Proposition 9.5.1A3A9384DC317

Proof. Embed $C$ in a local model and apply theorem 9.3.4. The base is a disc $\Delta '=\{ |z|<r\} $, and since the zeros of $\delta $ are isolated, we may take the parametrization polydisc so small that $\delta $ has no zero in $\Delta '\setminus \{ 0\} $. Let $C$ be the representative in this polydisc and $C^*=C\setminus \{ x\} =\pi ^{-1}(\Delta '\setminus \{ 0\} )$. Then $C^*$ is a covering of the punctured disc $\Delta '\setminus \{ 0\} $ with $q$ sheets, and it is connected: if $\delta (0)=0$, it is the covering locus, and if $\delta (0)\neq 0$, then $q=1$, because the fiber over $0$ is $\{ x\} $. We shall show that lifting a circle around the puncture permutes the $q$ points of a fiber cyclically, and then obtain $\gamma $ by lifting the exponential map to $C^*$.

We first recall two facts about coverings. A path in $\Delta '\setminus \{ 0\} $ lifts uniquely to $C^*$ once its initial point is prescribed: subdivide the path into finitely many pieces lying in discs over which the covering is trivial, and lift piece by piece. If two paths with the same endpoints are homotopic with fixed endpoints, their lifts with the same initial point have the same endpoint. Indeed, let $h\colon [0,1]^2\to \Delta '\setminus \{ 0\} $ be the homotopy, with $h(0,s)$ and $h(1,s)$ independent of $s$, and let $e(s)$ be the endpoint of the lift of $h(\cdot ,s)$ with the prescribed initial point. By uniform continuity there is an integer $m\geq 1$ such that $h$ maps each square $Q_{jl}=[(j-1)/m,j/m]\times [(l-1)/m,l/m]$ into a disc $D_{jl}$ over which the covering is trivial. A lift of a path in $D_{jl}$ stays in the sheet over $D_{jl}$ that contains its initial point, because these sheets are open and disjoint.

Fix $l$ and $s,s'\in [(l-1)/m,l/m]$. We show by induction on $j$ that the lifts of $h(\cdot ,s)$ and $h(\cdot ,s')$ over $[(j-1)/m,j/m]$ lie in one sheet over $D_{jl}$. For $j=1$ they start at the same point. If this holds for some $j<m$, the points of the two lifts at the parameter $j/m$ are joined by the lift, inside this sheet, of the path $h(j/m,\cdot )$ between $s$ and $s'$; this lift is a connected set lying over $D_{jl}\cap D_{j+1,l}$, so it lies in one sheet over $D_{j+1,l}$, which then contains both lifts over $[j/m,(j+1)/m]$. For $j=m$ we obtain two points in one sheet over $D_{ml}$ lying over the same point $h(1,s)=h(1,s')$, so $e(s)=e(s')$. Hence $e$ is constant on each interval $[(l-1)/m,l/m]$, and therefore on $[0,1]$.

Next, every loop based at a point $b\in \Delta '\setminus \{ 0\} $ is homotopic with fixed endpoints to the $k$th power of the circle around $0$ through $b$, for some integer $k$. Indeed, lifting the argument continuously along the loop, we write the loop as $\exp $ of a path in the half-plane $H=\{ \zeta :\operatorname{Re}\zeta <\log r\} $ from some $\zeta _0$ to $\zeta _0+2\pi \mathrm ik$; the straight-line homotopy to the segment from $\zeta _0$ to $\zeta _0+2\pi \mathrm ik$ stays in the convex set $H$. Let $\sigma $ be the permutation of the fiber over $b$ obtained by lifting the circle. Lifting any loop at $b$ therefore induces a power of $\sigma $. Since $C^*$ is connected, any two points of the fiber over $b$ are joined by a path in $C^*$, whose image is a loop at $b$; hence the powers of $\sigma $ act transitively on the $q$ points of the fiber, and $\sigma $ is a cycle of length $q$.

The map $\exp \colon H\to \Delta '\setminus \{ 0\} $ lifts to a map $\Phi \colon H\to C^*$ with $\pi \circ \Phi =\exp $: lift along paths from a base point $\zeta _0$, starting at a chosen point of the fiber; since $H$ is convex, the endpoint does not depend on the path. The map $\Phi $ is holomorphic, because locally it is $\exp $ followed by the inverse of $\pi $ on a sheet. The vertical segment from $\zeta $ to $\zeta +2\pi \mathrm i$ is mapped by $\exp $ onto the circle through $\mathrm e^\zeta $; so $\Phi (\zeta +2\pi \mathrm i)=\sigma _\zeta (\Phi (\zeta ))$, where $\sigma _\zeta $ is the permutation of the fiber over $\mathrm e^\zeta $ obtained by lifting this circle. By the preceding paragraph, applied with $b=\mathrm e^\zeta $, it is a cycle of length $q$; hence $\Phi (\zeta +2\pi \mathrm iq)=\Phi (\zeta )$. Therefore

does not depend on the branch of the logarithm; it is holomorphic, and $\pi (\gamma (t))=t^ q$. The coordinates of $\gamma $ are bounded, because $C$ lies in the bounded parametrization polydisc, so by the removable singularity theorem they extend holomorphically to $t=0$. Since $\pi $ is proper, $\gamma (t)$ stays in a compact subset of $C$ as $t\to 0$; hence $\gamma (0)\in C$, and $\gamma (0)=x$ because $\gamma (0)$ lies over $0$.

Let $\Delta =\{ |t|<r^{1/q}\} $. The map $\gamma \colon \Delta \to C$ is surjective: the points of $C$ over $z_0=\mathrm e^{\zeta }\neq 0$ are $\Phi (\zeta +2\pi \mathrm ik)=\gamma (\mathrm e^{(\zeta +2\pi \mathrm ik)/q})$, $0\leq k<q$, because $\sigma _\zeta $ is a $q$-cycle. It is injective: if $\gamma (t_1)=\gamma (t_2)$ with $t_1,t_2\neq 0$, then $t_1^ q=t_2^ q$, so $t_2=t_1\mathrm e^{2\pi \mathrm ik/q}$; writing $q\log t_1=\zeta $, we get $\gamma (t_2)=\Phi (\zeta +2\pi \mathrm ik)=\sigma _\zeta ^ k(\gamma (t_1))$, which forces $k\equiv 0$ modulo $q$. On $\Delta \setminus \{ 0\} $ the map $\gamma $ is a local biholomorphism into the manifold $C^*$, since $\pi \circ \gamma (t)=t^ q$ has nonzero derivative; being bijective, it is a biholomorphism of $\Delta \setminus \{ 0\} $ onto $C\setminus \{ x\} $.

Finally, let $X$ be a normal complex space of dimension one, and let $x\in X$; if $x$ is an isolated point, there is nothing to prove. Otherwise the germ $(X,x)$ has dimension one, and it is irreducible by the definition of normal points; let $\gamma \colon \Delta \to C$ be the map just constructed for a representative $C$ of this germ. Its inverse is holomorphic on the manifold $C\setminus \{ x\} $, which is contained in the regular locus, and it is bounded; so it is weakly holomorphic and, by normality, extends to a holomorphic function $\tau $ near $x$. Then $\tau \circ \gamma =\mathrm{id}$ on $\Delta \setminus \{ 0\} $, hence on $\Delta $, and $\gamma \circ \tau =\mathrm{id}$ on $C\setminus \{ x\} $, hence on $C$, because these maps are continuous and the punctured sets are dense. Thus $\gamma $ is an isomorphism of $\Delta $ onto a neighborhood of $x$, and $x$ is a regular point. $\square$