# Bijection is Open iff Inverse is Continuous

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## Contents

## 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 if and only if $f^{-1}$ is continuous.

## Proof

Let $f$ be a bijection.

Let $g := f^{-1}$.

By Bijection iff Inverse is Bijection we have that $g$ is a bijection and that $g^{-1} = f$.

Let $f$ be open.

Then by definition of open mapping:

- $\forall H \in \tau_1: f \sqbrk H \in \tau_2$

taking $H \in \tau_1$ by definition of open in $T_1$.

But $f = g^{-1}$ and so:

- $\forall H \in \tau_1: g^{-1} \sqbrk H \in \tau_2$

which is exactly the definition for $g$ to be continuous.

The argument works the other way.

Let $g$ be continuous.

Then by definition of continuous mapping:

- $\forall H \in \tau_1: g^{-1} \sqbrk H \in \tau_2$

But $g^{-1} = f$ and so:

- $\forall H \in \tau_1: f \sqbrk H \in \tau_2$

which is exactly the definition for $f$ to be open.

$\blacksquare$

## Also see

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions