Does free functor preserve monomorphism? The free functor is left adjoint to the forgetful functor so it preserves epimorphism. In the category of modules and algebras, it also preserves monomorphisms (the free functors being free modules and polynomial rings, respectively). Is it generally true that free functors preserve monomorphisms?
Edit: A free functor is a functor from Set to a concrete category C which assigns each set S to an object in C which is free over S. For example, for A-modules, it assigns a set S to a free A-module with basis S. More detailed definition can be found in Wikipedia.
 A: 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.  
A: 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.
A: 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.
