When is a fold monomorphic/epimorphic

If $\mathcal{C}$ has an initial object and the unique map $0 \to X$ is an epimorphism, then $\mathrm{fold}\ \xi : A \to X$ is an epimorphism. This follows from the fact that the composite $0 \to A \to X$ is an epimorphism. This condition seems too restrictive, but it is "almost necessary". Indeed, if $F(0) \simeq 0$ (for example, if $F$ is the identity functor), then $A \simeq 0$, so $\mathrm{fold}\ \xi : A \to X$ is an epimorphism if and only if $0 \to X$ is. It is hard to imagine some nice sufficient properties of $F$ which are not true for the identity functor.

Now, let's discuss when $\mathrm{fold}\ \xi$ is a monomorphism. Suppose that the following conditions hold:

• $\mathcal{C}$ is locally finitely presentable.
• $F$ preserves sufficiently long sequential colimits.
• The unique map $0 \to X$ is a monomorphism.
• The map $\xi : F(X) \to X$ is a monomorphism.
• $F$ preserves monomorphisms.

Then $\mathrm{fold}\ \xi : A \to X$ is a monomorphism. Indeed, if the second condition holds, then $A$ can be constructed as a colimit of a transfinite sequence $$ 0 \to F(0) \to F^2(0) \to F^3(0) \to \ldots$$ This colimit exists since $\mathcal{C}$ is locally presentable. The map $f^n : F^n(0) \to X$ is constructed as the composite $F^n(0) \xrightarrow{F(f^{n-1})} F(X) \xrightarrow{\xi} X$. Since $0 \to X$ and $\xi$ are monomorphisms and $F$ preserves monomorphisms, we can prove that $f^n$ is a monomorphism by induction. At a limit stage, we need to use the fact that $\mathcal{C}$ is locally finitely presentable. Such a category has a generator consisting of finitely presentable objects. Thus, to prove that a map $f^\alpha : F^\alpha(0) \to X$ is a monomorphism, we just need to show that for every pair of maps $g,h : S \to F^\alpha(0)$, where $S$ is finitely presentable, if $f^\alpha \circ g = f^\alpha \circ h$, then $g = h$. Since $S$ is finitely presentable, maps $g$ and $h$ factor through $F^\beta(0)$ for some $\beta < \alpha$. Now, it follows that $g = h$ by induction hypothesis. This shows that $\mathrm{fold}\ \xi$ is a monomorphism since it equals $f^\lambda : F^\lambda(0) \to X$ for some $\lambda$.