Kernel of an arrow that factors through a monic? Suppose an arrow $A\overset{f}{\rightarrow}B$ factors as $A\overset{q}{\rightarrow} J \overset{j}{\rightarrowtail}B$. When does $\ker f=\ker q$ and how can I prove it?
 A: Kevin Carlson's answer is assuming you are working in a category with a zero object, but it actually works the same way in the full generality of kernel pairs.
Suppose kernel pairs of $f$ and $q$ exist and let's denote them $k_1,k_2 \colon R \rightrightarrows A$ and $\ell_1,\ell_2 \colon S \rightrightarrows A$ respectively. Then, $fk_1=fk_2$ and $j$ is monic, so $qk_1=qk_2$, inducing an (unique) arrow $h \colon R \to S$ respecting $\ell_1h=k_1$ and $\ell_2h=k_2$. On the other hand, since $q\ell_1=q\ell_2$, then composing by $j$, we get $f\ell_1=f\ell_2$ inducing an (unique) arrow $g \colon S \to R$ respecting $k_1g = \ell_1$ and $k_2g=\ell_2$. The two following diagrams illustrate the situation :
 $\qquad$ 
Then you just have to check that $h$ and $g$ are inverse isomorphisms. This comes from uniqueness in the universal property of a kernel pair. The following diagram states that $gh$ is induced by the kernel pair $k_1,k_2$ on the morphisms $k_1,k_2$ : $\mathrm{id}_R$ also works, so by uniqueness $gh = \mathrm{id}_R$.

You can do the same for $hg$.
A: Always. The kernel of $f$ represents maps $z$ into $A$ which are killed by composition with $f$; but if $fz=jqz=0,$ then $qz=0$ and $z$ factors also through the kernel of $q$. In the other direction, if $qz=0$ then certainly $fz=0$.
