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In this question it was asked what relation, if any, exists between the notions of a monomorphism in a general category $\mathcal C$ and an injective map. Of course the question only makes sense if $\mathcal C$ is a concretizable category. If $\pi:\mathcal C\to Set$ is a faithful functor, we may regard a morphism $f:X\to Y$ of objects of $\mathcal C$ as an injection (relative to $\pi$) if $\pi(f)$ is injective in $Set$.

One then notices that this notion depends very much on the choice of concretization, and thus that this notion is not an intrinsic property of the category itself.

However, Emil Jerabek noticed that it is natural to ask the following: given a concretizable category, does it admit a concretization which preserves monomorphisms?

Apologies if this question is trivial, hopelessly too general, or of no interest to category theorists.

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An easy sufficient condition for monomorphism-preserving concretizability is the existence of a generating set (as opposed to a proper class). A generating set is a set $\mathcal S$ of objects such that, for any distinct parallel morphisms $f,g:A\to B$, there is a morphism $h:S\to A$ with $S\in\mathcal S$ and $fh\neq gh$. This says that the coproduct of the representable functors Hom$(S,-)$, over all $S\in\mathcal S$, is a faithful set-valued functor, i.e., a concretization. As a coproduct of representable functors, it preserves monomorphisms.

I would not expect this sufficient condition to be necessary. So I don't claim that this is a complete answer --- it just takes care of an easy situation, which happens to cover a lot of categories.

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    $\begingroup$ For what it's worth, it's straightforward to exhibit a monomorphism-preserving-concretizable category with no generating set; just take any groupoid which is not equivalent to a small groupoid, e.g. the category of sets and isomorphisms. $\endgroup$ Apr 11, 2012 at 17:46
  • $\begingroup$ Also, if this "forgetful" functor preserves wide pullbacks and the category is sufficiently complete, then it has a generating set. $\endgroup$ Jul 31, 2019 at 16:17

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