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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?

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  • $\begingroup$ One can construct small counterexample categories with, say, 3 objects. Would that be satisfactory? Or do you specifically want examples with "known" categories? $\endgroup$ – Arthur Apr 9 at 15:00
  • $\begingroup$ 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. $\endgroup$ – Sigurd Apr 9 at 15:05
  • $\begingroup$ 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. $\endgroup$ – Maik Pickl Apr 9 at 15:10
  • $\begingroup$ 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. $\endgroup$ – user647486 Apr 9 at 15:13
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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.

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

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    $\begingroup$ I like that. Simple and one of these "why haven't I thought about that" things. $\endgroup$ – Maik Pickl Apr 9 at 15:13
  • $\begingroup$ What is the definition of $F^{op}$? $\endgroup$ – Sigurd Apr 9 at 15:16
  • $\begingroup$ @Sigurd It is the same as $F$, but between dual (opposite) categories. $\endgroup$ – Oskar Apr 9 at 15:18
  • $\begingroup$ Ok I see, thanks. $\endgroup$ – Sigurd Apr 9 at 15:19
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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 $\{0,1\} \to \{\ast\}$.

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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.

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