TITLE: Is $\{(x,y,z,0)\in\mathbb{R}^4:x^4+y^2+z^2=1\}$ diffeomorphic to $S^2$?
QUESTION [7 upvotes]: I was working on Problem 5-1 of Smooth Manifolds by Professor John Lee, and it lead me to wanting to show that 
$\{(x,y,z,0)\in\mathbb{R}^4:x^4+y^2+z^2=1\}$ is diffeomorphic to $S^2$, and that is supposed to be $x^4$, not $x^2$. 
The diffeomorphism I suspect is $(x,y,z,0)\mapsto (\operatorname{sgn}(x)x^2,y,z)$. This is the only way I could think of making the map bijective. But I'm having a hard time telling if the inverse is smooth.
I think the inverse map is $(a,b,c)\mapsto (\sqrt{a},b,c,0)$ if $a\geq 0$, and $(a,b,c)\mapsto (-\sqrt{-a},b,c,0)$ if $a\leq 0$. This both seem smooth away from zero, but I'm not sure about smoothness at the origin.

REPLY [6 votes]: Another way to map $S^2$ diffeomorphically to $W^2=\{x^4+y^2+z^2=1\}\subset\mathbb{R}^3$ is by
$$(x,y,z)\mapsto(x,\sqrt{1+x^2}\cdot y,\sqrt{1+x^2}\cdot z).$$
This map is obviously smooth with a smooth inverse, and it is well defined as shown immediately.