Let where and are topological spaces.
Then is continuous if the preimage of every open set in is open in .
Lemma
Equivalently, is cts iff preimage of every closed set in is closed in .
Proof
Let be closed in . Then is open. So is open.
Now is the set of all values which are not mapped to
i.e. they are mapped to .
So indeed, which is closed.
On the other hand, suppose is closed for every closed .
Then for any open in , we have open which is exactly .