Counterexample: functor does not preserve monomorphisms I need to show that functors need not preserves mono's and epi's. For epi's, I have as counterexample the forgetful functor $F : \mathbf{Ring} \to \mathbf{Set}$. We have that $f: \mathbb{Z} \hookrightarrow \mathbb{Q}$ is an epi in $\mathbf{Ring}$, but it is not an epi in $\mathbf{Set}$, since it is not surjective. Does anyone know a simple counterexample for mono's? 
 A: The functor $F^{op}$ does not preserve monomorphisms.
A: The reason why it is more "difficult" to find a non mono-preserving functor in nature is that forgetful functors $\mathbf C \to \mathsf{Set}$ preserve (even create) finite limits when $\mathbf C$ is a category of algebraic structure. And these are usually our toy examples.
To find non mono-preserving functor, one need to keep clear of that situation. In order to do that, Arnaud D.'s answer focus on a free functor rather than a forgetful one. I propose to simply quit the algebraic world: consider the functor $\pi_0 : \mathsf{Top} \to \mathsf{Set}$ that associates to each topological space $X$ the set of its connected components. Then the mono $\{0,1\} \hookrightarrow [0,1]$ that includes the (discrete space of) endpoints into the interval gets mapped through $\pi_0$ to the non injective function $\{0,1\} \to \{\ast\}$.
A: Consider the category $C_1$ on the $2$ objects $\{1,2\}$ with identity morphisms, and a single morphism $1\to 2$. That morphism is mono.
Now consider the category $C_2$ on three objects $\{a,b,c\}$ with identity morphisms, as well as the four morphisms
$$
a\to b\\
a\to b\\
b\to c\\
a\to c
$$
The morphism $b\to c$ is not mono.
There is a functor $F$ from $C_1$ to $C_2$ given by $F(1)=b, F(2)=c$. It does not preserve the monomorphism.
A: For an example "in nature" : the abelianization functor $\mathbf{Grp}\to \mathbf{Ab}$ does not preserve monomorphisms. For example, the abelianization of the alternating group $A_n$ is trivial for $n\geq 5$, while the abelianization of a cyclic group $C_m$ is $C_m$ (since it's abelian). Every element in $A_n$ defines a cyclic subgroup, hence a monomorphism $C_m\to A_n$, but the abelianization of this morphism is $C_m\to 0$, which is not a monomorphism.
