Suppose the short exact sequence of groups $$\operatorname{Ker}f\rightarrowtail G\twoheadrightarrow G/\operatorname{Ker}f$$ has a right splitting $G/\operatorname{Ker}f\rightarrowtail G$. Then the intersection of $\operatorname{Ker}f$ with $G/\operatorname{Ker}f$ is trivial because any element of $\operatorname{Ker}f$ becomes the unit in $G/\operatorname{Ker}f$.
I'm trying to formulate this in terms of arrows and can't see the light... I think this should quickly follow in any finitely complete pointed regular category, but I just don't see how to use the zeros well.
How to categorically prove the intersection (pullback) of the subobjects $\operatorname{Ker}f$ and $G/\operatorname{Ker}f$ is zero?
Suppose you have a map $a:T\to G$ which factors through both $\ker f$ and $G/\ker f$. Since $a$ factors through $\ker f$, $fa=0$. But if $i:G/\ker f\to G$ is the splitting map and we factor $a=ib$ for some $b$, this means $0=fa=fib=b$ and so $a=ib=0$. That is, $a$ factors uniquely through the zero object.