Under what conditions the frame morphism induced by continuous function is surjective?

Question

Let $f\colon X\longrightarrow Y$ a continuous function between topological spaces. Then we have a contravariant functor $\Omega$ from the topological spaces to Frames, which sends each space $X$ in to its topology $\Omega(X)$ and each continuous function in to the inverse image between the topologies. Which conditions I have to ask to a continous function $f$ to guarantee that $\Omega (f)$ is surjective.

Answer 1

There isn't really a good answer to this, and in particular there is not any natural condition that is not just a restatement of the definition of $\Omega(f)$ being surjective.

However, there is one restatement of the definition that is very familiar: $\Omega(f)$ being surjective just means that $X$ has the subspace topology from $Y$ (with respect to the map $f$). In other words, a set $U\subseteq X$ is open iff there exists an open set $V\subseteq Y$ such that $f^{-1}(U)=V$. When $f$ is the inclusion of a subset of $Y$, this is just the usual definition of the subspace topology (but the definition really makes sense with respect to any map, not just inclusions of subsets). Note also that if $X$ is $T_0$, then $f$ must be injective for this to happen (since preimages under $f$ must distinguish any two points of $X$ in order for the subspace topology to be $T_0$). So if you restrict to $T_0$ spaces, this is not really more general than the ordinary notion of subspaces.