[proofplan] We verify the three defining properties of an [equivalence relation](/page/Equivalence%20Relation). Reflexivity follows from the identity map on each [open set](/page/Open%20Set), including the empty open set, since all pointwise conformality conditions are then either immediate or vacuous. Symmetry follows by applying the inverse theorem for conformal isomorphisms, and transitivity follows by composing the two conformal isomorphisms and checking both conformality and bijectivity. [/proofplan]

[step:Use identity maps to prove reflexivity] Let $U\in\mathcal{O}$. Define the identity map \begin{align*} \operatorname{id}_U:U\to U,\qquad z\mapsto z. \end{align*} This map is bijective. If $U\neq\varnothing$, then $\operatorname{id}_U$ is holomorphic on $U$ and satisfies \begin{align*} (\operatorname{id}_U)'(z)=1 \end{align*} for every $z\in U$, so it is conformal on $U$. If $U=\varnothing$, the same conclusion holds vacuously: the empty identity map is bijective and has no point at which a holomorphicity or nonvanishing derivative condition can fail. Thus $\operatorname{id}_U:U\to U$ is a conformal isomorphism, and therefore $U\sim U$. [/step]

[step:Invert a conformal isomorphism to prove symmetry]Let $U,V\in\mathcal{O}$ and suppose $U\sim V$. By definition of $\sim$, there exists a conformal isomorphism \begin{align*} f:U\to V. \end{align*} Since $U$ and $V$ are open subsets of $\mathbb{C}$ and $f:U\to V$ is a conformal isomorphism, the inverse theorem for conformal isomorphisms [citetheorem:8269] applies. Hence \begin{align*} f^{-1}:V\to U \end{align*} is a conformal isomorphism. Therefore $V\sim U$.[/step]

[guided]We must prove that the relation is symmetric, so we start with a conformal equivalence in one direction and produce one in the opposite direction. Assume $U,V\in\mathcal{O}$ and $U\sim V$. The definition of $\sim$ gives a conformal isomorphism \begin{align*} f:U\to V. \end{align*} The natural candidate for a map from $V$ back to $U$ is the inverse function $f^{-1}$. We are allowed to use it because a conformal isomorphism is, in particular, bijective. To know that this inverse is again conformal, we invoke the inverse theorem for conformal isomorphisms [citetheorem:8269]. Its hypotheses require open sets $U,V\subset\mathbb{C}$ and a conformal isomorphism $f:U\to V$; these are exactly the objects and assumption we have in scope. Therefore \begin{align*} f^{-1}:V\to U \end{align*} is a conformal isomorphism. This is precisely the assertion that $V\sim U$.[/guided]

[step:Compose conformal isomorphisms to prove transitivity] Let $U,V,W\in\mathcal{O}$ and suppose $U\sim V$ and $V\sim W$. Then there exist conformal isomorphisms \begin{align*} f:U\to V \end{align*} and \begin{align*} g:V\to W. \end{align*} Define \begin{align*} h:U\to W,\qquad z\mapsto g(f(z)). \end{align*} Since $f$ and $g$ are bijective, their composition $h=g\circ f$ is bijective: injectivity follows by applying injectivity of $g$ and then injectivity of $f$, and surjectivity follows by first choosing a preimage under $g$ and then a preimage under $f$. Since $f$ is conformal on $U$ and $g$ is conformal on $V$, the composition theorem for [conformal maps](/page/Conformal%20Maps) [citetheorem:8271] applies to the open sets $U,V,W$ and gives that $g\circ f$ is conformal on $U$. Hence $h:U\to W$ is a bijective [conformal map](/page/Conformal%20Map), so it is a conformal isomorphism. Therefore $U\sim W$. [/step]

[step:Collect the three properties] We have shown that for every $U\in\mathcal{O}$, $U\sim U$; for all $U,V\in\mathcal{O}$, $U\sim V$ implies $V\sim U$; and for all $U,V,W\in\mathcal{O}$, the conjunction of $U\sim V$ and $V\sim W$ implies $U\sim W$. Thus $\sim$ is reflexive, symmetric, and transitive on $\mathcal{O}$. Therefore $\sim$ is an equivalence relation on the class of open subsets of $\mathbb{C}$. [/step]