Let be $f:M\to N$ a continuous function where $M\subseteq X$. Then we know that every closed subset $S\subseteq N$ has a preimage $f^{-1}(S)$ such that there exists a closed set $Q\subseteq X$ with $f^{-1}(S)=Q\cap M$.
In the equivalent statement which is about open sets we are able to construct an open set $Q$ and by that we prove the statement regarding open sets. Now, I am wondering if it is also possible to show the above statement directly by constructing a closed set $Q$ (no use of contraposition or the equivalent statement with open sets)?