Lemma 12.5.4 (010C)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 12: Homological Algebra
Section 12.5: Abelian categories
Lemma 12.5.4 (cite)

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

$f$ is injective if and only if $f$ is a monomorphism,
$f$ is surjective if and only if $f$ is an epimorphism, and
$f$ is an isomorphism if and only if $f$ is injective and surjective.

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. For the proof of (3) note that an isomorphism is both a monomorphism and epimorphism, which by (1), (2) proves $f$ is injective and surjective. If $f$ is both injective and surjective, then $x = \mathop{\mathrm{Coim}}(f)$ and $y = \mathop{\mathrm{Im}}(f)$ whence $f$ is an isomorphism. $\square$

Comments (1)

Comment #366 by Fan on November 30, 2013 at 23:06

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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