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. $\square$

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 $\ker(f) = 0$. To show $f$ is a monomorphism it suffices to show that if $g: z \to x$ and $f \cdot g = 0$ then $g = 0$. This follows from the fact that $g$ factors through $\ker(f) = 0$.

Conversely, assume $f$ is a monomorphism. Then the canonical map $\ker(f) \to x$ and the zero map $0: \ker(f) \to x$ become identical when composed with $f$ on the left, so the canonical map is zero. Then the zero map $\ker(f) \to x$ can be factored through $\ker(f)$ in two ways: via the identity map and the zero map, so $1_{\ker(f)} = 0$, and hence $\ker(f) = 0$.

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