Let $\mathsf{C}$ be a an abelian category. Let $f \colon X \rightarrow Y$ be a morphism in $\mathsf{C}$. I am looking at the following two statements:

  1. The morphism $f$ is a monomorphism if and only if $\operatorname{ker}(f)\cong 0$.
  2. The morphism $f$ is an epimorphism if and only if $\operatorname{im}(f)\cong Y$.

The forward direction of 1. is easily proven in any category with kernels. By dropping the requirement that $\mathsf{C}$ is abelian (but leaving some mild assumptions on $\mathsf{C}$) the backward implication of 1. becomes false (c.f. If the kernel (cokernel) of a morphism $f$ is trivial, then is $f$ injective (surjective), or mono (epi)?). Is the backward implication of 1. true in any abelian category? Is 2. true in any abelian category? I would also be happy with a reference.

  • 2
    $\begingroup$ Yes, both statements are true in any abelian category. Once you have statement 1, its formal dual is that $f$ is an epimorphism iff $\text{coker}(f) = 0$. Now show that this condition is equivalent to $\text{im}(f) \cong Y$. $\endgroup$ Dec 26, 2022 at 6:58
  • 3
    $\begingroup$ Note that the second statement must not be interpreted as the fact that $\operatorname{im}(f)$ and $Y$ are isomorphic, but rather that the canonical morphism $\operatorname{im}(f)\to Y$ is an isomorphism. $\endgroup$ Dec 26, 2022 at 13:08

1 Answer 1


Suppose that $f$ is a monomorphism. It follows for the canonical morphism $k \colon \ker(f) \to X$ from the equalities $f ∘ k = 0 = f ∘ 0$ that $k = 0$. But $k$ is itself a monomorphism (by the universal property of the kernel), whence it further follows from $\newcommand{\id}{\mathrm{id}} k ∘ \id_{\ker(f)} = 0 ∘ \id_{\ker(f)} = 0 = k ∘ 0$ that $\mathrm{id}_{\ker(f)} = 0$. This shows that $\ker(f) = 0$.

Suppose conversely that $\ker(f) = 0$. If $g, h \colon K \to X$ are two morphisms with $f ∘ g = f ∘ h$, then $f ∘ (g - h) = 0$ and it follows that $g - h$ factors through $\ker(f)$. Since $\ker(f) = 0$, this means that $g - h = 0$ and thus $g = h$. This shows that $f$ is a monomorphisms.

This shows the first statement. We note that by duality, we have also proven the following statement:

  1. The morphism $f$ is an epimorphism if and only if $\newcommand{\coker}{\operatorname{coker}} \coker(f) = 0$.

The image of $f$ is defined as the kernel of its cokernel. More explicitly, $\newcommand{\im}{\operatorname{im}} i \colon \im(f) \to Y$ is the kernel of $c \colon Y \to \coker(f)$.

Let us recall two observations about kernels and equalizers:

  • For every morphism $g \colon A \to B$ in a category with zero morphisms, the kernel $\ker(g) \to A$ is precisely the equalizer of $g$ and the zero morphism.

  • Given any two morphisms $g, h \colon A \to B$ in an arbitrary category, we have $g = h$ if and only if the equalizer $\mathrm{eq}(g, h)$ exists and $\mathrm{eq}(g, h) \to A$ is an isomorphism.

We now find that $f$ is an epimorphism if and only if $\coker(f) = 0$ (by statement 3), if and only if $c \colon Y \to \coker(f)$ is the zero morphism (because $c$ is an epimorphism, dually to the argumentation for $\ker(f)$ in the first paragraph), if and only if $\ker(c) \to Y$ is an isomorphism (by the above two observations), if and only if $i \colon \im(f) \to Y$ is an isomorphism (by definition of the image).

This shows the second statement.


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