  We prove that if a continuous function $f : X \to f(X)$ takes open sets into
elements of the Boolean algebra generated by open and closed subsets in $f(X)$,
then there exist
  $X_n \subset X,$ $(n \in \omega)$ such that $f$ is open on every $X_n$ and
$f(X_n)$ cover $ f(X).$
