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.

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 $\mathcal C^{op}$).

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.

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.