A branched projection of a smooth hypersurface

The germ $W=y^2-x$ is irreducible in ${\mathcal{O}}_{{\mathbb{C}}^2,0}$. Indeed, by [F1] it is a Weierstrass polynomial of degree $2$; if $W=GH$ with $G,H$ Weierstrass polynomials of positive degree, then both have degree $1$, so $G=y-a(x)$ and $H=y-b(x)$ with $a,b\in {\mathcal{O}}_{\mathbb{C},0}$; comparing coefficients gives $a+b=0$ and $ab=-x$, hence $a^2=x$, contradicting [F3] because the order of $a^2$ is even and that of $x$ is $1$. Therefore $W$ is irreducible in the polynomial ring and, by [F2], in the germ ring; in particular $W$ is reduced, since a germ divisible by the square of an irreducible germ is a product of two nonunits.

Every point of $X$ is regular. One has $dW=(-1,2y)\mathrm{\ne }0$ at every point. Its local germ is reduced: a squared nonunit factor would make both the value and every first derivative vanish at that point by the product rule. Hence [F6] applies to $W$ and every point is regular; the singular locus of $X$ is empty. Equivalently $X$ is the graph $x=y^2$.

In the product of [F5], every slice $y^2=x$ has both roots in $D$, because $\mathrm{\mid }y{\mathrm{\mid }}^2=\mathrm{\mid }x\mathrm{\mid }<\epsilon <r^2$; thus the fixed projection is surjective. For a compact $K\subset V$, its preimage is the closed bounded subset $\{(x,y):x\in K,y^2=x\}$ of ${\mathbb{C}}^2$, entirely inside $V\times D$, and is compact by [F5]. Thus the projection is proper. At $x\mathrm{\ne }0$ its two roots are distinct and ${\mathrm{\partial }}_{y}W=2y\mathrm{\ne }0$; the implicit-function theorem of [F5] gives two disjoint local holomorphic sheets. They exhaust each nearby fibre, since every slice has exactly two roots. Finally [F4] gives $D_W=0^2-4(-x)=4x$, so [F5] gives $B_{\pi }=\{0\}$.

Thus $B_{\pi }=\{0\}$ is nonempty while $\mathrm{Sing}(X_W)=\mathrm{\varnothing }$ by step 2.1. The projection is branched over $0$ because the two roots of the slice coincide there by [F4], although its fibre is the regular point $(0,0)$. Hence $B_{\pi }\mathrm{\ne }\pi (\mathrm{Sing}(X_W))$, refuting the claim.