Proposition 9.8.32312A839D109

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. $\square$