Proposition

Suppose $f: X \rightarrow Y$ is a bijective continuous map. Show that the following are equivalent:

  1. $f$ is a homeomorphism.
  2. $f$ is open.
  3. $f$ is closed.

Solution

$1 \rightarrow 2$

$f$ is a homeomorphism, so it must be open.

$2 \rightarrow 3$

Let $C \subset X$ be closed. Then $f(X \setminus C)$ is open in $Y$ because $f$ is open. Since $f$ is bijective, $f(X \setminus C) = f(X) \setminus f(C)$. Thus $f(C)$ is closed in $Y$.

$3 \rightarrow 1$

Let $U \subset X$ be open. Then $f(X \setminus U)$ is closed in $Y$ because $f$ is closed. Since $f$ is bijective, $f(X \setminus U) = f(X) \setminus f(U)$. Thus $f(U)$ is open in $X$.