Open and Closed Maps on Topological Spaces

Open and Closed Maps on Topological Spaces

We have looked quite a bit at continuous maps on topological spaces, so we will turn out attention to looking at other types of maps on topological spaces - in particular, open maps and closed maps which we define below.

Definition: Let $X$ and $Y$ be topological spaces. A map $f : X \to Y$ is called an Open Map if for every open set $U \subseteq X$ we have that the image $f(U) \subseteq Y$ is open. Furthermore, $f$ is said to be a Closed Map if for every closed set $C \subseteq X$ we have that the image $f(C) \subseteq Y$ is closed.

If $f$ is an open/closed map then we sometimes just say that $f$ is open/closed.

The concept of continuity of a map on topological spaces lies closely to open/closed maps when $f : X \to Y$ is a bijection. We prove two simple results in the following theorem regarding open/closed bijective maps on topological spaces.

Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be a bijective map. Then:
a) If $f$ is open then the inverse map $f^{-1} : Y \to X$ is continuous.
b) If $f$ is closed then the inverse map $f^{-1} : Y \to X$ is continuous.
  • Proof of a): Suppose that $f : X \to Y$ is a bijective map and that $f$ is open. Since $f$ is bijective, the inverse map $f^{-1} : Y \to X$ exists, and since $f$ is open we have that for every open set $U$ in $X$ that then $f(U)$ is open in $Y$, i.e., for every open set in the codomain of $f^{-1}$ we have that the inverse image of $f^{-1}$ of $U$ (which is just $f(U)$ is open, which implies that $f^{-1}$ is continuous.
  • Proof of b) Similarly, if $f : X \to Y$ is a bijective map and $f$ is closed, then since $f$ is bijective, the inverse map $f^{-1} : Y \to X$ exists, and since $f$ is closed we have that for every closed set $C$ in $X$ that then $f(C)$ is closed in $Y$, i.e., for every closed set in the codomain of $f^{-1}$ we have that the inverse image of $f^{-1}$ of $C$ (which is just $f(C)$ is closed, which implies that $f^{-1}$ is continuous.
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