PomeLemma 9.4.1.F56FCBEF9522 Let $F\colon Y\to X$ be a bimeromorphic holomorphic map of complex spaces. Then $F^{-1}(B)$ is nowhere dense in $Y$ for every nowhere dense analytic subset $B\subset X$. Consequently, the set $E$ in the definition may be enlarged by any nowhere dense analytic subset of $X$; if $U\subset X$ is the complement of a nowhere dense analytic subset, then $F^{-1}(U)$ is dense in $Y$; and every nonempty open subset of $Y$ contains a nonempty open set that $F$ maps isomorphically onto an open subset of $X$.
Lemma 9.4.1F56FCBEF9522
Proof. Let $E$ be as in the definition, and let $G\subset F^{-1}(B)$ be open in $Y$. Then $G\setminus F^{-1}(E)$ is an open subset of $Y\setminus F^{-1}(E)$ contained in $F^{-1}(B\setminus E)$, which corresponds under the isomorphism $Y\setminus F^{-1}(E)\to X\setminus E$ to the nowhere dense subset $B\setminus E$ of $X\setminus E$; so $G\setminus F^{-1}(E)$ is empty, and $G$, an open subset of the nowhere dense set $F^{-1}(E)$, is empty as well. In particular $F$ is also an isomorphism over $X\setminus (E\cup B)$, and $F^{-1}(E\cup B)=F^{-1}(E)\cup F^{-1}(B)$ is nowhere dense, as the union of two closed nowhere dense sets; so $E$ may be replaced by $E\cup B$. If $U=X\setminus B$, then $F^{-1}(U)$ is the complement of the closed nowhere dense set $F^{-1}(B)$, hence dense. Finally, a nonempty open set $G\subset Y$ meets the dense open set $Y\setminus F^{-1}(E)$, and $F$ maps the nonempty open set $G\setminus F^{-1}(E)$ isomorphically onto an open subset of $X\setminus E$. $\square$
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