Is there a category $\mathcal{C}$ with zero object and a morphism $f:A\to B$ such that $0:A\to A$ is a kernel but $f$ is not monic?
If $\mathcal{C}$ is preadditive then $\ker f = 0 \iff f$ is monic. This is because $fg = fg'$ implies $f(g-g') = 0$ and hence $g-g' = 0 \circ h = 0$ since $0$ is a kernel hence $g = g'$. But in general only $\impliedby$ should hold. Is there a counter-example?
Let $\mathcal{C}$ be the category of pointed sets, whose zero object is the singleton set $\{*\}$. Let $f: (S, s_0) \to (T, t_0)$ be a morphism such that $f^{-1}(t_0) = \{s_0\}$. Then $f$ has kernel $\{*\}$, as any map $g: (X, x_0) \to (S, s_0)$ equalizing $f$ and the zero map must have $g(X) = \{s_0\}$ and hence factors uniquely through $\{*\}$.
However, $f$ is not generally a monomorphism. Here's a counterexample: let $(S, s_0) = (T, t_0) = (\{1, 2, 3\}, 1)$ and let $f(1) = 1, f(2) = f(3) = 2$. Let $(X, x_0) = (\{1, 2\}, 1)$, and define $g, g': (X, x_0) \to (S, s_0)$ by $g(1) = g'(1) = 1$ and $g(2) = 2, g'(2) = 3$. Then $fg = fg'$, but $g \neq g'$.
