Let $U : \mathfrak{T} \to \mathfrak{S}$ be the underlying functor from the category of topological spaces (with continuous maps) to the category of sets. It is a result that a morphism $j : X_0 \to X$ in $\mathfrak{T}$ is a topological embedding if and only if $j$ is a monomorphism and for any $f : Y \to X$ in $\mathfrak{T}$, a factorization $U(j)g_0 = U(f)$ in $\mathfrak{S}$ implies $jf_0 = f$ in $\mathfrak{T}$, with $g_0 = U(f_0)$. An exercise asks to dualize the latter property, which is phrased in purely categorical language, but I'm not sure what dualization does to the functor $U$. The mantra for dualizing is "apply the result to the opposite category and reinterpret in the original category". But $U$ isn't a (covariant) functor $\mathfrak{T}^{opp} \to \mathfrak{S}$, which makes the "apply to the opposite category" step confusing. Does that matter? Or is it enough to just treat $U$ as a mapping of objects/morphisms in $\mathfrak{T}$ or $\mathfrak{T}^{opp}$ to $\mathfrak{S}$, and not worry about its functoriality properties when dualizing?
My best guess for the dual would be the following: A morphism $j : X \to X_0$ such that $j$ is an epimorphism and for all $f : X \to Y$ in $\mathfrak{T}$, a factorization $g_0U(j) = U(f)$ in $\mathfrak{S}$ implies $f_0j = f$ in $\mathfrak{T}$, with $g_0 = U(f_0)$. This feels right, but I'm not sure. Can't concepts only be dualized when they are meaningful in any category? Wouldn't the fact that this concept involves a specific functor prohibit it from being "universal" enough to be dualized?