Image of Interior of Set under Homeomorphism is Interior of Image

Theorem

Let $X$ and $Y$ be topological spaces.

Let $f : X \to Y$ be a homeomorphism.

Let $A \subseteq X$.

Then:

$\paren {f \sqbrk A}^\circ = f \sqbrk {A^\circ}$

where:

$A^\circ$ is the interior of $A$ in $X$

$\paren {f \sqbrk A}^\circ$ is the interior of $f \sqbrk A$ in $Y$.

First, we show that $f \sqbrk {A^\circ} \subseteq \paren {f \sqbrk A}^\circ$.

Let $y \in f \sqbrk {A^\circ}$.

Then there exists $x \in A^\circ$ such that $y = \map f x$.

Since $x \in A^\circ$, there exists an open set $U \subseteq A$ in $X$ such that $x \in U$.

Then $y = \map f x \in f \sqbrk U \subseteq f \sqbrk A$.

By definition of a homeomorphism, $f$ is an open map.

So $f \sqbrk U$ is an open set in $Y$ contained in $f \sqbrk A$ and containing $y$.

So from the definition of interior, we have $y \in \paren {f \sqbrk A}^\circ$.

Hence $y \in f \sqbrk {A^\circ}$ implies that $y \in \paren {f \sqbrk A}^\circ$.

So $f \sqbrk {A^\circ} \subseteq \paren {f \sqbrk A}^\circ$.

$\Box$

Now we show that $\paren {f \sqbrk A}^\circ \subseteq f \sqbrk {A^\circ}$.

Let $y \in \paren {f \sqbrk A}^\circ$.

Then there exists an open set $V \subseteq f \sqbrk A$ in $Y$ such that $y \in V$.

We then have $\map {f^{-1} } y \in f^{-1} \sqbrk V$.

Since $f$ is continuous, $f^{-1} \sqbrk V$ is open in $X$, is contained in $A$ and contains $\map {f^{-1} } y$.

So we have $\map {f^{-1} } y \in A^\circ$.

Since $y = \map f {\map {f^{-1} } y}$, we obtain $y \in f \sqbrk {A^\circ}$.

Hence $y \in \paren {f \sqbrk A}^\circ$ implies $y \in f \sqbrk {A^\circ}$.

So $\paren {f \sqbrk A}^\circ \subseteq f \sqbrk {A^\circ}$.

Hence we obtain $\paren {f \sqbrk A}^\circ = f \sqbrk {A^\circ}$.

$\blacksquare$