# Bijection is Open iff Inverse is Continuous

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## Contents

## Theorem

Let $T_1 = \left({S_1, \tau_1}\right), T_2 = \left({S_2, \tau_2}\right)$ 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:

- $\forall H \in \tau_1: f \left({H}\right) \in \tau_2$

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

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

- $\forall H \in \tau_1: g^{-1} \left({H}\right) \in \tau_2$

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:

- $\forall H \in \tau_1: g^{-1} \left({H}\right) \in \tau_2$

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

- $\forall H \in \tau_1: f \left({H}\right) \in \tau_2$

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

$\blacksquare$

## Also see

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 1$: Functions