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Let $M,N,K$ be three monoids (or even groups, if you like) and let $N \to K$ be an injective homomorphism. Then, the induced morphism $M \sqcup N \to M \sqcup K$ is also injective.

This is easy to prove using the normal form for the elements in the coproduct $M \sqcup N$ (these are products of elements of $|M| \setminus \{1\}$ or $|N| \setminus \{1\}$, alternating). Is there also a proof which merely relies on the universal property of a coproduct? I ask this just out of curiosity. Of course, not every category has the property that monomorphisms are stable under coproducts. So the proof will involve specific monoids and applies the universal property to homomorphisms into them, I guess.

Notice: Surjective homomorphisms of monoids are precisely the regular epimorphisms, and these are stable under coproducts, since colimits commute with colimits. Since limits do not commute with colimits, the question about monomorphisms is not formal.

Notice: The universal property implies that the coproduct inclusions $M \xrightarrow{i_M} M \sqcup N \xleftarrow{i_N} N$ are injective, in fact split monomorphisms.

Martin Brandenburg
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    If a proof only relied on the universal property of the coproduct, wouldn’t that proof be reusable to prove that all coproducts preserve monomorphisms? – k.stm Jan 23 '16 at 07:56
  • @k.stm: See my edit. – Martin Brandenburg Jan 23 '16 at 07:58
  • Ah, so you want to avoid a set-theoretic construction of the monoid coproduct, is that it? – k.stm Jan 23 '16 at 08:00
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    @k.stm: I want to avoid the usual description of the underlying set of the coproduct of monoids. – Martin Brandenburg Jan 23 '16 at 08:03
  • I mean, the explicit construction of the free product is exactly a "specific monoid" you can use...what sort of "specific monoid" would you find more satisfying than that? – Eric Wofsey Jan 23 '16 at 08:04
  • @EricWotsey One point is, though, that it strongly depends on the notion of a set and that it has to be done for every pair or triple of $M$, $N$ or $M$, $N$, $K$. Whereas, if you could do the proof only using a really specific monoid, the proof can be seen in a rather self-contained axiomatic framework of the category of monoids (which is always nicer and hopefully even simpler). – k.stm Jan 23 '16 at 08:08
  • @EricWofsey: Yes, you may use this monoid, but your reasoning probably uses the normal form for the elements of the underlying set, and my question is exactly if one can avoid this. – Martin Brandenburg Jan 23 '16 at 08:21
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    One might hope for a categorical proof along the lines of "in every category satisfying [some categorical condition], coproducts will preserve monomorphisms," then verifying that monoids satisfy this condition. – Qiaochu Yuan Jan 23 '16 at 09:26

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