A sufficient condition for a surjection between topological space to be continuous Let $f:~X\mapsto Y$ be a surjection between two topological spaces $X$ and $Y$. Suppose that the following condition holds:
$$(f(A))^\circ\subseteq f(A^\circ),~\forall A\subseteq X.$$
Is it true that $f$ is continuous?

Take an open $U$ in $Y$. Consider the set $G$ of functions $g:U\to f^{-1}(U)$ such that $f\circ g=id_U$ (i.e. sections of $f\mid_{f^{-1}(U)}:f^{-1}(U)\to U$).
First notice that $\bigcup\limits_{g\in G}g(U)=f^{-1}(U)$ (use set theory only).
Then take $A=g(U)$ for some $g\in G$. Then $f$ is bijective on $A$. Moreover, by the hypothesis,
$$
U=f(A)^\circ\subseteq f(A^\circ),
$$
so $f(A^\circ)=f(A)$. Hence $A=A^\circ$ is open.
This implies $f^{-1}(U)$ is open as well.
Therefore $f$ must be continuous.

If any error occurs, please point it out. Thanks in advance, and hope this helps.
P.S. Thanks to yoyo for the modification.
