Corollary 9.7.3390229269CD2

Proof. The composite is proper. Let $E_ X\subset X$ and $E_ Y\subset Y$ be nowhere dense analytic subsets as in the definition of bimeromorphic maps for $f$ and $g$. The image $f(E_ Y)$ is analytic by theorem 9.7.2, and it is nowhere dense: an open set $G\subset f(E_ Y)$ satisfies $G\setminus E_ X\subset f(E_ Y\setminus f^{-1}(E_ X))$, which corresponds under the isomorphism $f\colon Y\setminus f^{-1}(E_ X)\to X\setminus E_ X$ to the nowhere dense set $E_ Y\setminus f^{-1}(E_ X)$; so $G\subset E_ X$, and $G=\varnothing $. Hence $B=E_ X\cup f(E_ Y)$ is a nowhere dense analytic subset of $X$. The map $f$ is an isomorphism over $X\setminus B$, and $f^{-1}(X\setminus B)$ does not meet $E_ Y$, so $f\circ g$ is an isomorphism over $X\setminus B$. Finally, $f^{-1}(B)$ is a nowhere dense analytic subset of $Y$, and $(f\circ g)^{-1}(B)=g^{-1}(f^{-1}(B))$ is nowhere dense in $Z$, by two applications of lemma 9.4.1. $\square$