Any category with zero object, every kernel is monomorphism.

Question

This is the problem 5.53 of Rotman, Homological Algebra.

In any category having a zero object calling it $0$, prove that every kernel is monomorphism.

Let $f:A\to B$ and $\ker(f):K\to A$ be the kernel. Any $g:X\to K$, $f\circ \ker(f)\circ g=0$. In sum, I have

$\begin{array} KK & \stackrel{\ker(f)}{\longrightarrow} & A & \stackrel{f}{\longrightarrow} B\\ \uparrow{g}\\ X \end{array}$

I can choose $X=0$ here. If there is $\ker(\ker(f)):K'\to K$, then there is a unique $0$-map $0_{K'}:0\to K'$. Since $0$ is zero object, I can reverse the arrow. However, it is not clear why the diagram commutes for $K'\to 0\to K$ which is $0\in\mathrm{Hom}(K',K)$ compatible with $K'\to K$.

1. How do I know that I can reverse arrow?

2. Is there a way to show $K'$ being both initial and final in the category?

Answer 1

The kernel is the equaliser of $f$ and the zero map from $A$ to $B$ (that factoring through the zero object). All equaliser maps (in any category) are monic.

Answer 2

This answer is purely based on universal property of $Ker(f)$. It is essentially proving equalizers are monic.

Take any $g,h:X\to K$ such that $Ker(f)\circ g=Ker(f)\circ h$ calling it $l_1=l_2:X\to A$. So $f\circ l_i=0$. By universal property of $Ker(f)$, the map $l_i$ factors through $K$ uniquely. Hence $g=h$.