Is a map from a closed surface to 3-space differentiable at the pre-image of a branch point?

Question

Let $S$ be a closed surface and let $f$ be a map from $S$ to $\mathbb{R}^3$ such that $f(S)$ is a generic surface. The singularity set of $S$ contains double points, triple points and branch points. The branch point has a neighborhood homeomorphic to a cone over a figure eight. Is the map $f$ differentiable at the pre-image of a branch point? I think no, because we can not assign an orientation normal vector to a branch point, right?

Answer 1

Consider $$ f: \Bbb S^2 \to \Bbb R^3 : (x, y, z) \mapsto (x^3, y, z) $$ The domain of $f$ is a surface; the image of $f$ is also a surface. The singularity set of $f$ contains all points of the sphere where $x = 0$, because the rank of the differential at those points is $1$. So the claim that the singularity set of $S$ consists of double, triple and branch points is mistaken.

If that statement -- that the singularity set consists of those things -- was meant to be an additional hypothesis, the answer is still "no". The Steiner surface, given by $$ f(x, y, z) = (xy, yz, zx) $$ on the unit sphere, is clearly differentiable at every point (it's polynomial!), but the image surface contains several points whose neighborhoods look like a cone on a figure-8.