# 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?

• One can construct small counterexample categories with, say, 3 objects. Would that be satisfactory? Or do you specifically want examples with "known" categories? – Arthur Apr 9 at 15:00
• Yes that would be satisfactory. An example of known categories is also fine, but only if it's simple like the one above, otherwise it will be inaccessible. – Sigurd Apr 9 at 15:05
• The Wikipedia page about monomorphisms has an example of a monomorphism in the category of divisible groups that is not injective. Just apply the forgetful functor to sets. – Maik Pickl Apr 9 at 15:10
• You can take in any category with an object $X$ that has monomorphism $m$ to some other object and add an arrow $e:X\to X$ such that $f\circ e=f$ for all morphisms $f$ going from $X$ except $X$'s identity $1_X$. Define $e\circ 1_X=1_X\circ e=e$. The monomorphism $m$ wouldn't distinguish $e$ from $1_X$ anymore. The function that includes the original category into the modified one wouldn't preserve the property of being a monomorphism. – user647486 Apr 9 at 15:13

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.

The functor $$F^{op}$$ does not preserve monomorphisms.

• I like that. Simple and one of these "why haven't I thought about that" things. – Maik Pickl Apr 9 at 15:13
• What is the definition of $F^{op}$? – Sigurd Apr 9 at 15:16
• @Sigurd It is the same as $F$, but between dual (opposite) categories. – Oskar Apr 9 at 15:18
• Ok I see, thanks. – Sigurd Apr 9 at 15:19

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 $$\{0,1\} \to \{\ast\}$$.

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.