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


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

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

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