[proofplan] We unpack the definition of a conformal isomorphism: it gives a bijective [conformal map](/page/Conformal%20Map) from $U$ onto $V$, and its inverse is conformal by the defining isomorphism property, equivalently by [citetheorem:8269]. Since [conformal maps](/page/Conformal%20Maps) between open subsets of $\mathbb C$ are holomorphic, and holomorphic maps are continuous, both $f$ and $f^{-1}$ are continuous. A bijection with continuous inverse is precisely a homeomorphism. [/proofplan] [step:Unpack the conformal isomorphism structure] Since $f:U\to V$ is a conformal isomorphism, $f$ is bijective from $U$ onto $V$ and is conformal on $U$. Define the inverse map \begin{align*} g:V\to U,\qquad g=f^{-1}. \end{align*} By the inverse property for conformal isomorphisms, equivalently [citetheorem:8269], the map $g:V\to U$ is also a conformal isomorphism. In particular, $g$ is conformal on $V$. [guided] The first point is to separate the topological requirements from the analytic hypotheses. A homeomorphism is a bijective continuous map whose inverse is continuous, so we must produce exactly three pieces of information: bijectivity of $f$, continuity of $f$, and continuity of $f^{-1}$. Because $f:U\to V$ is a conformal isomorphism, the word "isomorphism" supplies bijectivity from $U$ onto $V$. Thus every point $v\in V$ has a unique preimage in $U$. We may therefore define the inverse map \begin{align*} g:V\to U,\qquad g=f^{-1}. \end{align*} The inverse map is not merely a set-theoretic inverse: by the inverse property for conformal isomorphisms, equivalently [citetheorem:8269], $g:V\to U$ is itself a conformal isomorphism. Hence $g$ is conformal on $V$. This is the analytic input needed later to prove that the inverse map is continuous. [/guided] [/step] [step:Use holomorphicity to obtain continuity of both maps] A conformal map between open subsets of $\mathbb C$ is holomorphic at every point of its domain. Hence $f:U\to V\subset\mathbb C$ is holomorphic on $U$, and $g:V\to U\subset\mathbb C$ is holomorphic on $V$. By the standard theorem that holomorphic maps are continuous (citing a result not yet in the wiki: Holomorphic maps are continuous), $f$ is continuous as a map from $U$ to $\mathbb C$, and $g$ is continuous as a map from $V$ to $\mathbb C$. Since $V$ and $U$ carry the subspace topologies inherited from $\mathbb C$, continuity of $f:U\to\mathbb C$ together with $f(U)\subset V$ implies continuity of $f:U\to V$, and continuity of $g:V\to\mathbb C$ together with $g(V)\subset U$ implies continuity of $g:V\to U$. [/step] [step:Conclude that the conformal isomorphism is a homeomorphism] We have shown that $f:U\to V$ is bijective, that $f$ is continuous, and that its inverse map $f^{-1}=g:V\to U$ is continuous. By the definition of a homeomorphism, $f:U\to V$ is a homeomorphism with respect to the subspace topologies inherited from $\mathbb C$. [/step]