[proofplan] We prove both implications from the definition of a homeomorphism as a bijective continuous map whose inverse is continuous. In the forward direction, the inverse map carries closed subsets of $X$ to closed preimages in $Y$, and those preimages are exactly the images under $f$. In the reverse direction, the hypothesis that $f$ is closed turns complements of open subsets of $X$ into closed subsets of $Y$, which proves continuity of the inverse map. [/proofplan] [step:Show that a homeomorphism is continuous and closed] Assume that $f$ is a homeomorphism. By definition, $f$ is continuous, and its inverse map \begin{align*} g: Y \to X \end{align*} is continuous, where $g(y)$ is the unique point $x \in X$ such that $f(x) = y$. Let $C \subset X$ be closed in $X$. Then $X \setminus C$ is open in $X$. Since $g$ is continuous, $g^{-1}(X \setminus C)$ is open in $Y$. Hence its complement $Y \setminus g^{-1}(X \setminus C)$ is closed in $Y$. By the definition of preimage under $g$, \begin{align*} Y \setminus g^{-1}(X \setminus C) = g^{-1}(C). \end{align*} Since $g = f^{-1}$, we have $g^{-1}(C) = f(C)$. Therefore $f(C)$ is closed in $Y$. Since $C \subset X$ was arbitrary, $f$ is closed. [/step] [step:Use closedness of the bijection to prove continuity of the inverse] Assume that $f$ is continuous, closed, and bijective. Define the inverse map \begin{align*} g: Y \to X \end{align*} by letting $g(y)$ be the unique point $x \in X$ satisfying $f(x) = y$. Let $O \subset X$ be open in $X$. Then $X \setminus O$ is closed in $X$, so $f(X \setminus O)$ is closed in $Y$ because $f$ is closed. We claim that \begin{align*} g^{-1}(O) = Y \setminus f(X \setminus O). \end{align*} Indeed, for any $y \in Y$, surjectivity of $f$ gives a point $x \in X$ with $y = f(x)$, and the definition of $g$ gives $g(y) = x$. Thus $y \in g^{-1}(O)$ exactly when $x \in O$, which is exactly when $x \notin X \setminus O$. By injectivity of $f$, this is equivalent to $y \notin f(X \setminus O)$. This proves the claimed identity. Since $f(X \setminus O)$ is closed in $Y$, its complement $Y \setminus f(X \setminus O)$ is open in $Y$. Therefore $g^{-1}(O)$ is open in $Y$. Since $O \subset X$ was arbitrary and open, $g$ is continuous. [guided] We need to prove that the inverse map is continuous. Since continuity between topological spaces is tested by preimages of open sets, we start with an arbitrary open subset $O \subset X$ and prove that its preimage under the inverse map is open in $Y$. Because $f$ is bijective, the inverse map is well-defined: \begin{align*} g: Y \to X \end{align*} where $g(y)$ is the unique point $x \in X$ such that $f(x) = y$. The word "unique" uses both parts of bijectivity: surjectivity gives existence of such an $x$, and injectivity gives uniqueness. Let $O \subset X$ be open in $X$. Its complement $X \setminus O$ is closed in $X$. The hypothesis that $f$ is closed says that the image of every closed subset of $X$ is closed in $Y$, so $f(X \setminus O)$ is closed in $Y$. The key point is to identify the preimage $g^{-1}(O)$ with the complement of this [closed set](/page/Closed%20Set). We claim that \begin{align*} g^{-1}(O) = Y \setminus f(X \setminus O). \end{align*} To prove this identity, take any $y \in Y$. Since $f$ is surjective, there exists $x \in X$ such that $y = f(x)$. By definition of $g$, this point satisfies $g(y) = x$. Now $y \in g^{-1}(O)$ if and only if $g(y) \in O$, which is equivalent to $x \in O$. This is equivalent to $x \notin X \setminus O$. Because $f$ is injective, the condition $x \notin X \setminus O$ is equivalent to saying that $y = f(x)$ is not in $f(X \setminus O)$. Indeed, if $y$ were equal to $f(x')$ for some $x' \in X \setminus O$, then injectivity would force $x' = x$, contradicting $x \notin X \setminus O$. Hence \begin{align*} y \in g^{-1}(O) \iff y \in Y \setminus f(X \setminus O). \end{align*} Since $y \in Y$ was arbitrary, the claimed identity follows. Now $f(X \setminus O)$ is closed in $Y$, so its complement $Y \setminus f(X \setminus O)$ is open in $Y$. Therefore $g^{-1}(O)$ is open in $Y$. Since this holds for every open subset $O \subset X$, the inverse map $g: Y \to X$ is continuous. [/guided] [/step] [step:Conclude the equivalence] In the forward direction, we proved that every homeomorphism is continuous and closed. In the reverse direction, assuming that $f$ is continuous, closed, and bijective, we proved that its inverse map $g = f^{-1}: Y \to X$ is continuous. Therefore $f$ is a bijective continuous map with continuous inverse, so $f$ is a homeomorphism. This proves the desired equivalence. [/step]