Proposition

A map between topological spaces is continuous if and only if the preimage of every closed subset is closed.

Solution

Let $f: X \rightarrow Y$ be given.

For any subset $C \subset Y$ and for any $x \in X$,

\[\begin{align*} x \in f^{-1}(Y \setminus C) &\iff f(x) \in Y \setminus C \\ &\iff f(x) \notin C \\ &\iff x \notin f^{-1}(C) \\ &\iff x \in Y \setminus f^{-1}(C). \end{align*}\]

Therefore, $f^{-1}(Y \setminus C) = X \setminus f^{-1}(C)$.

Now, suppose that $f$ is continous.

Let $C \subset Y$ be closed. Since $C$ is closed, $Y \setminus C$ is open. Since $f$ is continuous, $f^{-1}(Y \setminus C) = X \setminus f^{-1}(C)$ is open, so $f^{-1}(C)$ is closed.

Suppose, on the other hand, that the preimage of every closed subset is closed. Let $V \subset Y$ be open. Then $Y \setminus V$ is closed. $f^{-1}(Y \setminus V)$ is closed, and thus $X \setminus f^{-1}(V)$ is closed. This implies that $f^{-1}(V)$ is open. Therefore, $f$ is continuous.