Counterexample: functor does not preserve monomorphisms

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

An example from algebraic topology.

For $n\in \mathbb{Z}_{\geq 0}$, consider the $n$-th singular homology functor $H_n\colon \mathsf{Top}\rightarrow \mathsf{Ab}$ which maps a topological space $X$ to $H_n(X)$, the $n$-th homology group of the singular chain complex on $X$. On morphisms it acts by post-composition on representatives, i.e. $H_n(f)([c])=[f\circ c]$.

In general, this functor does not preserve monomorphisms. Consider for instance the inclusion $\iota\colon \mathbb{S}^1\rightarrow \mathbb{D}^2$, where we include the $1$-circle as the boundary of the $2$-disk. Now, $H_1(\mathbb{S}^1)\cong \mathbb{Z}$ and $H_1(\mathbb{D}^2)\cong 0$, by Hurewicz theorem. In particular, the group homomorphism $H_1(\iota)$ isn't monic.

Put differently: If the singular homology functor did preserve monomorphisms, the connecting homomorphism in the long exact sequence of a pair of spaces would be trivial. Given a subspace $A$ of a topological space $X$, the boundary of any relative $n$-cycle $c\in H_n(X,A)$ would be the boundary of an $n$-chain in $A$. We would obtain a short exact sequence $0\rightarrow H_n(A)\rightarrow H_n(X)\rightarrow H_n(X,A)\rightarrow 0$ for any $n\in \mathbb{Z}_{\geq 0}$. This is false in general.

Answer 6

By definition, any additive functor $F\colon \mathcal{C}\rightarrow \mathcal{D}$ between abelian categories that is not left exact. For a prototypical example consider, with $R$ a unital ring, the functor $-\otimes_R M\colon \mathsf{R}\text{-mod} \rightarrow \mathsf{R}\text{-mod}$ for any non-flat $R$-module $M$. For instance, you could set $R=\mathbb{Z}$ and $M=\mathbb{Z}/n\mathbb{Z}$ for $n\geq 2$. Then the monomorphism $\mathbb{Z}\xrightarrow{n\cdot} \mathbb{Z}$ is sent to the zero morphism $\mathbb{Z}/n\mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$.