For a domain $U=\{\, (u,v) \in \mathbb{R}^2 \mid -\pi<u<\pi,\ 0<v<1 \,\}$ we have a mapping $X \colon U \to \mathbb{R}^3$ defined by $X(u,v) = (\sin u, \sin 2u, v)$.
The resulting surface is quite straightforward and looks like an infinity symbol when intersected with planes orthogonal to $z$-axis. See Wolframalpha.
This mapping is also injective - that is noticeable when letting $u$ gain values from $-\pi$ to $\pi$.
But what is an inverse continuous map $X^{-1} \colon X(U) \to U$ then?
If $\sin u,\sin 2u\geq 0$ we must have $u\in[0,0.5\pi]$. If $\sin u\geq0\geq\sin2u$ we have $u\in(0.5\pi,\pi)$. Similarly, if $\sin u,\sin 2u\leq 0$ we must have $u\in[-0.5\pi,0]$ and so on. So the inverse appears to be given as $$ X^{-1}(x,y,z)= \begin{cases} (\sin^{-1}x,z) &\text{ if }x\cdot y\geq 0\\ (\operatorname{sgn} x\cdot\pi-\sin^{-1}x,z) &\text{ otherwise.} \end{cases} $$