Bijection is Open iff Closed

Theorem
Let $T_1 = \left({X_1, \tau_1}\right), T_2 = \left({X_2, \tau_2}\right)$ be topological spaces.

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

Then $f$ is open $f$ is closed.

Proof
Let $f$ be a bijection.

Suppose $f$ is an open mapping.

From the definition of open mapping:
 * $\forall H \in \tau_1: f \left({H}\right) \in \tau_2$

As $f$ is a bijection:
 * $f \left({X_1 \setminus H}\right) = f \left({X_1}\right) \setminus f \left({H}\right) = X_2 \setminus f \left({H}\right)$

By definition of closed set:
 * $X_1 \setminus H$ is closed in $T_1$

and as $f$ is an open mapping:
 * $f \left({X_1 \setminus H}\right) = X_2 \setminus f \left({H}\right)$

is closed in $T_2$.

Hence by definition $f$ is a closed mapping.

A similar argument demonstrates that if $f$ is closed then it is open.

Also see

 * Bijection is Open iff Inverse is Continuous