Prove $φ(R^2)$ where $φ(u,v)=(v\cos u,v\sin u,bu)$ is the smooth function.

Question

Prove $φ(R^2)$ is a smooth surface and $φ(u,v)=(v\cos u,v\sin u,bu)$ $R^2\rightarrow R^3$ and b>0 constant. Its rank $Dφ$ is 2, so I'm ok with that part. Only thing to prove is that it has an continuous inverse so $φ$ can be an acceptable parametrization. Can i just argue the topology definition . I know the existence of the inverse hence if the pre image of the inverse of open sets of$ R^2$ are open in $R^3$ then the inverse is continuous.But still that needs proof

Answer 1

Let $M := \varphi(\Bbb R^2)$. Then $\varphi : \Bbb R^2\to M$ is obviously onto and you proved that it is also injective. Furthermore, the rank of the Jacobian is $2$ in each point (which you also proved). Now, the inverse of $\varphi$ is given by $$ \psi(x,y,z) = \left(\frac z b\,,\,x\cos\frac z b+y\sin\frac z b\right),\quad (x,y,z)\in M. $$ This is easily verifyable by computing $\psi(\varphi(u,v))$. So, $\varphi^{-1}$ is obviously continuous.