9.8 Connected fibers and resolution modelsF0CABED1A01F

Proof. Choose a nowhere dense analytic subset $A\subset X$ such that $f^{-1}(A)$ is nowhere dense and $f$ is an isomorphism over $X\setminus A$. Suppose that the compact fiber over $x$ is disconnected, and write it as the union of two disjoint nonempty compact sets. Since $Y$ is Hausdorff, they have disjoint open neighborhoods $V_1$ and $V_2$. Since $f$ is proper, hence closed, $U=X\setminus f(Y\setminus (V_1\cup V_2))$ is an open neighborhood of $x$ with $f^{-1}(U)\subset V_1\cup V_2$. As $(X,x)$ is irreducible, corollary 9.4.9 allows us to shrink $U$ so that $U\setminus A$ is connected. Each set $V_ i\cap f^{-1}(U)$ is open and nonempty, so it meets the dense set $Y\setminus f^{-1}(A)$. Therefore $V_1$ and $V_2$ split $f^{-1}(U\setminus A)$ into two disjoint nonempty open sets. This contradicts the connectedness of $U\setminus A$, which is isomorphic to $f^{-1}(U\setminus A)$. $\square$

Proof. Let $f\colon Y\to X$ be finite and bimeromorphic, with $X$ normal, and let $A\subset X$ be a nowhere dense analytic subset such that $f$ is an isomorphism over $X\setminus A$ and $f^{-1}(A)$ is nowhere dense. Normal spaces are unibranch, so the finite fibers of $f$ are connected by theorem 9.8.1, hence consist of at most one point. The image of $f$ is closed, because $f$ is proper, and contains the dense set $X\setminus A$; so $f$ is a continuous closed bijection, that is, a homeomorphism. Let $x\in X$ and $y=f^{-1}(x)$. Choose a neighborhood $V$ of $y$, relatively compact in the domain of a local model of $Y$, so that the coordinate functions of the model are bounded on $V$. Composed with $f^{-1}$, they are bounded holomorphic functions on $f(V)\setminus A$, where $f^{-1}$ is holomorphic, and $f(V)$ is an open neighborhood of $x$. By lemma 9.4.4 they extend holomorphically to $f(V)$, and the extensions are the continuous functions $(\text{coordinates})\circ f^{-1}$. Hence $f^{-1}$ is holomorphic.

For the second assertion, over the dense open set $S\setminus \phi ^{-1}(X_{\mathrm{sing}})$ there is a unique lift $\psi _0=\nu ^{-1}\circ \phi $, because $\nu $ is an isomorphism over $X_{\mathrm{reg}}$. We extend it by the argument used for the uniqueness in the proof of theorem 9.4.6. Let $s\in S$ and $x=\phi (s)$, choose disjoint neighborhoods $W_ l$ of the finitely many points of $\nu ^{-1}(x)$, relatively compact in domains $W'_ l$ of local models of $\widetilde X$, and a neighborhood $U$ of $x$ with $\nu ^{-1}(U)\subset \bigcup _ lW_ l$, which exists by properness. Since $S$ is normal, hence unibranch, corollary 9.4.9 gives a neighborhood $V$ of $s$ with $\phi (V)\subset U$ such that $V\setminus \phi ^{-1}(X_{\mathrm{sing}})$ is connected. Then $\psi _0$ maps this set into a single $W_ l$. The coordinate functions of $W'_ l$ composed with $\psi _0$ are bounded and extend holomorphically to $V$, by lemma 9.4.4; the extended map takes values in $\overline{W_ l}\subset W'_ l$, and it satisfies the equations of the local model, which hold on a dense subset of $V$. This gives a holomorphic lift on $V$, with $\nu \circ \psi =\phi $ by density. Lifts agree on the dense set where $\psi _0$ is defined, so these local lifts glue, and the lift is unique. $\square$

Proof. We take for $S$ the image of the critical set of $f$; the main point is that a component of $S$ of codimension one would contain a point over which $f$ is finite, hence an isomorphism by corollary 9.8.2.

Step 1: the set $S$. We may work over each connected component of $M$, of dimension $n$ say; every connected component of its inverse image has dimension $n$ as well, because it contains a nonempty open set mapped isomorphically onto an open subset of $M$, by lemma 9.4.1. The map $f$ is surjective, since its image is closed and dense. The critical set of $f$, where the Jacobian determinant vanishes in local coordinates, is analytic, and its image $S$ is analytic by theorem 9.7.2. Over $M\setminus S$ the map $f$ is proper and, by proposition 5.2.4(1), a local biholomorphism, so its fibers are discrete and finite. They are connected by theorem 9.8.1, and nonempty; so they are single points, and $f$ is a bijective local biholomorphism, hence an isomorphism, over $M\setminus S$. Since $f$ is an isomorphism over a dense open subset of $M$, it has no critical points over this set, so $S$ does not meet it and is nowhere dense.

Step 2: $S$ has codimension at least two. Suppose that $S$ has an irreducible component $D_0$ of codimension one. Choose a point of $(D_0)_{\mathrm{reg}}$ outside the other components of $S$, and a coordinate polydisc $U$ centered at it in which $D_0\cap U=S\cap U=\{ h=0\} $ for one of the coordinate functions $h$; then $D_0\cap U$ is a connected manifold. Let $A\subset M$ be a nowhere dense analytic subset such that $f^{-1}(A)$ is nowhere dense and $f$ is an isomorphism over $M\setminus A$. The function $h\circ f$ does not vanish identically on any connected component $V$ of $f^{-1}(U)$: otherwise $f$ would map the nonempty open set $V\setminus f^{-1}(A)$ isomorphically onto an open subset of $M$ contained in the hypersurface $D_0$, which is impossible. By proposition 9.5.6, $E=f^{-1}(D_0\cap U)$ is pure of dimension $n-1$.

Let $K\subset D_0\cap U$ be a compact neighborhood of the chosen point in $D_0$. Only finitely many irreducible components $E_1,\ldots ,E_ k$ of $E$ meet the compact set $f^{-1}(K)$, and by theorem 9.7.2 each image $f(E_ j)$ is an analytic subset of $D_0\cap U$. If its dimension is $n-1$, then $f(E_ j)$ contains an open subset of the connected manifold $D_0\cap U$ and therefore equals it, since proper analytic subsets of connected manifolds are nowhere dense (lemma 9.2.8); we then call $E_ j$ dominating. Otherwise $f(E_ j)$ has dimension at most $n-2$.

For a dominating $E_ j$, the maximal rank of $f$ restricted to the connected manifold $(E_ j)_{\mathrm{reg}}$, as a map into the coordinate hyperplane $D_0\cap U$, is $n-1$. Otherwise $f((E_ j)_{\mathrm{reg}})$ would be $\mathcal H^{2n-2}$-null by lemma 9.3.15(1); so would $f((E_ j)_{\mathrm{sing}})$, by the same lemma, since $(E_ j)_{\mathrm{sing}}$ has dimension at most $n-2$ by corollary 9.3.13(1) and is therefore, by (2) of the same corollary, a finite union of complex submanifolds of dimension at most $n-2$; and $f(E_ j)$ would have dimension at most $n-2$ by lemma 9.3.15(2).

Let $R_ j$ be the union of $(E_ j)_{\mathrm{sing}}$ and of the regular points where this rank is less than $n-1$. It is analytic. Indeed, near a point of $E_ j$, let $e$ be a local equation of the irreducible hypersurface $E_ j$ as in lemma 9.3.19, and let $f'$ be the map formed by the $n-1$ coordinates of $f$ in the coordinate hyperplane $D_0\cap U$. At a regular point $p$ of $E_ j$ we have $\mathrm de(p)\neq 0$ and $T_ pE_ j=\ker \mathrm de(p)$, so the rank of $f'|_{E_ j}$ at $p$ is $n-1$ exactly when the Jacobian determinant of $(e,f')$ does not vanish at $p$. Hence $R_ j$ is, near this point, the union of the analytic set $(E_ j)_{\mathrm{sing}}$ and of the zero set in $E_ j$ of this Jacobian determinant. It is different from $E_ j$, because the rank $n-1$ is attained on $(E_ j)_{\mathrm{reg}}$; so it is nowhere dense in the irreducible set $E_ j$ and has dimension at most $n-2$, by corollary 9.3.17(2), and $f(R_ j)$ is analytic of dimension at most $n-2$, by theorem 9.7.2 and lemma 9.3.15(3).

Choose a point $a$ in the interior of $K$ outside the finitely many sets $f(E_ j)$, $E_ j$ not dominating, and $f(R_ j)$, $E_ j$ dominating. Let $p\in f^{-1}(a)$. Since $f^{-1}(a)\subset E\cap f^{-1}(K)$, the point $p$ lies on some of the components $E_1,\ldots ,E_ k$. Each of them that contains $p$ is dominating, and $p\notin R_ j$, so $f|_{E_ j}$ is a local biholomorphism at $p$ onto an open subset of $D_0$, by proposition 5.2.4(1) in charts, and $p$ is isolated in $f^{-1}(a)\cap E_ j$. The components of $E$ not containing $p$ form a locally finite family of closed sets, so near $p$ the set $f^{-1}(a)$ lies in the union of the components containing $p$, and $p$ is isolated in $f^{-1}(a)$. Thus $f^{-1}(a)$ is discrete, and being compact, it is finite.

Let $p_1,\ldots ,p_ l$ be the points of $f^{-1}(a)$, and choose disjoint coordinate balls $B_1,\ldots ,B_ l$ around them. By properness there is a neighborhood $U_0$ of $a$ with $f^{-1}(U_0)\subset \bigcup _ iB_ i$. The fibers of $f$ over $U_0$ are compact analytic subsets of these balls, hence finite by corollary 9.5.3. Thus $f$ is finite and bimeromorphic over the manifold $U_0$, which is normal, and it is an isomorphism over $U_0$ by corollary 9.8.2. In particular $f$ has no critical points over $U_0$, which contradicts $a\in S$. Therefore $S$ has codimension at least two.

Step 3: strict transforms. Let $D$ be an irreducible hypersurface. Then $S\cap D$ has dimension at most $n-2$, so $D\setminus S$ is dense in $D$ by corollary 9.3.17(2). The closure $D'$ of $f^{-1}(D\setminus S)=f^{-1}(D)\setminus f^{-1}(S)$ is analytic by lemma 9.3.7, and $f$ is an isomorphism from $D'\setminus f^{-1}(S)=f^{-1}(D\setminus S)$ onto $D\setminus S$. The set $f(D')$ is closed, because $f$ is proper, it contains the dense subset $D\setminus S$ of $D$, and it is contained in the closed set $D$; so $f(D')=D$.

The set $Q=f^{-1}(D_{\mathrm{reg}}\setminus S)$ is a connected manifold of dimension $n-1$, because $D_{\mathrm{reg}}\setminus S$ is connected by lemma 9.2.9; it is open in $D'$ and dense in $D'$, since $D_{\mathrm{reg}}\setminus S$ is open and dense in $D\setminus S$. Hence $Q$ consists of regular points of $D'$ and lies in a single connected component of $(D')_{\mathrm{reg}}$, whose closure, an irreducible component of $D'$ by corollary 9.3.16, contains $\overline Q=D'$; so $D'$ is irreducible, of dimension $n-1$. Finally, if $S_1$ is another analytic subset of codimension at least two off which $f$ is an isomorphism, then $D\setminus (S\cup S_1)$ is dense in $D\setminus S$ and in $D\setminus S_1$, by corollary 9.3.17(2), so $f^{-1}(D\setminus (S\cup S_1))$ is dense in both $f^{-1}(D\setminus S)$ and $f^{-1}(D\setminus S_1)$, and the two closures agree. $\square$

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. $\square$

Proof. After reducing to the case of a manifold, we blow up the image of the divisor in $X$ as long as it has codimension at least two; each step lowers the order of vanishing along the divisor of the Jacobian determinant of the model.

Step 1: reduction to a manifold. Let $\widehat X\to X$ be a resolution given by theorem 9.6.6, a finite sequence of blowups with smooth centers, since $X$ is compact. Let the prime divisor be represented by $(M_0,F_0)$, and let $M$ be a common smooth model of $M_0$ and $\widehat X$, with $E\subset M$ the strict transform of $F_0$ (proposition 9.8.3); then $(M,E)$ represents the same divisor, as we see on the common smooth model $M$. A finite sequence of blowups with smooth centers over $\widehat X$, composed with $\widehat X\to X$, is such a sequence over $X$, and smooth models of $\widehat X$ are smooth models of $X$, by corollary 9.7.3. Hence we may assume that $X$ is a compact complex manifold, and we are given a proper bimeromorphic map $f\colon M\to X$ from a compact manifold and an irreducible hypersurface $E\subset M$.

Since $E$ is irreducible, hence connected, it lies in a connected component $M_ E$ of $M$, and $f(M_ E)$ lies in a connected component $X_ E$ of $X$. These components have the same dimension $n$, because $M_ E$ contains a nonempty open set mapped isomorphically onto an open subset of $X$ (lemma 9.4.1); moreover $n\geq 1$ and $\dim E=n-1$, since $E$ is a nonempty hypersurface. If $n=1$, the center $f(E)$ considered below is a point of $X_ E$, hence a hypersurface.

Step 2: the order of the Jacobian. The Jacobian determinant $J_ f$ of $f$ in local coordinates is holomorphic and determined up to multiplication by holomorphic units. It does not vanish identically near any point, because $f$ is an isomorphism over a dense open subset of $X$ whose inverse image is dense in $M$ (lemma 9.4.1). Its order of vanishing $\operatorname {ord}_ EJ_ f$ along $E$ is the largest $k$ such that $J_ f$ is divisible by the $k$th power of a local equation of $E$ near points of $E_{\mathrm{reg}}$; definition 6.12.3 uses the same notion of order along a hypersurface. It is a finite nonnegative integer $k$, since a function divisible by all powers of a local coordinate $y$ has vanishing Taylor coefficients in $y$ and vanishes near the point. It does not depend on the point of $E_{\mathrm{reg}}$, which is a connected manifold by corollary 9.3.17: for each integer $j$, the set of points of $E_{\mathrm{reg}}$ at which the order is at least $j$ is open, and it is closed by the identity theorem on $E_{\mathrm{reg}}$, since a holomorphic function vanishing on $E$ near a regular point is divisible by a local equation of $E$.

For composable maps the chain rule gives

because orders of vanishing along $E$ add under multiplication. Replacing $M$ by a common smooth model $h\colon M'\to M$ and $E$ by its strict transform $E'$ does not change $k$: by proposition 9.8.3, $h$ is an isomorphism near generic points of $E'$, so $\operatorname {ord}_{E'}J_ h=0$ and $\operatorname {ord}_{E'}(J_ f\circ h)=k$.

Step 3: one principalization lowers the order. The center $Z=f(E)$ is an analytic subset of $X$ by theorem 9.7.2, of dimension at most $\dim E=n-1$ by lemma 9.3.15(3). It is irreducible: if $Z=Z_1\cup Z_2$ with analytic subsets $Z_1,Z_2$, then the irreducible set $E$ is the union of its analytic subsets $E\cap f^{-1}(Z_1)$ and $E\cap f^{-1}(Z_2)$, so it lies in one of them, and $Z$ equals $Z_1$ or $Z_2$. The same argument shows that images of irreducible analytic sets under proper holomorphic maps are irreducible. Suppose first that $Z$ is a hypersurface, and let $S\subset X$ be as in proposition 9.8.3 for $f$. The strict transform $Z^ M$ of $Z$ in $M$ is an irreducible hypersurface containing $f^{-1}(Z\setminus S)$. The set $E\cap f^{-1}(Z\setminus S)$ is open in $E$, and it is nonempty, because $f(E)=Z$ is not contained in $S$, which has codimension at least two. So $E\cap Z^ M$ is an analytic subset of $E$ with interior points in $E$, and $E\subset Z^ M$ by corollary 9.3.17(2); then $E=Z^ M$ by corollary 9.3.17(3). Hence $(X,Z)$ represents the given divisor, as we see on the common smooth model $M$.

Suppose now that $Z$ has codimension at least two. Its ideal $\mathcal I_ Z$ is coherent by theorem 9.3.10, and it is not identically zero on any component of $X$, since $Z$ is nowhere dense. Let $\sigma =\sigma _1\circ \cdots \circ \sigma _ t\colon X'\to X$ be the finite sequence of blowups with centers $C_ i\subset X_{i-1}$ and transformed ideals $\mathfrak a_ i$, $\mathfrak a_0=\mathcal I_ Z$, given by theorem 9.6.7; it is proper and bimeromorphic by corollary 9.7.3, so $X'$ is a smooth model of $X$. Let $h\colon M'\to M$ and $f'\colon M'\to X'$ be the maps from a common smooth model of $M$ and $X'$, so that $f\circ h=\sigma \circ f'$, let $E'$ be the strict transform of $E$, let $\rho _ i\colon M'\to X_ i$ be the induced maps, and let $Z_ i=\rho _ i(E')$; thus $Z_0=Z$.

Let $r\geq 2$ be the codimension of $Z$, which is pure by corollary 9.3.17(1). We claim that for some $i$ the center $C_ i$ contains $Z_{i-1}$, and that the connected component of $C_ i$ containing $Z_{i-1}$ has codimension $r$. The sets $Z_ j$ are irreducible, as images of the irreducible set $E'$, and $\sigma _{j+1}(Z_{j+1})=Z_ j$. Put $\tau _ j=\sigma _1\circ \cdots \circ \sigma _ j$. Suppose that $Z_{l-1}\not\subset C_ l$ for all $l\leq j$. We show by induction on $j$ that there are a nowhere dense analytic subset $A_ j\subset Z$ and an open set $N_ j\subset X\setminus A_ j$ containing $Z^\circ _ j=Z\setminus A_ j$ such that $\tau _ j$ is an isomorphism over $N_ j$ and, on $\tau _ j^{-1}(N_ j)$,

For $j=0$ take $A_0=\varnothing $ and $N_0=X$.

Assume the assertion for some $j<t$, and suppose that $Z_ j\not\subset C_{j+1}$. On $\tau _ j^{-1}(N_ j)$ we have $C_{j+1}\subset V(\mathfrak a_ j)=\tau _ j^{-1}(Z^\circ _ j)\subset Z_ j$. By corollary 9.3.17(2), $C_{j+1}\cap Z_ j$ is nowhere dense in $Z_ j$. Its image $\tau _ j(C_{j+1}\cap Z_ j)\subset Z$ is analytic by theorem 9.7.2, and it is nowhere dense in $Z$: its part in $Z^\circ _ j$ corresponds under the isomorphism $\tau _ j$ to the nowhere dense part of $C_{j+1}\cap Z_ j$ in the open subset $\tau _ j^{-1}(Z^\circ _ j)$ of $Z_ j$, and $A_ j$ is nowhere dense. Let $A_{j+1}=A_ j\cup \tau _ j(C_{j+1}\cap Z_ j)$ and $N_{j+1}=N_ j\setminus A_{j+1}$. Then $\tau _ j^{-1}(N_{j+1})$ does not meet $C_{j+1}$, so $\sigma _{j+1}$ is an isomorphism over it and does not change the ideal there: $\mathfrak a_{j+1}=\mathcal O_{X_{j+1}}(E_{j+1})\mathfrak a_ j\mathcal O_{X_{j+1}}=\mathcal I_ Z\mathcal O_{X_{j+1}}$ on $\tau _{j+1}^{-1}(N_{j+1})$. Moreover $Z_{j+1}\cap \tau _{j+1}^{-1}(N_{j+1})=\sigma _{j+1}^{-1}(Z_ j\cap \tau _ j^{-1}(N_{j+1}))=\tau _{j+1}^{-1}(Z^\circ _{j+1})$, because $\sigma _{j+1}(Z_{j+1})=Z_ j$ and $\sigma _{j+1}$ is injective there. This completes the induction.

Since $A_ j$ is nowhere dense in $Z$, the set $Z^\circ _ j$ is nonempty, and so is $\tau _ j^{-1}(Z^\circ _ j)$. If $Z_{l-1}\not\subset C_ l$ held for all $l\leq t$, the induction would reach $j=t$, and $\mathfrak a_ t$ would be the ideal of the nonempty set $\tau _ t^{-1}(Z^\circ _ t)$ on $\tau _ t^{-1}(N_ t)$, whereas $\mathfrak a_ t$ is the unit ideal. So there is a least index $i$ with $Z_{i-1}\subset C_ i$, and the assertion holds for $j=i-1$. On $\tau _{i-1}^{-1}(N_{i-1})$ we have $Z_{i-1}\subset C_ i\subset V(\mathfrak a_{i-1})=\tau _{i-1}^{-1}(Z^\circ _{i-1})\subset Z_{i-1}$, so $C_ i$ coincides there with a copy of the open subset $Z^\circ _{i-1}$ of $Z$, which is pure of codimension $r$. The set $Z_{i-1}$ is connected, being the closure of the connected manifold $(Z_{i-1})_{\mathrm{reg}}$, so it lies in one connected component of the manifold $C_ i$, and this component has codimension $r$.

By (9.6.4) and proposition 9.6.3, near points over this component the Jacobian determinant of $\sigma _ i$ is, in any local coordinates, a unit times the $(r-1)$st power of a local equation of the exceptional divisor $E_ i=\sigma _ i^{-1}(C_ i)$. Since $Z_ i=\rho _ i(E')$ lies over $Z_{i-1}\subset C_ i$, the function $J_{\sigma _ i}\circ \rho _ i$ vanishes on $E'$; it does not vanish identically near any point, because $J_{\sigma _ i}$ vanishes only on $E_ i$, and $\rho _ i^{-1}(E_ i)$ is nowhere dense, $\rho _ i$ being proper and bimeromorphic as the composite of $f'$ with $\sigma _{i+1}\circ \cdots \circ \sigma _ t$ (corollary 9.7.3). Hence it has positive order along $E'$. The Jacobians of the other blowups composed with the corresponding maps $\rho _ j$ are holomorphic, so they have nonnegative order, and by the chain rule $J_\sigma \circ f'=\prod _ j(J_{\sigma _ j}\circ \rho _ j)$. Thus

Applying (9.8.7) to $f\circ h=\sigma \circ f'$ and using Step 2, we obtain

Step 4: conclusion. The pair $(M',E')$ represents the given prime divisor, as we see on the common smooth model $M'$, and $X'$ is a compact manifold. We repeat Step 3 with $X'$, $f'$ and $E'$ in place of $X$, $f$ and $E$. Each repetition in which the center has codimension at least two lowers the nonnegative integer $k$ by at least one, so after at most $k$ repetitions the center is a hypersurface on the current model, which is obtained from $X$ by a finite sequence of blowups with smooth centers, and it represents the given prime divisor. Finally, by lemma 9.8.5(2), principalizing the ideal of this hypersurface by theorem 9.6.7 gives a further sequence of blowups with smooth centers, finite because the model is compact, on which the strict transform of the hypersurface is smooth and represents the same prime divisor. $\square$