Homeomorphisms Between Topological Spaces

Table of Contents

Definition: Let

$E$

and

$F$

be topological spaces. A Homeomorphism between

$E$

and

$F$

is a bijective function

f : E \to F

such that both

$f$

and

f^{-1}

are continuous. If such a homeomorphism between

$E$

and

$F$

exists, then

$E$

and

$F$

are said to be Homeomorphic.

Proposition 1: Let

$E$

and

$F$

be topological spaces and let

A \subseteq E

. Suppose that

f : E \to F

is a homeomorphism. Then:
(1)

f(\mathrm{int}(A)) = \mathrm{int} (f(A))

.
(2)

f(\overline{A}) = \overline{f(A)}

Proof of (1): Let $y \in f(\mathrm{int}(A))$ . Then there exists an $x \in \mathrm{int}(A)$ with $$f(x) = y$$ . Let $$U$$ be an open set in $$E$$ such that $x \in E \subseteq A$ . Then $f(x) \in f(E) \subseteq f(A)$ . But since $$f$$ is a homeomorphism, $$f$$ is an open map, and since $$E$$ is open, so if $$f(E)$$ . Thus $f(x) \in \mathrm{int} (f(A))$ .

Now let $y \in \mathrm{int} f(A)$ . Then there exists an open set $$V$$ in $$F$$ with $y \in F \subseteq f(A)$ . So $f^{-1}(y) \in f^{-1}(F) \subseteq A$ . But since $$f$$ is a homeomorphism, $$f$$ is continuous, and thus $f^{-1}(F)$ is an open set, which shows that $f^{-1}(y) \in \mathrm{int} (A)$ . Thus $y \in f(\mathrm{int}(A))$

Proof of (2): Let $y \in f(\overline{A})$ . Let $$V_y$$ be a neighbourhood of $$y$$ . Then there exists an open set $$U_y$$ such that $y \in U_y \subseteq V_y$ . Then $f^{-1}(y) \in f^{-1}(U_y) \subseteq f^{-1}(V_y)$ . Since $$f$$ is continuous, $f^{-1}(U_y)$ is an open set and thus $f^{-1}(V_y)$ is a neighbourhood of $f^{-1}(y)$ . But since $y \in f(\overline{A})$ we have that $f^{-1}(y) \in \overline{A}$ and so:

(1)

\begin{align} \quad f^{-1}(V_y) \cap A \neq \emptyset \end{align}

Hence $V_y \cap f(A) \neq \emptyset$ . Since this holds true for all neighbourhoods $$V_y$$ of $$y$$ in $$F$$ , we have that $y \in \overline{f(A)}$ .

Now let $y \in \overline{f(A)}$ . Let $$V_y$$ be a neighbourhood of $f^{-1}(y)$ . Then there exists an open set $$U_y$$ in $$E$$ such that $f^{-1}(y) \in U_y \subseteq V_y$ . Since $$f$$ is an open map, $$f(U_y)$$ is an open set. So $y \in f(U_y) \subseteq f(V_y)$ . Thus $$f(V_y)$$ is a neighbourhood of $$y$$ . Since $y \in \overline{f(A)}$ we have that:

(2)

\begin{align} \quad f(V_y) \cap f(A) \neq \emptyset \end{align}

Therefore $V_y \cap A \neq \emptyset$ . Since this holds true for all neighbourhoods $$V_y$$ of $f^{-1}(y)$ in $$E$$ , we have that $f^{-1}(y) \in \overline{A}$ . Thus $y \in f(\overline{A})$ .

Proposition 2: Let

$E$

and

$F$

be topological spaces. Suppose that

$E$

and

$F$

are homeomorphic. Then:
(1)

$E$

is Hausdorff if and only if

$F$

is Hausdorff.
(2)

$E$

is first countable if and only if

$F$

is first countable.

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