[proofplan] Use the inverse map of the homeomorphism. Since $f$ is a homeomorphism, its inverse map $g:Y\to X$ is continuous. The image of an [open set](/page/Open%20Set) $U\subset X$ under $f$ is exactly the preimage $g^{-1}(U)$, so it is open in $Y$. For a [closed set](/page/Closed%20Set) $C\subset X$, apply the open-set result to $X\setminus C$ and use bijectivity to identify $f(X\setminus C)$ with $Y\setminus f(C)$. [/proofplan]

[step:Introduce the continuous inverse map] Since $f:X\to Y$ is a homeomorphism, $f$ is bijective and its inverse map \begin{align*} g:Y\to X \end{align*} is continuous. Here $g(y)$ denotes the unique point $x\in X$ such that $f(x)=y$. [/step]

[step:Rewrite the image of an open set as a preimage under the inverse]Let $U\subset X$ satisfy $U\in\tau_X$. We prove that $f(U)=g^{-1}(U)$. If $y\in f(U)$, then there exists $x\in U$ such that $y=f(x)$. Since $g=f^{-1}$ as an inverse map, $g(y)=x\in U$, so $y\in g^{-1}(U)$. Conversely, if $y\in g^{-1}(U)$, then $g(y)\in U$. Since $f(g(y))=y$, we have $y\in f(U)$. Hence $f(U)=g^{-1}(U)$. Because $g:Y\to X$ is continuous and $U$ is open in $X$, the preimage $g^{-1}(U)$ is open in $Y$. Therefore $f(U)$ is open in $Y$.[/step]

[guided]Let $U\subset X$ be open, meaning $U\in\tau_X$. The direct image $f(U)$ is not usually controlled by continuity of $f$ itself, because continuity is defined by preimages of open sets. The point of using the inverse map is that the image under $f$ becomes a preimage under $g=f^{-1}$. We first verify the identity \begin{align*} f(U)=g^{-1}(U). \end{align*} Take $y\in f(U)$. By definition of image, there exists $x\in U$ such that $y=f(x)$. Since $g$ is the inverse map of $f$, applying $g$ gives $g(y)=x$. Thus $g(y)\in U$, so $y\in g^{-1}(U)$. For the reverse inclusion, take $y\in g^{-1}(U)$. By definition of preimage, $g(y)\in U$. Since $f$ and $g$ are inverse maps, $f(g(y))=y$. Therefore $y$ is the image under $f$ of the point $g(y)\in U$, so $y\in f(U)$. The two inclusions give $f(U)=g^{-1}(U)$. Now use continuity of \begin{align*} g:Y\to X. \end{align*} By the definition of continuity between topological spaces, the preimage under $g$ of every open subset of $X$ is open in $Y$. Since $U\in\tau_X$, it follows that $g^{-1}(U)\in\tau_Y$. Therefore $f(U)\in\tau_Y$.[/guided]

[step:Use complements and bijectivity to prove the closed-set statement] Let $C\subset X$ be closed in $X$. Then $X\setminus C$ is open in $X$. By the open-set part already proved, $f(X\setminus C)$ is open in $Y$. We claim that \begin{align*} f(X\setminus C)=Y\setminus f(C). \end{align*} If $y\in f(X\setminus C)$, then $y=f(x)$ for some $x\in X\setminus C$. If $y\in f(C)$ also held, then $y=f(c)$ for some $c\in C$. Injectivity of $f$ would give $x=c$, contradicting $x\notin C$ and $c\in C$. Hence $y\in Y\setminus f(C)$. Conversely, let $y\in Y\setminus f(C)$. Since $f$ is surjective, there exists $x\in X$ such that $y=f(x)$. If $x\in C$, then $y\in f(C)$, contrary to $y\in Y\setminus f(C)$. Thus $x\in X\setminus C$, and hence $y\in f(X\setminus C)$. Therefore $Y\setminus f(C)=f(X\setminus C)$ is open in $Y$. By the definition of closed subset, $f(C)$ is closed in $Y$. [/step]