# Let $\mathcal{A}$ be an Abelian category (as defined in Stacks), then all monomorphisms are kernels.

I'm struggling to prove this statement, using the definitions below (I'm assuming the proof for the statement about epimorphisms is analogous). I know that a morphism $f:x\to y$ is monic if and only if it has zero kernel, however, when drawing the diagram, that gives me maps to the left of $f:x\to y$, while, letting $z=\mathrm{coker}(f)$, presuming $f:x\to y$ to be the kernel of $y\to z$, I want to have that for any $g:w\to y$ s.t. $w\to y\to z = 0$, I want $\exists!\ w\to x$. Unfortunately, $w$ is in the wrong position to use the universal property either of $\ker\ f$ or of $\mathrm{coker}\ f$. How would this be done?

Definitions:

A category $\mathcal{A}$ is said to be abelian if

1. $\forall\ x,y,z\in\mathcal{A}$, $\mathrm{Mor}(x,y)$ is an abelian group, and $\mathrm{Mor}(x,y)\times\mathrm{Mor}(y,z)\xrightarrow{\mathrm{comp}}\mathrm{Mor}(x,z)$ is bilinear (this condition is called preadditivity)
2. $\mathcal{A}$ has a zero object, as well as finite direct sums – using preadditivity, this tells us that we have products and coproducts, which are the same in a preadditive category (this condition + preadditivity is known as additivity)
3. For every morphism $x\xrightarrow{f}y$, $\ker\ f$ and $\mathrm{coker}\ f$ exist. We also define $\mathrm{im}\ f=\ker(\mathrm{coker}\ f)$, $\mathrm{coim}\ f=\mathrm{coker}(\ker\ f)$).
4. For every morphism $x\xrightarrow{f}y$, $\mathrm{im}\ f\cong\mathrm{coim}\ f$.
• The obvious guess is that $f=\ker(\operatorname{coker}f)$, when $f$ is a monomorphism. – egreg Jul 26 '16 at 19:31
• That was my guess. I'm struggling to prove that $f$ is the kernel, though. – Monstrous Moonshine Jul 26 '16 at 19:33

You haven't used the fourth axiom, which is key here (without it the statement is false). If $f : x \to y$ is monic, it has trivial kernel, and so $\text{coim}(f)$ is just $x$. By axiom 4, this is also $\text{im}(f)$. This says precisely that $f$ is the kernel of its cokernel.
For a counterexample with axiom 4 dropped, consider the category of Hausdorff topological vector spaces over $\mathbb{R}$. Here the monomorphisms are the injective maps, but the kernels are the injective maps with closed image. It's a nice exercise to figure out what the image and coimage are here.