In an Abelian category, every Kernel is Kernel of Cokernel

Question

I am trying to prove that if $\phi : A \rightarrow B$ is the kernel of a morphism, then it is the Kernel of its Cokernel.

Here's my work so far: Let $\phi : A \rightarrow B$ be kernel of $j : B \rightarrow Z$. Since A is abelian category, $\phi$ has cokernel $i : B \rightarrow K$. We know that $j \circ \phi = 0$ so that $j$ factors through $i$. My image essentially looks like: $A \xrightarrow{\phi} B \xrightarrow{j} Z$ with $K$ below arrow from $B$ to $Z$ so that we have $B \xrightarrow{i} K \xrightarrow{\exists!} Z$.

I am not sure how to continue with this since I can not figure out how to factor through $\phi$.

Answer 1

Everything you wrote looks fine. Now, as you mention, $j$ is the composition of $i$ and the map $K \to Z.$ So any map $\psi: C \to B$ equalizing $i$ and $0$ certainly equalizes $j$ and $0,$ and thus factors uniquely through $\ker(j) = A.$