Constructing a functor that sends a monic(epic) morphism to a non-monic(epic) morphism

The title says it all. Given two categories $$C$$ and $$D$$, I am asked to construct a funtor $$F$$ such that $$F(f)$$ is not monic (resp. epic) for a monic (resp. epic) morphism $$f\in C$$.

My work: There is a hint that tells me to look at Monoids as a category with one element. I know that a functor from these two categories is an homomorphism from these two monoids and that the morphisms are regarded as elements of the monoid but when it comes to actually craft a functor and a morphism I have no success. I tried to consider $$(\mathbb{Z},.)$$ as a monoid and using as functor (to itself) the trivial homomophism but given a monic morphism $$f$$ isnt $$F(f)$$ always monic? I'm getting this because $$F(f)\circ g_1=F(f)\circ g_2=1.g_1=1.g_2$$.

• Constant functors? – Randall Dec 18 '18 at 2:03
• That's what I thought but either I made a mess in the second part of my attempt or this thought is wrong. – 2ndYearFreshman Dec 18 '18 at 2:08

I think it's much easier to find a counterexample if you look at posets, rather than monoids. Indeed in a poset every arrow is trivially a monomorphism and an epimorphism; so all you have to do is take a functor from a poset to a category, in such a way that not every arrow is sent to a monomorphism/epimorphism. For example, you can define a functor from the poset $${0\leq 1}$$ to the category of sets by sending $$0,1$$ to the set $$\{x,y\}$$, and the unique arrow $$0\to 1$$ to the function $$f$$ defined by $$f(x)=x=f(y)$$, which is neither a mono nor an epi; this is pretty much the smallest existing counterexample.