Suppose that $U: \mathsf{Top} \to \mathsf{Set}$ is the forgetful functor. I believe that for a topological quotient map $\pi:R \to S$, and a map $\phi:U(S)\to U(T)$, we have that $\phi \circ U(\pi)$ is the $U$-image of some arrow iff $\phi$ is.
I think the same thing should be true in various categories for $\pi$ a quotient map. If we're talking about $\mathsf{Ring}$ or $\mathsf{Grp}$, basically what we're doing is relying on the property that a relation holds in $S$ iff it holds for all its preimages in $R$. And similarly in $\mathsf{Top}$, we use that a set is open in $S$ iff its preimage is open in $R$.
Since a quotient of $R$ is just an equivalence class of epimorphisms below $R$, it seems that this property should have a categorical interpretation. The Wikipedia article on quotient categories didn't shed much light. Since I expressed these in terms of the forgetful functor, I wonder if this is generalizable outside concrete categories somehow? Does one need just that $\pi$ is an epimorphism, or some stronger property? Is the property I mention the essential property that is necessary, or is there something more fundamental?
The question seems to be: Let $U : C \to D$ be a faithful functor and let $\pi : X \to Y$ be a morphism in $C$. Assume that $\phi : U(Y) \to U(Z)$ is a morphism in $D$ and that $ \phi \circ U(\pi) = U(f)$ for some morphism $f : X \to Z$ (it is unique since $U$ is faithful). When is there a morphism $\overline{f} : Y \to Z$ (automatically unique) with $\phi = U(\overline{f})$? This would imply $\overline{f} \circ \pi = f$. Let us say that $\pi$ witnesses $U$-lifts if this is possible for all $\phi$.
In fact, let $(X,TX \to X)$, $(Y,TY \to Y)$, $(Z,TZ \to Z)$ be three $T$-algebras, $\pi : X \to Y$ a homomorphism of $T$-algebras and $\phi : Y \to Z$ be a morphism in $D$ such that $\phi \circ \pi$ is a homomorphism of $T$-algebras. This means that the outer rectangle in $$\begin{array}{c} TX & \to & TY & \to & TZ \\ \downarrow && \downarrow && \downarrow \\ X & \to & Y & \to & Z\end{array} $$ commutes. If $X \to Y$ is an epimorphism of $T$-algebras, it is an epimorphism in $D$. Since $T=UF$ (where $F$ is left adjoint to $U$), $T$ preserves epimorphisms, so that $TX \to TY$ is an epimorphism. The square on the left hand commutes since $\pi$ is homomorphism of $T$-algebras. Now a diagram chase shows that the square on the right commutes, i.e. that $\phi$ is a homomorphism of $T$-algebras.
Examples. The forgetful functors $\mathsf{Ab} \to \mathsf{Set}$ and $\mathsf{Grp} \to \mathsf{Set}$ are monadic and preserve epimorphisms. The forgetful functor $\mathsf{Ring} \to \mathsf{Set}$ is monadic, but it does not preserve epimorphisms. In fact, there are non-surjective epimorphisms of rings which do not witness $U$-lifts. For example, consider $\mathbb{Z} \to \mathbb{Q}$. In order to simplify the notation, I will omit $U$. Let $R$ be a ring, and let $\phi : \mathbb{Q} \to R$ be any map on the underlying sets which satisfies $\phi(z)=z \cdot 1$ for $z \in \mathbb{Z}$, but does some crazy things on $\mathbb{Q} \setminus \mathbb{Z}$. Then $\phi|_{\mathbb{Z}}$ is a ring homomorphism, but $\phi$ is not a ring homomorphism. The forgetful functor $\mathsf{Top} \to \mathsf{Set}$ is not monadic, but it preserves epimorphisms. For the last two examples, we may use the following:
In fact, let $\pi$ be a regular epimorphism, i.e. the coequalizer of two morphisms $a,b : P \to X$. We have $U(f) U(a) = \phi U(\pi) U(a) = \phi U(\pi) U(b) = U(f) U(b)$ and hence $fa=fb$. Therefore, there is a unique morphism $\overline{f} : Y \to Z$ such that $\overline{f} \pi = f$. It follows $U(\overline{f}) U(\pi) = U(f) = \phi U(\pi)$. Since $U(\pi)$ is an epimorphism, we get $U(\overline{f})=\phi$, as desired.
Examples. Surjective ring homomorphisms witness lifts for $\mathsf{Ring} \to \mathsf{Set}$. Continuous quotient maps witness lifts for $\mathsf{Top} \to \mathsf{Set}$.