Is the projection of a smooth submanifold a submanifold?

Question

Given $S\subset \mathbb{R}^{n+1}$ a submanifold, i.e. for all $x\in S$ there exists an open $A\ni x$, open $B\ni 0$ in $\mathbb{R}^{n+1}$ and smooth diffeomorphism $f:A\rightarrow B$ such that $f(A\cap S)=(\mathbb{R}^d\times\{0\})\cap B$ for $d$ is the dimension of $S$, such that $(x_1,...,x_n,,x_{n+1}),(x_1,...,x_n,x_{n+1}')\in S\Rightarrow x_{n+1}=x_{n+1}'$, is the projection of $S$ onto the first $n$ coordinates also a submanifold?

Intuition tells me yes with the last condition used to prevent self-intersections, but I can't seem to be able to construct the diffeomorphism required, mostly because I can't explicitly write $x_{n+1}$ smoothly in terms of the other coordinates when on $S$. This seems like an elementary question too yet the literature says very little about projecting submanifolds.

Answer 1

No. Imagine a 2D sheet of paper floating perpendicularly above the $x$-axis of the plane and centered at the origin. Twist the lower end 90 degrees clockwise parallel to the $y$-axis, and the upper end 90 degrees counterclockwise parallel to the $y$-axis, leaving the vertical line above the origin fixed. The projection down of this looks something like two triangles joined at one vertex (at the origin).

Edit due to injectivity condition: Still don't think it suffices. Imagine an open line segment in $\mathbb{R}^3$ bent into something that projects onto the $(x,y)$-plane as a figure-eight, but where the cross-point in the projection onto the $(x,y)$-plane is below the 'holes' at the open endpoints, and one point on the interior of the line segment (i.e. the curve is arranged so the endpoint holes and this interior point all differ only in the $z$-coordinate, and the rest of the open line segment forms the remainder of the figure-eight).

Answer 2

The answer is most definitely no. Think of a sheet of paper twisting in $3$-space. The sheet is a $2$-manifold, and its projection will mostly also be a $2$-manifold (locally), but for some angles it will be reduced to a line/point.

EDIT: Added an image of what the projection typically looks like. The projection is the shaded cone.

Answer 3

Here's a simple counterexample. Let $S\subset\mathbb R^3$ be the curve parametrized by $\gamma(t) = (t^3, t^2, t)$. Its projection onto the $(x,y)$-plane is the cusp curve defined by $y^3 = x^2$, which is not a smooth submanifold. The projection is injective because $t_1^3 = t_2^3$ implies $t_1=t_2$.