Given a functor $F : \mathcal C \to \mathcal C$ with initial algebra $\alpha : FA \to A$, and another algebra $\xi : FX \to X$, we obtain a unique morphism $\mathsf{fold}~\xi : A \to X$ such that $\mathsf{fold}~\xi \circ \alpha = \xi \circ F(\mathsf{fold}~\xi)$.


I am looking for a sufficient condition - ideally phrasable as $P(F) \wedge Q(\xi)$ rather than $R(F, \xi)$ - which guarantees that:


1. $\mathsf{fold}~\xi$ is monomorphic,
2. $\mathsf{fold}~\xi$ is epimorphic.


(I'm looking for separate conditions for 1 and for 2.)


I would expect:


1. $F$ preserves monomorphism and $\xi$ is monomorphic
2. $F$ preserves epimorphism and $\xi$ is epimorphic


but I can't seem to prove this.


EDIT: I'm also interested in the dual question (which, mathematically, of course just has dual answers). I mention this anyway, because Valery Isaev's answer uses local finite presentability, which is something that might not have been answered had I asked the dual question. So feel free to use concepts that are only dually familiar. In particular, I'm willing to assume that $F^{op}$ is polynomial (or similar in a less powerful category, e.g. $FX$ is a coproduct of powers of $X$ in $\mathcal C^{op}$).



When is a fold monomorphic/epimorphic
Suppose $({\cal E,M})$ is a factorization system on $\cal C$ such that $\cal M$ consists of monomorphisms and $F$ preserves $\cal E$-morphisms. If $X$ is an $F$-algebra, define an *$\cal M$-subalgebra* of $X$ to be a morphism $Y\to X$ of $F$-algebras that is an $\cal M$-morphism. Then if $A$ is the initial $F$-algebra, the map $A\to X$ lies in $\cal E$ if and only if $X$ has no proper $\cal M$-subalgebras.


In the "if" direction, note that the assumption that $F$ preserves $\cal E$-morphisms implies that $({\cal E,M})$-factorizations lift to $F$-algebras. In particular, if we factor the unique $F$-algebra morphism $A\to X$ as $A\to Z \to X$, then the unique lifting property of $F A \to F Z$ (which is in $\cal E$) against $Z\to X$ (which is in $\cal M$) makes $Z$ an $F$-algebra and the two maps $F$-algebra maps. In particular, $Z$ is an $\cal M$-subalgebra of $X$. Thus, if $X$ has no proper $\cal M$-subalgebras, then $Z\cong X$ and hence $A\to X$ is in $\cal E$.


Conversely, if $A\to X$ is in $\cal E$, suppose $Z\to X$ is an $\cal M$-subalgebra. Then $A\to X$ factors through $Z$ (since $A$ is the initial $F$-algebra), but then $Z\to X$ must be an isomorphism since it is an $\cal M$-morphism that is factored through by some $\cal E$-morphism.


In particular, if $\cal C$ admits (epi, strong mono) factorizations, then $A\to X$ is epi if and only if $X$ has no proper strong-subalgebras.


**EDIT:** Note that if $X$ has no proper $\cal M$-subalgebras, then $\xi : F X \to X$ is in $\cal E$. For we can $({\cal E,M})$-factor it as $F X \to Z\to X$ and show that $Z$ becomes an $\cal M$-subalgebra. The assumption on $X$ then implies $Z\cong X$, hence $\xi\in\cal E$.


On the other hand, as Valery pointed out, taking $F=\rm Id$ shows that the converse can't hold. Any endo-epimorphism $X\to X$ (e.g. $(+1) : \mathbb{Z}\to\mathbb{Z}$) gives an $F$-algebra $\xi$ that is an epimorphism, but $0 = A \to X$ will rarely be epi.


