Union of holomorphic atlases is holomorphic atlas.

Question

Let $S$ be a surface with open subsets $V$ and $W$ such that $s = V \cup W$. Suppose that $V$ and $W$ have holomorphic atlases $\Phi$ and $\Psi$ such that the holomorphic atlases $\Phi|_{V \cap W}$ and $\Psi|_{V \cap W}$ on $V \cap W$ are compatible. Show that $\Phi \cup \Psi$ is a holomorphic atlas on $S$.

Answer 1

Let $S$ be a surface with open subsets $V$ and $W$ such that $S = V \cup W$. Suppose that $V$ and $W$ have holomorphic atlases $\Phi$ and $\Psi$ such that the holomorphic atlases $\Phi|_{V \cap W}$ and $\Psi|_{V \cap W}$ are compatible. Now, consider the collection $\theta = \Phi \cup \Psi$ of charts on $S$. Since $\Phi$ covers $V$ and $\Psi$ covers $W$, it follows that $\theta$ covers $S$ and hence is an atlas for $S$. It remains to check that $\theta$ is holomorphic. Since $\Phi$ and $\Psi$ are holomorphic atlases for $V$ and $W$, it suffices to check that, for $\phi_\alpha \in \Phi$ and $\psi_\beta \in \Psi$, where $\phi_\alpha: U_\alpha \to V_\alpha$ and $\psi_\beta: U_\beta \to V_\beta$, the transition function$$\theta_{\alpha\beta} = \phi_\alpha \circ \psi_\beta^{-1}: \psi_\beta\left(U_\alpha \cap U_\beta\right) \to \phi_\alpha\left(U_\alpha \cap U_\beta\right)$$is holomorphic in the usual sense. Now, note that $U_\alpha \cap U_\beta \subset V \cap W$. Thus, by the compatibility of $\Phi$ and $\Psi$, the function $\phi_\alpha \circ 1_{V \cap W} \circ \psi_\beta^{-1}$, where $1_{V \cap W}: V \cap W \to V \cap W$ is identity map on $V \cap W$, is holomorphic. This immediately implies that $\theta_{\alpha\beta}$ is holomorphic, as we have$$\phi_\alpha \circ 1_{V \cap W} \circ \psi_\beta^{-1} = \phi_\alpha \circ \psi_\beta^{-1}.$$Thus, $\theta$ is a holomorphic atlas on $S$.