Proposition

Suppose $f: X \rightarrow Y$ is a homeomorphism and $U \subset X$ is an open subset. Show that $f(U)$ is open in $Y$ and the restriction $f\mid_U$ is a homeomorphism from $U$ to $f(U)$.

Solution

• By this post, $f\mid_U$ is continuous.
• $f \mid_U: U \rightarrow f(U)$ is surjective.
• $f \mid U$ is injective since $f$ is injective.
• Let $V \cap f(U) \subset Y \cap f(U)$ be an open set in $Y \cap U$ where $V$ is an open subset of $Y$. By the definition of a subspace topology, every open set can be expressed in this form. $f\mid_U^{-1}(V \cap f(U)) = U \cap f^{-1}(V \cap f(U)) = U \cap f^{-1}(V) \cap U = U \cap f^{-1}(V)$. Since $V$ is open in $Y$ and $f$ is continuous, $f^{-1}(V)$ is open in $X$. Therefore, $f^{-1}(V) \cap U$ is open in $U$.

Hence, $f\mid_U$ is a homeomorphism.