Inverse function of $(x,y) \mapsto (\cos x \cosh y, \sin x \cosh y,y)$

Question

Consider the continuous function $$ F \colon \mathbb{R}^2 \rightarrow \mathbb{R}^3 : (x,y) \mapsto (\cos x \cosh y, \sin x \cosh y,y). $$ How can I find the inverse function $F^{-1} \colon D \subset \mathbb{R}^3 \rightarrow \mathbb{R}^2$? (My goal is to show that $F$ restricted on an appropriate domain $D$ is a homeomorphism, i.e. $F$ is bijective, $F$ and $F^{-1}$ are continuous.)

Answer 1

The map $F$ is neither injective nor surjective: It cannot be injective since it is periodic in $x$, and it cannot be surjective on dimensional grounds.

To get an inverse map nevertheless we restrict the domain to the strip $R:=\>]{-\pi},\pi[\>\times{\mathbb R}$ and put $F(R)=:S$, a surface in three-space: $$F:\>R\to S,\qquad(x,y)\mapsto(u,v,w):=(\cos x\cosh y,\sin x\cosh y,y)\ .$$ The map $F$ considered in this way is a homeomorphism between $R$ and $S$. The inverse map takes the form $$F^{-1}:\quad S\to R,\qquad(u,v,w)\mapsto(x,y):=\bigl({\rm Arg}(u,v), w\bigr)\ .$$ Here $${\rm Arg}:\>\bigl({\mathbb R^2}\setminus{\rm negative}\ x{\rm -axis}\bigr) \to\>]{-\pi},\pi[\>,\qquad (u,v)\mapsto {\rm Arg}(u,v)$$ is the principal value of the argument, in short: the "polar angle". If $u>0$ then ${\rm Arg}(u,v)=\arctan{v\over u}$.

Answer 2

$$(u,v,w)=(\cos x \cosh y, \sin x \cosh y,y)$$

can be inverted by

$$(x,y)=(\arctan\frac uv,w)$$

under the condition $\cosh w=\sqrt{u^2+v^2}$ but this does not define a bijection, unless you restrict the domain of $x$.