a map in $R$-Mod is injective iff it is a monomorphism

Question

I would like to see a detailed argument of this fact:

a map in $R$-Mod is injective iff it is a monomorphism

Answer 1

Monomorphism are preserved by right adjoint functors and reflected by faithful functors. The forgetful functor $\mathsf{Mod}_R \rightarrow \mathsf{Set}$ satisfies both properties.

Edit the down-to-earth way is to note that a monomorphism $f:M \rightarrow N$ in $\mathsf{Mod}_R$ is precisely a morphism satisfying $\ker f = 0$ and that this is equivalent to $f$ being injective. Indeed applying the monomorphism property to the two morphisms $R \overset{0}{\rightarrow} M$ and $R \overset{m}{\rightarrow} M, 1 \mapsto m$ for any element $m \in \ker f$ shows $\ker f = 0$. As $f$ not injective implies $\ker f \neq 0$ by taking the difference of two distinct elements mapping to the same, we find that monomorphisms are injective. The converse direction feels cleanest by using the faithful argument, but can be done in a similarly elementary way.