Continuity Defined from Closed Sets

Theorem
Let $T_1$ and $T_2$ be topological spaces.

Let $f: T_1 \to T_2$ be a mapping.

Then $f$ is continuous iff for all $V$ closed in $T_2$, $f^{-1} \left({V}\right)$ is closed in $T_1$.

Proof
First we show the following.

Let $W \in T_2$.

We note that $f^{-1} \left({T_2}\right) = T_1$.

Hence, from Mapping Preimage of Set Difference, we have that $f^{-1} \left({T_2 \setminus W}\right) = T_1 \setminus f^{-1} \left({W}\right)$.


 * Suppose the condition on closed sets holds.

Let $U$ be open in $T_2$.

Then $T_2 - U$ is closed in $T_2$.

By hypothesis, $f^{-1} \left({T_2 - U}\right) = T_1 \setminus f^{-1} \left({U}\right)$ is closed in $T_1$.

So $f^{-1} \left({U}\right)$ is open in $T_1$.

This is true for any $U \in T_2$, so $f$ is continuous.


 * Now let $f$ be continuous.

Let $V$ be closed in $T_2$.

Then $T_2 \setminus V$ is open in $T_2$.

As $f$ is continuous, $f^{-1} \left({T_2 - V}\right) = T_1 \setminus f^{-1} \left({V}\right)$ is open in $T_1$.

So $f^{-1} \left({V}\right)$ is closed in $T_1$.