# Monomorphisms in a concrete category

Let $\mathcal{C}$ be a concrete category, i.e., a category which admits a faithful functor $C:\mathcal{C}\rightarrow \mathsf{Set}$.

It is certainly not the case that $f$ a mono in $\mathcal{C}$ implies that $C(f)$ is a mono (i.e. injective) in $\mathsf{Set}$. However, given $\mathcal{C}$ concrete, is there some concretization $C:\mathcal{C}\rightarrow \mathsf{Set}$ such that $C(f)$ is injective for every mono $f$? If this were the case, to what degree would such a concretization be unique?

The example which I have in mind is the category with two objects $A,B$ and one non-trivial arrow $f:A\rightarrow B$. For any $m,n\ \mathbb{N}$, any choice of $m$ element set $C(A)$, any choice of $n$ element set $C(B)$, and any choice of map $C(f):C(A)\rightarrow C(B)$, $C$ will be a concretization. It is clear that in general $C(f)$ will fail to be injective; however, it is also clear that there are many choices for which it will be. This suggests that the existence result might be true, but that in general such a choice of $C$ will be highly non-unique.

The dual question for epis I don't even bother asking because this fails even for canonical concretizations (e.g. the inclusion $\mathbb{Z}\hookrightarrow \mathbb{Q}$ is an epi in $\mathsf{Ring}$).

• +1. But I don't agree with "This suggests that the existence result might be true". You have looked at a small toy example. Arbitrary categories are very far from that. Actually I doubt that every concrete category has a mono-preserving forgetful functor to sets, but it could be very hard to disprove the existence of such a functor. I would start to think about some specific examples first before asking the general question. Consider the category $\mathsf{Div}$ of divisible abelian groups. It has the monomorphism $\mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$ which is not injective. How to concretize? – Martin Brandenburg Nov 2 '14 at 8:16
• If $\mathcal{C}$ is small then there is certainly a concretisation with the desired property: $C \mapsto \coprod_{C'} \mathrm{Hom}(C', C)$. More generally it suffices to have a small separating family. – Zhen Lin Nov 2 '14 at 10:56
• Note that your required property is true if you can find a concretization with a left adjoint, that is a free construction. (Right adjoints preserve limits, hence monos.) It arises quite often with categories of some sort of algebraic objects. – Pece Nov 2 '14 at 14:34
• @MartinBrandenburg I agree. This is because, in addition to this artificial example, I also had in mind a lot of concrete examples like $\mathsf{Set}$, $\mathsf{Top}$, $\mathsf{Ring}$, $\mathsf{Grp}$, etc. Yesterday, when I went reply to this part of your comment, I went to double-check that I wasn't mistaken about any of these example, and that's when I came across the divisible abelian groups counterexample . . . (cont.) – Jonathan Gleason Nov 3 '14 at 20:27
• @MartinBrandenburg . . . Realizing that my idea that "Monomorphisms in canonically concrete categories are injective." was just plain wrong, I wrote this example into an answer without even finishing reading your comment! How silly of me. My apologies =P – Jonathan Gleason Nov 3 '14 at 20:28