention how I came about this question. I was trying to show that if $X$ is a topological space and $F, G$ are sheaves of sets on $X$, then a morphism of sheaves $\eta: F \rightarrow G$ is a monomorphism if and only if the induced maps on stalks are injective (**).

There is probably a way to prove this without using the fact that the components of $\eta$ are monic (i.e. injective set functions in this case), but I already know from an argument using the Yoneda Lemma that in this particular case because my target category is the category of sets, $\eta$ is a monomorphism if and only if each of its components are (and the proof of (**) is quite elementary, once I know this fact).

Also, I am not aware of a proof of the fact that $\eta$ is an epimorphism iff every component of $\eta$ is an epimorphism, even when the target category is the category of sets. So, is this statement true when the target category is the category of sets?

(Later edit: After starting a bounty on this question, I realized that it might not be possible to give the weakest possible conditions on the categories $C$ and $D$, like I am looking for. In that case, any known conditions on $C$ and $D$ for which my question holds will also suffice.)

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One direction is easy: whenever all components $\tau_c$, $c \in C$ of a natural transformation $\tau$ are epi/mono, the same is true of $\tau$. Indeed, supposing for example that the $\tau_c$'s are all mono, and that $\tau\sigma = \tau\sigma'$, then for each $c$ we have $\tau_c\sigma_c=\tau_c\sigma'_c$, hence $\sigma_c=\sigma'_c$. Thus $\sigma=\sigma'$ and $\tau$ is mono. The other direction seems much more difficult and I doubt there is a simple characterization... –  Bruno Joyal Nov 16 '11 at 6:48
    
I don't think there is a simple characterization. In fact I don't think this statement is valid for any categories $C$ and $D$. I think one needs to put restrictions on $C$ and $D$. For example when $C$ is a small category and $D$ is the category of sets, then one can show that a natural transformation is a monomorphism implies each of its components are monic (i.e. injective). I think you use the Yoneda lemma to show this. What I am looking for is the weakest possible restrictions on $C$ and $D$ for which this statement holds. –  Rankeya Nov 16 '11 at 7:35
    
Okay, so I tagged Algebraic Geometry even though this is not strictly a question of Algebraic Geometry, because I was motivated by the question while trying to solve a problem which has relevance in Algebraic Geometry. Also, I think someone else who has solved the problem that motivated my question might have had a similar sort of question. –  Rankeya Nov 16 '11 at 18:16
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The following mathoverflow question and answer are relevant: mathoverflow.net/questions/52846/… –  Ittay Weiss May 16 '12 at 5:28
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1 Answer

up vote 3 down vote accepted
+50

I claim that, if $\mathcal{D}$ admits arbitrary coproducts, then for any $c \in \mathcal{C}, d \in \mathcal{D}$, there is an object $h_{c,d} \in \mathrm{Fun}(\mathcal{C}, \mathcal{D})$ such that $\hom(h_{c,d}, G) = \hom(d, Gc)$. Given such an object $h_{c,d}$, it follows that a monomorphism of functors $G \to G'$ must have the property that $\hom(d, Gc) \to \hom(d, G'c)$ is injective for each pair of objects $d, c$, meaning that you have a pointwise monomorphism (i.e. $Gc \to G'c$ is a monomorphism). To get such a functor, you imitate the Yoneda lemma: let $h_{c,d}(c') = \hom(c, c') \times d$ (since $\mathcal{D}$ admits arbitrary coproducts, it is tensored over sets and the preceding makes sense). Now given an arbitrary functor $G$ and given $d \to Gc$, one extends to $h_{c,d}(\cdot) = \hom(c, \cdot) \times d \to G$ in the natural way. I won't write out the details, but if I'm not being silly then what I claimed is true.

Now to the question on epimorphisms. Ideally, I want to claim that there is a functor $h'_{c,d}$ such that mapping into it from another functor $G$ is the same as giving a map $Gc \to d$. In other words, if we take the functor from $\mathrm{Fun}(\mathcal{C}, \mathcal{D})$ to $\mathcal{D}$ given by "evaluation" at $c$, then there is a right adjoint to this evaluation (previously the $h_{c,d}$ construction showed that there was a left adjoint). If there is such a right adjoint, then we're done.

In fact, let's state the whole argument more categorically:

A left adjoint preserves epimorphisms. A right adjoint preserves monomorphisms.

This is not hard at all (if $(F, G)$ is an adjunction and $c \to c'$ is an epimorphism, then we have to show that $Fc \to Fc'$ is one too; that is, $\hom(Fc', d) \to \hom(Fc, d)$ is always an injection of sets. But this is $\hom(c,' Gd) \to \hom(c, Gd)$).

So the point of the above discussion is that if $\mathcal{D}$ admits coproducts, then the evaluation functor at $c$, $\mathrm{Fun}(\mathcal{C}, \mathcal{D}) \to \mathcal{D}$ is a right adjoint. Thus it sends "global" monomorphisms of functors to monomorphisms by evaluation at each $c \in \mathcal{C}$, and thus "global" monomorphisms are "pointwise" epimorphisms. (The converse, as you observed, is easy.) Now we can try to the same thing for epimorphisms. As above, the main point is to construct a right adjoint to the evaluation functor. This can be done whenever $\mathcal{D}$ has all products. Namely, given $c ,d$, we consider the functor $c \mapsto d^{\hom(c, c')}$. One can check that this has the analogous universal property to $h_{c,d}$ above, and that it defines a left adjoint to the evaluation functor. So, if the category $\mathcal{D}$ has arbitrary products, then epimorphisms of natural transformations are pointwise epimorphisms.

One can ask the more general question: if $F: \mathcal{C} \to \mathcal{C}'$ is a functor, when is $F_*: \mathrm{Fun}(\mathcal{C}', \mathcal{D}) \to \mathrm{Fun}(\mathcal{C}, \mathcal{D})$ a left or right adjoint? The answer is that if $\mathcal{D}$ is cocomplete and complete, it's both. Since the "lower star" preserves limits and colimits (which are calculated pointwise, after all). Some specific examples are the lower star and upper star from pre(!)sheaf theory, and the skeleton and coskeleton functors on simplicial sets.

I'm not sure what the answer to your question is without such assumptions on $\mathcal{D}$.

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Okay. It is possible I am making a conceptual error, but a sheaf of sets on a topological space $X$ is a contravariant functor from the category $C_X$, whose elements are open subsets of $X$ and whose only maps are inclusion maps, to the category $Sets$ of sets. Now, $Sets$ has arbitrary products right? But, why is it true then that you can have a morphism of sheaves of sets on $X$ which is epi, but whose components are not all epi as in this question math.stackexchange.com/questions/58306/… ? –  Rankeya Nov 20 '11 at 11:53
    
A morphism of sheaves of sets on $X$ is a natural transformation right? –  Rankeya Nov 20 '11 at 11:55
    
Apart from the point of doubt, which I hope you will clarify, your answer is precisely what I was looking for. –  Rankeya Nov 20 '11 at 12:22
    
@Rankeya: A sheaf is not simply a contravariant functor; it's a contravariant functor which satisfies a descent (gluing) condition. A presheaf is simply a contravariant functor. An epimorphism of presheaves is a pointwise epimorphism. (Sheaves are a "reflective subcategory" of presheaves and the inclusion functor is a right adjoint, but not a left adjoint.) –  Akhil Mathew Nov 20 '11 at 14:20
    
Thanks. I was making an obvious error in regarding morphisms of sheaves and of presheaves as being the same. –  Rankeya Nov 20 '11 at 15:24
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