Let $M\subset\mathbb{R}^3$ be a (compact) convex body and denote $\pi:(x, y, z)\to(x,y)$ the projection to the $xy$-plane. The image $\pi(M)$ is a convex shape on $\mathbb{R}^2$, and has a boundary $B=\partial\pi(M)$. Is it true that:
There is a continuous curve $C\subset \pi^{-1}(B)\cap M$ such that the projection $\pi |_C: C\to B$ is a homeomorphism?
I have a feeling that this is the case, but couldn't prove it, or couldn't find any online statements. Could anybody prove or disprove? I tried to construct counter-examples but either did not find any one.
Here is a counterexample. Let $C_1 = \{ (x,\sqrt{1-x^2},1) \, : \, -1 \le x \le 1 \}$ and let $C_2 = \{(x,-\sqrt{1-x^2},0) \, : \, -1 \le x \le 1\}$. Let $M$ be the convex hull of $C_1 \cup C_2$. Then $\pi(M)$ is the unit disk and $B$ is the unit circle. If a point $(x,y) \in B$ has $y > 0$, then the only pre-image lies in $C_1$, and if $y < 0$, then the only pre-image lies in $C_2$. Therefore, there cannot be a curve $C$ in $M$ such that $\pi$ restricted to $C$ is a homeomorphism onto $B$.