Bijection is Open iff Closed

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ 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:
 * $\forall H \in \tau_1: f \sqbrk H \in \tau_2$

As $f$ is a bijection:
 * $f \sqbrk {S_1 \setminus H} = f \sqbrk {S_1} \setminus f \sqbrk H = S_2 \setminus f \sqbrk H$

By definition of closed set:
 * $S_1 \setminus H$ is closed in $T_1$

and as $f$ is an open mapping:
 * $f \sqbrk {S_1 \setminus H} = S_2 \setminus f \sqbrk H$

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
 * Open Mapping is not necessarily Closed Mapping