# Monomorphisms and epimorphisms in abelian categories

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.

• 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 $\text{im}(f) \cong Y$. Dec 26, 2022 at 6:58
• 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. Dec 26, 2022 at 13:08

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.