[proofplan] We prove both directions by unpacking the definitions of continuity and homeomorphism. In the forward direction, continuity of $f$ gives preservation of openness under preimage, while continuity of the inverse map gives the reverse implication after identifying $V$ with the inverse-map preimage of $f^{-1}[V]$. In the reverse direction, the assumed equivalence immediately gives continuity of $f$, and bijectivity lets us rewrite every inverse-map preimage under $f^{-1}:Y\to X$ as an image $f(U)$ whose openness is forced by the same equivalence. [/proofplan]

[step:Distinguish the inverse map from preimage notation] Since $f:X\to Y$ is bijective, define its inverse map \begin{align*} g:Y\to X \end{align*} by the rule that $g(y)$ is the unique point $x\in X$ satisfying $f(x)=y$. Thus \begin{align*} g\circ f=\operatorname{id}_X \end{align*} and \begin{align*} f\circ g=\operatorname{id}_Y. \end{align*} Throughout the proof, $f^{-1}[V]$ denotes the preimage of a subset $V\subset Y$ under $f$, while $g$ denotes the inverse function. [/step]

[step:Use continuity of both directions to prove the open-set equivalence]Assume first that $f$ is a homeomorphism. Then $f:X\to Y$ is continuous and $g:Y\to X$ is continuous. Let $V\subset Y$ be arbitrary. If $V\in\tau_Y$, then continuity of $f$ gives $f^{-1}[V]\in\tau_X$. Conversely, assume $f^{-1}[V]\in\tau_X$. Since $g$ is continuous and $f^{-1}[V]$ is open in $X$, the preimage $g^{-1}[f^{-1}[V]]$ is open in $Y$. We now identify this set with $V$. For $y\in Y$, \begin{align*} y\in g^{-1}[f^{-1}[V]] \iff g(y)\in f^{-1}[V] \iff f(g(y))\in V \iff y\in V. \end{align*} Hence $g^{-1}[f^{-1}[V]]=V$, so $V\in\tau_Y$. Therefore, for every $V\subset Y$, \begin{align*} V\in\tau_Y \iff f^{-1}[V]\in\tau_X. \end{align*}[/step]

[guided]Assume that $f$ is a homeomorphism. By definition, this means that $f:X\to Y$ is bijective, $f$ is continuous, and the inverse map $g:Y\to X$ is continuous. Fix an arbitrary subset $V\subset Y$. We must prove both implications in the equivalence \begin{align*} V\in\tau_Y \iff f^{-1}[V]\in\tau_X. \end{align*} First suppose $V\in\tau_Y$. Since $f:X\to Y$ is continuous, the preimage under $f$ of every open subset of $Y$ is open in $X$. Applying this to the [open set](/page/Open%20Set) $V$, we obtain \begin{align*} f^{-1}[V]\in\tau_X. \end{align*} Now suppose instead that $f^{-1}[V]\in\tau_X$. We want to prove that $V$ is open in $Y$. The available continuity statement in this direction is the continuity of the inverse map $g:Y\to X$: the preimage under $g$ of every open subset of $X$ is open in $Y$. Since $f^{-1}[V]$ is open in $X$, continuity of $g$ gives \begin{align*} g^{-1}[f^{-1}[V]]\in\tau_Y. \end{align*} It remains to show that this open set is actually $V$. Let $y\in Y$. By definition of preimage under $g$, \begin{align*} y\in g^{-1}[f^{-1}[V]] \iff g(y)\in f^{-1}[V]. \end{align*} By definition of preimage under $f$, \begin{align*} g(y)\in f^{-1}[V] \iff f(g(y))\in V. \end{align*} Since $g$ is the inverse map of $f$, we have $f(g(y))=y$. Therefore \begin{align*} f(g(y))\in V \iff y\in V. \end{align*} Combining these equivalences gives \begin{align*} y\in g^{-1}[f^{-1}[V]] \iff y\in V. \end{align*} Since this holds for every $y\in Y$, the two subsets of $Y$ are equal: \begin{align*} g^{-1}[f^{-1}[V]]=V. \end{align*} Because the left-hand side is open in $Y$, $V$ is open in $Y$. This proves the equivalence for the arbitrary subset $V\subset Y$.[/guided]

[step:Use the open-set equivalence to prove continuity of $f$] Assume now that, for every subset $V\subset Y$, \begin{align*} V\in\tau_Y \iff f^{-1}[V]\in\tau_X. \end{align*} Let $V\in\tau_Y$. Applying the forward implication in the displayed equivalence gives $f^{-1}[V]\in\tau_X$. Since this holds for every open subset $V$ of $Y$, the map $f:X\to Y$ is continuous. [/step]

[step:Use bijectivity to prove continuity of the inverse map] Let $U\in\tau_X$ be arbitrary, and define the subset $V\subset Y$ by \begin{align*} V=f(U)=\{f(x)\in Y:x\in U\}. \end{align*} Because $f$ is injective, its preimage satisfies \begin{align*} f^{-1}[V]=U. \end{align*} Indeed, if $x\in f^{-1}[V]$, then $f(x)=f(u)$ for some $u\in U$, and injectivity gives $x=u\in U$; the reverse inclusion follows directly from the definition of $V$. Since $U\in\tau_X$ and $f^{-1}[V]=U$, the assumed equivalence gives $V\in\tau_Y$. Finally, using the inverse map $g:Y\to X$, we have \begin{align*} g^{-1}[U]=f(U)=V. \end{align*} Thus $g^{-1}[U]$ is open in $Y$ for every open subset $U\subset X$, so $g:Y\to X$ is continuous. We have shown that $f$ is bijective, $f$ is continuous, and its inverse map is continuous; hence $f$ is a homeomorphism. [/step]