Lemma 9.8.55488C1DAFED2

Proof. We begin with two remarks used in the proof of (1). First, two maps from a smooth model $M_3$ to a smooth model $M_1$ that commute with the maps to $X$ agree. Indeed, let $U\subset X$ be the complement of a nowhere dense analytic subset over which both models are isomorphisms; the two maps agree over $U$, and the part of $M_3$ over $U$ is dense, by lemma 9.4.1.

Second, strict transforms are compatible with composition. Let $g\colon P\to N$ and $h\colon N\to M$ be proper bimeromorphic maps between complex manifolds, so that $h\circ g$ is proper and bimeromorphic by corollary 9.7.3, and let $D\subset M$ be an irreducible hypersurface, with strict transform $D^ N$ in $N$. We claim that the strict transform of $D^ N$ under $g$ is the strict transform $D^ P$ of $D$ under $h\circ g$. Let $S_ h,S_{hg}\subset M$ and $S_ g\subset N$ be analytic subsets of codimension at least two as in proposition 9.8.3 for $h$, $h\circ g$ and $g$, and let $T=S_ h\cup S_{hg}\cup h(S_ g)$, which is analytic of codimension at least two by theorem 9.7.2 and lemma 9.3.15(3).

The set $O=D^ N\cap h^{-1}(M\setminus T)=h^{-1}(D\setminus T)$ is open in $D^ N$, nonempty, and does not meet $S_ g$; its complement $D^ N\cap h^{-1}(T)$ is an analytic subset of the irreducible set $D^ N$ different from $D^ N$, hence nowhere dense by corollary 9.3.17(2). So $O$ is dense in $D^ N\setminus S_ g$, and since $g$ is an isomorphism over $N\setminus S_ g$, the strict transform of $D^ N$ under $g$ is the closure of $g^{-1}(O)=(h\circ g)^{-1}(D\setminus T)$. By the independence of the choice of $S_{hg}$ in proposition 9.8.3, applied with $T$, this closure is also $D^ P$.

Now let $(M_ i,F_ i)$, $i=1,2,3$, be pairs such that $(M_1,F_1)$ and $(M_2,F_2)$ are identified on a common smooth model $N_{12}$, and $(M_2,F_2)$ and $(M_3,F_3)$ on $N_{23}$. Let $P$ be a common smooth model of $N_{12}$ and $N_{23}$, and let $p_ i\colon P\to M_ i$ be the composite maps; by the first remark, the two composite maps $P\to M_2$, through $N_{12}$ and through $N_{23}$, coincide. By the second remark, the strict transforms of $F_1$ and $F_2$ under $p_1$ and $p_2$ are the strict transforms in $P$ of their strict transforms in $N_{12}$, which agree; similarly for $F_2$ and $F_3$. Hence $F_1$ and $F_3$ have the same strict transform in $P$, and the identification is transitive; it is reflexive, with $M_1$ itself as common smooth model, and symmetric by definition. Since $p_ i$ maps the strict transform of $F_ i$ onto $F_ i$ (proposition 9.8.3), the images of $F_1$ and $F_3$ in $X$ both equal the image of their common strict transform in $P$; so the image of $F$ in $X$ depends only on the prime divisor.

We now prove (2). Let $(M,F)$ represent a prime divisor. The ideal $\mathcal I_ F$ is coherent by theorem 9.3.10, and it is not identically zero on any component of $M$, because $F$ is nowhere dense. Let $g\colon M'\to M$ be as in (2), so that $\mathcal I_ F\mathcal O_{M'}=\mathcal O_{M'}(-H)$ with $H$ an effective divisor whose support has simple normal crossings; $g$ is proper, it is an isomorphism off $g^{-1}(F)$, and $g^{-1}(F)$, the support of $H$, is nowhere dense, so $g$ is bimeromorphic. The strict transform $F'$ of $F$ is an irreducible hypersurface contained in the support of $H$, so by corollary 9.3.17(4) and (3) it is one of its irreducible components, which are smooth. The pair $(M',F')$ represents the same prime divisor, as we see on the common smooth model $M'$, which is a smooth model of $X$ by corollary 9.7.3. $\square$