If the domain has only monos, every conservative functor reflects split epis I'm reading a paper on accessible categories where the authors remark that when the domain has only monos, any conservative, accessible functor reflects split epimorphisms as well. This seemed to be quite a simple result (for there was no proof nor reference), but I'm struggling to find a proof. The reference in question is Remark 3.5 in this paper. Could someone give me a hint or tell me if any additional hypotheses are needed?

EDIT: The functor $\mathcal{P}(X) \to \text{Set}$ taking the subset $A$ to the set $X/A$ where $A$ is collapsed to a point and the inclusion $A \subset B$ to the obvious map $X/A \to X/B$ seems to give a counterexample.
 A: A condition that would work is to require the functor to only have monos in its image. That is: if $F: \mathcal K \to \mathcal L$ is conservative and every arrow in its image is mono then $F$ reflects split epis. This is simply because then every split epi in its image will also be mono and hence iso, which is reflected because $F$ is conservative.
In the comments you mention how the problematic remark is used in 4.12 in the paper you linked. Indeed the relevant functor there does have only monos in its image (and similar for the use of the remark in 4.11). We will use the notation of 4.12 in the linked paper. The situation can be summarised in the diagram below, which is a pullback of categories:
$$
\require{AMScd}
\begin{CD}
\mathcal{K} @>H>> \mathbf{Set}\\
@A \bar{G} AA @AA G A \\
\mathcal{L} @>>\bar{H}> \mathbf{Emb}(\mathbf{Set})
\end{CD}
$$
Here $\mathbf{Emb}(\mathbf{Set})$ is the category of sets and injective functions, $G$ is the inclusion functor and $H$ is a faithful functor (anything else is irrelevant for this answer).
We need to show that $\bar G(f)$ is a mono for any arrow $f$ in $\mathcal L$. If not, then there are distinct $g$ and $h$ in $\mathcal K$ such that $\bar G(f)g = \bar G(f)h$. By faithfulness of $H$ we then have that $H \bar G(f)$ is not mono, as witnessed by $H(g)$ and $H(h)$. However, $H \bar G(f) = G \bar H(f)$ is in the image of $G$ and is thus mono because $G$ has only monos in its image by definition. We this arrive at a contradiction, and conclude that $\bar G(f)$ is mono.
