A map $m : A \rightarrow B$ is regular monic if $(A, m)$ is an equalizer of some parallel pair $B \rightrightarrows C$.

$m$ is split monic if it admits a left inverse, i.e. a map $f$ with $fm = 1_A$.

Theorem: Split monic $\Rightarrow$ regular monic $\Rightarrow$ monic

Proof: Assume $m$ is split monic, and let $n$ be its left inverse. Then, take the parallel pair $(\id, m \circ n) : B \rightarrow B$. We claim $m$ is the equalizer of this parallel pair. Note that it is certainly a cone, since $m \circ n \circ m = m \circ \id = m$. Suppose that $(A', m')$ were another cone such that $m' = m \circ n \circ m'$. Define $\varphi : A' \rightarrow A$ as $\varphi = n \circ m'$. Then certainly the appropriate triangle commutes: $m \circ \varphi = m \circ n \circ m' = m'$. Suppose $\phi$ were another such map, so that $m \circ \phi = m'$. But composing on the left by $n$ implies that $n \circ m \circ \phi = n \circ m'$ and then $\phi = n \circ m' = \varphi$. So, $\varphi$ is unique and $(A, m)$ is the equalizer of the parallel pair, making $m$ regular monic.

Suppose $m$ is regular monic, and that it is the equalizer of $B \xrightarrow{f,g} C$. Take arbitrary maps $s, t: A' \rightarrow A$ with $m \circ s = m \circ t$. Then, of course, since $(A', m \circ s)$ and $(A', m \circ t)$ are (the same) cone over $B \xrightarrow{f,g} C$, we have a unique map $\varphi : A' \rightarrow A$ with $m \circ \varphi = m \circ s = m \circ t$. Taking $\varphi = s$ and $\varphi = t$ make this true, so we must have $\varphi = s = t$. So $m$ is monic.

An example of a monic which is not split occurs in $\mathbf{Ab}$. Consider the inclusion $i : \mathbb{Z} \rightarrow \mathbb{Q}$. Being inclusion, it's clearly monic. However, its a simple exercise in group theory to show there are no group homomorphisms from $\mathbb{Q} \rightarrow \mathbb{Z}$ in general, so there can certainly be no left inverse, so $i$ is not split monic.