$f=\text{ker}(g)$ for a short exact sequence in an abelian category. I am a rank amateur when it comes to category theory, and have gotten myself stuck on what should be a simple exercise.
A short exact sequence in an (abelian), category is a sequence of morphisms:
$$
0\rightarrow A\xrightarrow{f}B\xrightarrow{g}C\rightarrow0
$$
such that $\text{ker}(f)=0$$, \text{im}(f)=\text{ker}(g)$, and $\text{im}(g)=0$.
I want to show that for such a sequence, $f=\text{ker}(g)$.
The obvious way to proceed is to show that $f$ satisfies the universal property of the kernel, and then appeal to uniqueness.
To do so, I must first show that $g\circ f=0$, and then show that given a morphism $e:A'\to B$ with $g\circ e=0$, there exists a unique morphism $e':A'\to A$ such that $e=f\circ e'$.
I have been able to use the epi-mono factorisation that abelian-ness gives us to show that $g\circ f=0$, but I have been unable to define the required map. Any hints or help would be much appreciated.
EDIT:
Thanks to the comment of D. Brogan, I have been able to make a little more progress.
Using the categorical definition of image and the fact that the image and coimage are isomorphic in an abelian category, we have:
$$
\text{im}(f)=\text{ker}(\text{coker}(f))\cong\text{coker}(\text{ker}(f))=\text{coker(0)}
$$
So it will be enough to show that $\text{coker}(0)=f$. I have tried to appeal to the universal property of cokernels to do so, but have been unable to conclude.
 A: $f$ is monic (since $\ker(f) = 0$) and this implies $f = \text{im}(f)$ (since $f = \ker(\text{coker}(f))$)
A: There is something kind of subtle here that I didn't realize when I made my comment, so I'll flesh it out right now for my own sake. When defining the image of a morphism $f:A\to B$ in an abelian category, we usually say $\mathrm{im}(f)$ is defined to be either $\mathrm{coker}(\mathrm{ker}(f))$ or $\mathrm{ker}(\mathrm{coker}(f))$ and we leave it to the exercises to show that these two definitions are the same. The thing is that when we say $\mathrm{ker}(f)$ for example, we are not referring to an object but instead another morphism $K\to A$ with a universal property. Similarly $\mathrm{coker}(f)$ is really a morphism $B\to C.$ So then what do we mean when we say $\mathrm{coker}(\mathrm{ker}(f))\cong \mathrm{ker}(\mathrm{coker}(f))?$ Since one is a morphism out of $A$ and the other is a morphism into $B,$ this isn't immediately clear. The answer is that the objects in question are isomorphic and the morphisms are the ones induced by the universal property of the other.
For example, denote the object $\mathrm{coker}(\mathrm{ker}(f))$ by $C$ and the object $\mathrm{ker}(\mathrm{coker}(f))$ by $K.$ Then $A\to C$ is the morphism attached to the object $C.$ By the universal property of the cokernal there is a unique map $C\to B$ since $\mathrm{\ker}(f)\to A\to B$ is zero. This map $C\to B$ is isomorphic to $K\to B.$ Similarly we can get a map $A\to K$ which is isomorphic to $A\to C.$
Okay now that all that is out of the way, here's what to do. Since $\mathrm{ker}(f)=0,$ the identity $1_A:A\to A$ satisfies the universal property of $\mathrm{coker}(\mathrm{ker}(f))=\mathrm{coker(0\to A)}.$ Since $0\to A\to B$ is zero, by the universal property of the cokernal, there is a unique map $A\to B$ making the diagram commute. But this is just $f:A\to B$ since our cokernal is just the identity. As stated in the middle paragraph, after composing with an isomorphism this is exactly the map $\mathrm{ker}(\mathrm{coker}(f))\to B.$ Hence $f=\mathrm{im}(f).$
