Monomorphism under a functor

Question

Let $\mathcal{C}$ be a category. Let $X,Y$ be object in $\mathcal{C}$ and $f \in Hom_{\mathcal{C}}(X,Y)$. I want to know the condition on $F$ for which $F(f)$ is a monomorphism.

It's clear that if $F$ is fully faithful and dense then $F(f)$ is a monomorphism.

If I consider $i:\mathbb{Z} \to \mathbb{Q}$ the canonical map, this is an epimorphism in $Ring$ but it's not an epimorphism in $Set$, and so it is a monomorphism in $Ring^{op}$ but not in $Set^{op}$. The forgetful functor $U^{op}: Ring^{op} \to Set^{op}$ sends a monomorphism to a morphism that it's not a monomorphism.

So it's not sufficient $F$ is faithful.

Now the question is: is it sufficient that $F$ is fully faithful? I think the answer is no but I don't find a counterexample.

Answer 1

It is unfortunately not sufficient, as shown by the inclusion of the category generated by 2 objects and one arrow $A \longrightarrow B$ into the category with three objects $C \rightrightarrows A \longrightarrow B$ (where both composites from $C \to B$ are equal).

One condition which is often easier to satisfy in practice is that $F$ preserves finite limits. The condition that a morphism be a monomorphism can be encoded as a certain square being a pullback and so preserving finite limits suffices to preserve monomorphisms.

Answer 2

$\require{AMScd}$Being a monomorphism is a finite limit property: a morphism $m : E\to B$ in a category $\cal C$ is a monomorphism if and only if the square $$\begin{CD} E @= E \\ @| @VVmV\\ E @>>m> B \end{CD}$$ is a pullback (a "fibered product" if you like geometry). Then, since every functor preserves identities, the image of such a pullback under a functor $F : \cal C\to D$ is $$\begin{CD} FE @= FE \\ @| @VVFmV\\ FE @>>Fm> FB \end{CD}$$ and $Fm$ is a monomorphism if and only if this square remains a pullback.