Homeomorphisms Between Topological Spaces

# Homeomorphisms Between Topological Spaces

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 $y \in f(\mathrm{int}(A))$

- So we conclude that $f(\mathrm{int}(A)) = \mathrm{int} (f(A))$. $\blacksquare$

**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:

\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:

\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 $y \in f(\overline{A})$.

- So we conclude that $f(\overline{A}) = \overline{f(A)}$. $\blacksquare$

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. |