PomeCorollary 5.15.16.DF37542BAB03 Let $\Omega \subseteq \mathbb {C}^ n$, $\Omega '\subseteq \mathbb {C}^{n'}$ be two domains and $F\colon \Omega '\rightarrow \Omega $ be a holomorphic map. Then $F$ is $\mathcal{F}$-continuous.
Corollary 5.15.16DF37542BAB03
Proof. It suffices to show that the inverse image $F^{-1}(U)$ of each $\mathcal{F}$-open subset $U\subseteq \Omega $ is $\mathcal{F}$-open. By theorem 5.15.11, after possibly shrinking $\Omega $ and $\Omega '$, we may assume that $U$ has the form $\{ x\in \Omega :\psi (x)>0\} $, where $\psi \in \mathrm{PSH}(\Omega )$. Clearly, we may assume that $F^{-1}(U)$ is non-empty. Then $F^*\psi \in \mathrm{PSH}(\Omega ')$ by proposition 5.13.17, we find that
\[ F^{-1}(U)=\{ y\in \Omega ': F^*\psi (y)>0\} \]
is $\mathcal{F}$-open. $\square$
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