Monomorphisms in functor categories

Question

Let’s assume $C$ is a small category and $D$ is a locally small category. Now, let $F, G$ be two objects in $D^C$ and let $m:F\longrightarrow G$ be a morphism in $D^C$. Then, I guess we have m is a monomorphism if and only if for each $c\in Ob(C)$, $m_c: F(c)\longrightarrow G(c)$ is a monomorphism. One direction is easy to prove, but I do not know the other one is true or not. If it is true, then what is the proof? If it is not true, then what about if D be an abelian category? Mainly, I was trying to show that it is true if D is an abelian category, but I couldn’t.

Answer 1

This result is not true in general. Consider the category $C$ generated by the graph with edges $x\rightrightarrows y\to z,w\to y$, where there is a unique morphism $x\to z$. There is a unique natural transformation between the functors of domain $0\to 1$ with images $w\to y$ and $y\to z$, which I claim is a monomorphism, even though $y\to z$ is not a monomorphism in $C$. The reason is that no functor $F:(0\to 1)\to C$ admitting a natural transformation into $w\to y$ can include $x$ in its image, since there are no maps into $w$ from any object mapping into $x$. Thus $F$ factors through the subcategory $w\to y\to z$ of $C$, in which subcategory our natural transformation does have monomorphic legs.

As was discussed in the comments, the result is true in any abelian category (or even just a category with pullbacks.) It is also true in any category admitting coproducts up to the size of its hom-sets. Roughly, both these conditions guarantee that you never have the situation above where nothing maps to both $x$ and $w$, but I don’t know whether assuming precisely that that never happens is sufficient. For the coproduct condition, the precise argument is that then if $\alpha:F\to G$ and $\alpha_c:F(c)\to G(c)$ is not a monomorphism, then from a witness to non-monomorphicity of $\alpha_c$, $f,g:x\to F(c)$, we can construct a witness $c_!f,c_!g:c_!x\to F$, using the left Kan extension functor $c_!$. This sends $x$ to the functor $y\mapsto \coprod_{\mathrm{Hom}(x,y)} x$, and sends $f$ to the natural transformation such that $(c_!f_y)_k=F(k)\circ f$, for any $k:x\to y$.