Is $\{(x,y,z,0)\in\mathbb{R}^4:x^4+y^2+z^2=1\}$ diffeomorphic to $S^2$?

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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.

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Since they both are smooth convex sets, the "central projection" maps: $$\pi:C\to S,\qquad \pi(x,y,z)=\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}},$$ $$\pi^{-1}:S\to C,\qquad \pi^{-1}(x,y,z)=\sqrt{\frac{2}{(1-x^2)+\sqrt{1-2x^2+5x^4}}}\;(x,y,z)\tag{1}$$ are differentiable as a consequence of the Dini's theorem, since the "scale ratio" function is the inverse of a smooth map. For instance, the second "scale ratio" is found by solving $$ k^4 x^4 + k^2 (1-x^2) = 1.$$ It is not too difficult to check "by hand", too, that the two maps given in $(1)$ are smooth.

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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.