The Stacks project

Lemma 12.5.4. Let $f : x \to y$ be a morphism in an abelian category $\mathcal{A}$. Then

  1. $f$ is injective if and only if $f$ is a monomorphism, and

  2. $f$ is surjective if and only if $f$ is an epimorphism.

Proof. Proof of (1). Recall that $\mathop{\mathrm{Ker}}(f)$ is an object representing the functor sending $z$ to $\mathop{\mathrm{Ker}}(\mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(z, x) \to \mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(z, y))$, see Definition 12.3.9. Thus $\mathop{\mathrm{Ker}}(f)$ is $0$ if and only if $\mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(z, x) \to \mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(z, y)$ is injective for all $z$ if and only if $f$ is a monomorphism. The proof of (2) is similar. $\square$

Comments (1)

Comment #366 by Fan on

This Lemma actually does not require the full strength of Abelian. Pre-Abelian (ker and coker exists) suffices:

(1) Assume . To show is a monomorphism it suffices to show that if and then . This follows from the fact that factors through .

Conversely, assume is a monomorphism. Then the canonical map and the zero map become identical when composed with on the left, so the canonical map is zero. Then the zero map can be factored through in two ways: via the identity map and the zero map, so , and hence .

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  • 8 comment(s) on Section 12.5: Abelian categories

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