Difficult category theory: kernels In a category with a zero object ($0$ = initial and terminal) and zero morphisms are unique $A \to 0 \to B$ for every $A,B$, define the kernel of a map as the equalizer of $(f,0)$ and cokernel dually. I try to prove that if $f = \ker(g)$ then $f = \ker(\operatorname{coker}(f))$. So set $c = \operatorname{coker}(f)$. We have the diagrams $$A \xrightarrow{f} B \xrightarrow{g} C$$ $$A' \xrightarrow{\ker c} B \xrightarrow{c} C'$$ 1. $f$ equalizes $(g,0)$: $g f = 0$ 2. $c$ coequalizes $(f,0)$: $c f = 0$ 3. $\ker c$ equalizes $(c,0)$: $c \ker(c) = 0$ By (2) and $0 = 0 f$ we get that $f$ claims to equalize $(c,0)$ so there exists a universal map $u : A \to A'$ such that $\ker(c) u = f$, therefore $f$ equalizes $(c,0)$ and by uniqueness of equalizers $f = \ker(c)$. But I don't know if this is right, am I making some mistakes and can anyone show me a simple proof instead? Thank you

What you've written is mostly correct, but I'm not very convinced by your 'uniqueness of equalizers' argument. In any case, it could be expressed more neatly. What I'd do is as follows.

Let $f : A \to B$ be the kernel of $g : B \to C$. Let $c = \operatorname{coker}(f) : B \to Q$. We want to show that $f=\ker(c)$, which amounts to showing that $f$ equalizes $c$ and $0$. So suppose $p : P \to B$ has $cp=0$.

Since $cf=gf=0$ and $c$ is a coequalizer, there is a unique $q : Q \to C$ with $qc=g$. But then $0=q0=qcp=gp$, so $gp=gf=0$ and so there is a unique $u : P \to A$ with $p=fu$. Thus $f$ equalizes $c$ and $0$, so $f = \ker(c)$, as desired.

This is illustrated in the following makeshift commutative diagram.

$$\begin{array}{ccccccccc} &&P&&&&&& \\\ & \overset{u}{\swarrow} & & \overset{p}{\searrow} &&&&& \\\ A && \xrightarrow{f} && B && \xrightarrow{g} && C \\\ &&&&& \underset{c}{\searrow} && \underset{q}{\
earrow} & \\\ &&&&&& Q & \end{array}$$