[proofplan] The argument is a direct comparison between openness of $f$ and continuity of the inverse map. In the forward direction, a homeomorphism has a continuous inverse, and the image of an [open set](/page/Open%20Set) under $f$ is the preimage of that open set under the inverse function. In the reverse direction, openness of the bijection $f$ implies that every open subset of $X$ has open preimage under the inverse function, so the inverse is continuous. [/proofplan] [step:Use continuity of the inverse to prove that a homeomorphism is open] Assume that $f: X \to Y$ is a homeomorphism. By the definition of homeomorphism, $f$ is continuous and its inverse function $g: Y \to X$, defined by $g(y)=f^{-1}(y)$ for $y \in Y$, is continuous. Let $U \in \tau_X$. We claim that $f(U)=g^{-1}(U)$, where $g^{-1}(U)$ denotes the preimage of $U$ under the map $g$. If $y \in f(U)$, then there exists $x \in U$ such that $y=f(x)$, and therefore $g(y)=x \in U$, so $y \in g^{-1}(U)$. Conversely, if $y \in g^{-1}(U)$, then $g(y) \in U$, and since $f(g(y))=y$, we have $y \in f(U)$. Hence $f(U)=g^{-1}(U)$. Since $U$ is open in $X$ and $g$ is continuous, $g^{-1}(U)$ is open in $Y$. Therefore $f(U)$ is open in $Y$. As this holds for every $U \in \tau_X$, the map $f$ is open. Thus every homeomorphism is continuous and open. [/step] [step:Use openness of the bijection to prove continuity of the inverse] Assume conversely that $f: X \to Y$ is bijective, continuous, and open. Because $f$ is bijective, the inverse function $g: Y \to X$, defined by $g(y)=f^{-1}(y)$ for $y \in Y$, exists. To prove that $g$ is continuous, let $U \in \tau_X$. As in the previous step, the preimage of $U$ under $g$ satisfies \begin{align*} g^{-1}(U)=f(U). \end{align*} Because $f$ is open and $U$ is open in $X$, the set $f(U)$ is open in $Y$. Hence $g^{-1}(U)$ is open in $Y$ for every $U \in \tau_X$, so $g$ is continuous. [guided] We need to prove that the inverse function is continuous. Since $g: Y \to X$ is a map between topological spaces, continuity of $g$ means that for every open set $U \in \tau_X$, the preimage $g^{-1}(U)$ is open in $Y$. Let $U \in \tau_X$. The key point is to identify the set $g^{-1}(U)$ with the image $f(U)$. The notation here has two meanings: $g^{-1}(U)$ is a preimage under the map $g$, while $g$ itself is the inverse function of $f$. We verify the equality carefully. If $y \in g^{-1}(U)$, then $g(y) \in U$. Since $g$ is the inverse function of $f$, we have $f(g(y))=y$, so $y$ is the image under $f$ of the point $g(y) \in U$. Therefore $y \in f(U)$. Conversely, if $y \in f(U)$, then there exists $x \in U$ such that $y=f(x)$. Applying the inverse function gives $g(y)=g(f(x))=x$, so $g(y) \in U$. Therefore $y \in g^{-1}(U)$. Thus \begin{align*} g^{-1}(U)=f(U). \end{align*} Now the openness hypothesis on $f$ applies exactly to the open set $U \in \tau_X$: it says that $f(U) \in \tau_Y$. Hence $g^{-1}(U) \in \tau_Y$. Since this holds for every open set $U$ in $X$, the inverse function $g$ is continuous. [/guided] We have shown that $f$ is continuous by hypothesis and that its inverse function $g=f^{-1}:Y \to X$ is continuous. Since $f$ is bijective, this is precisely the definition of $f$ being a homeomorphism. This proves the reverse implication and completes the equivalence. [/step]