Does free functor preserve monomorphism?

Solution 1:

Almost all monomorphisms in $\mathbf{Set}$ are split (hence are preserved by any functor whatsoever), the exceptions being maps with empty domain. So it's just a question of what happens with those maps.

We consider an adjunction $$F \dashv U : \mathcal{C} \to \mathbf{Set}$$ where the "forgetful" functor $U : \mathcal{C} \to \mathbf{Set}$ reflects monomorphisms. (If $U$ is faithful, then $U$ reflects monomorphisms. In particular this holds when $U$ is monadic, e.g. $U : R\mathbf{-Mod} \to \mathbf{Set}$, $U : \mathbf{CRing} \to \mathbf{Set}$ etc.) Now consider $U F(\emptyset \to X)$ for a non-empty set $X$. There are two cases:

• If $U F \emptyset = \emptyset$, then $U F (\emptyset \to X)$ is an injective map (trivially), so $F (\emptyset \to X)$ is a monomorphism in $\mathcal{C}$.
• If $U F \emptyset \ne \emptyset$, then there exists a map $X \to U F \emptyset$, hence a morphism $F X \to F \emptyset$ by adjoint transposition. But $F \emptyset$ is the initial object in $\mathcal{C}$, so there is a unique morphism $F \emptyset \to F X$; thus the composite $F \emptyset \to F X \to F \emptyset$ must be the identity, i.e. the morphism $F (\emptyset \to X)$ is split monic.

So $F : \mathbf{Set} \to \mathcal{C}$ indeed preserves all monomorphisms.

Solution 2:

Yes, because monics in $\mathcal{Set}$ are split. Edit: This only holds with non-empty sets. That leaves this answer morally correct but not necessarily technically correct.

A monomorphism in set is an inclusion, and thus will have a left inverse (the inverse image of a point will either be a singleton or empty, and we can define the inverse arbitrarily when the inverse image is empty). Explicitly, let $m: X\to Y$ be monic. Then we can produce $r:Y\to X$ such that $r \circ m = \operatorname{id}_X$. Conversely, any map with a left inverse can be canceled on the left, so any such map must be monic. We have shown that in the category $\mathcal{Set}$, every monic is split monic, and that in an arbitrary category, a split monic is monic.

If $F$ is a functor $m$ is a split monic, then $F(r\circ m)=F(r)\circ F(m) = \operatorname{Id}_{F(X)}$, and so split monics are preserved by arbitrary functors (covariant) functors. In particular, free functors preserve monics.

Solution 3:

Let $U : C \to \mathsf{Set}$ be a functor and $F : \mathsf{Set} \to C$ left adjoint to $U$. The question is if $F$ preserves monomorphisms.

If $f : X \to Y$ is a monomorphism in $\mathsf{Set}$ and $X \neq \emptyset$, then $f$ is a split monomorphism (just take preimages on $f(X)$, and on $Y \setminus f(X)$ map everything to a chosen element of $X$), hence also $F(f)$. Now let $X=\emptyset$. If $Y=\emptyset$, then $f$ is an isomorphism, and hence also $F(f)$. Otherwise $f$ factors over $\{\star\}$, and $\{\star\} \to Y$ is mapped to a monomorphism. Hence, the question reduces to the case $Y=\{\star\}$.

Is $F(\emptyset) \to F(\{\star\})$ a monomorphism? Note that $F(\emptyset)$ is the initial object of $C$ and $F(\{\star\})$ is the free $C$-object on one generator. If $U$ is monadic, then this turns out to be true, see the argument on p.89-90 in Linton, Coequalizers in categories of algebras. (Edit: As Zhen Lin points out, the same argument works when $U$ reflects monomorphisms). If $C$ is a category of algebraic structures in the sense of universal algebra, and $U$ is the forgetful functor, then $U$ is monadic. For this Beck's monadicity criterion is quite useful. This answers the question in the affirmative in a lot of cases.

It also holds in a lot of other cases, for example $\mathsf{Top}$ (here $F$ assigns to a set the corresponding discrete space).

I have tried to find counterexamples, but didn't succeed so far.