Let $(f(t),0,g(t)),t\in [0,L]$ be a smooth curve satisfying $f(0)=f(L)$.Denote the solid formed by revolving the curve around the $z$-axis as (by) $M$. Suppose $G:M\to \Bbb{S}^2$ is a continuous map mapping every $u\in M$ to a unit normal vector to $M$.
Is $G$ injective?
Is $G$ surjective?
Remark: a unit normal to hypersurface $M$ at $x$ is a vector $v\in (T_x{M})^{\bot}$ such that $|v|=1$,($T_xM$ being the tangent space of M at x).
I am currently stuck showing it is not necessarily injective but I am not quite sure I know what to do. I had some ideas referring me to repetitive patterns in the original curve. For example,I wondered what if $M$ contains two vectors that share the same tangent space and what it can say. But eventually, I cannot presume $G$ maps $x$ to a normal correspondent to that orthogonal to $TxM$. In other words, I am quite lost here. Could you help me out?
A counterexample for injectivity for a closed curve is the torus, which may be given by revolving a circle that does not intersect the $z$-axis around the $z$-axis. Given any direction, there are two places with a parallel normal vector (on opposite sides). Another example for a non-closed curve is to take $g(t)=t$, $f(t)$ a quadric with several extrema ($t(L-t)(1-3t(L-t))$ will do), and then you have multiple places where the tangent is parallel.
Surjectivity is related to smoothness of the surface: if the surface formed is smooth at every point, given any plane, you can slide it towards the surface until it is tangent at a point $P$, say, and then $P$ has the required normal. To show smoothness is required, one can consider the lemon, given by revolving a smaller-than-semicircular arc: it is easy to see that the normals to the curve only cover an annulus on the sphere, missing regions near the poles.