Is every split mono regular?

Question

Let $f : A \to B$ be a split mono: $(\exists g : B \to A)(g \circ f = \text{id}_A)$.

(It it clear that $f$ is in particular a monomorphim.) I want to show that it is regular, i.e., there is an object $C$ and parallel arrows $k, l : B \to C$ such that both $k \circ f = \text{id}_A$, $l \circ f = \text{id}_A$, ánd, for any other object $D$ and morphism $d : D \to B$, there is a unique morphism $h : D \to A$ such that $d = f \circ h$. By monicity of $f$, it suffices to prove existence.

Of course, the likely candidates are $C = A, l = g$. If this is all to fit into an equaliser diagram, that would be to say that $k$ is also a left-inverse of $f$. Now these are of course not unique (take a look at matrices for example), which is a shame, because I have already proved that in an equaliser is an isomorphism iff the parallel maps coincide.

So we could distinguish two cases: $g = k$ and $g \neq k$. I think the first case would be very easy, but the fact that I don't see that second, makes my think the case distinction is unnecessary.

Can someone guide in the right direction?

Concretely: I want to find a morphism $k : B \to A$ that is also a left-inverse of $f$, ánd that satisfies, $k \circ d = g \circ d$, out of which to create a map $h : D \to A$ that makes $ d = f \circ h$.

Answer 1

Every split mono is regular. Consider $f : A \to B$ and $g : B \to A$ such that $g \circ f = 1_A$. Then $f$ is the equaliser of $1_B, f \circ g : B \to B$.

For we see that $(f \circ g) \circ f = f \circ (g \circ f) = f \circ 1_A = f = 1_B \circ f$.

Now suppose we have $h : C \to B$ such that $(f \circ g) \circ h = 1_B \circ h$. Then $f \circ g \circ h = h$. Note that for all $j : C \to A$, if $f \circ j = h$, then $j = 1_A \circ j = (g \circ f) \circ j = g \circ (f \circ j) = g \circ h$. And note that $f \circ (g \circ h) = (f \circ g) \circ h = h$. In other words, $g \circ h$ is the unique $j$ such that $f \circ j = h$.