How to prove that $\phi=f:S\rightarrow \mathbb{S}^2$ is a bijection without using $h$?

Question

If $f:\mathbb{R}^3\rightarrow \mathbb{R}$ given by $f(a,b,c)=e^{a^2}+e^{b^2}+e^{c^2}$ defining $h:\mathbb{R}\rightarrow \mathbb{R}$ by $h(t)= f(ta, tb, tc)$, we have that $h'(t)>0$ give us all conditions to define a bijection (diffeomorphism) $\phi=f:S\rightarrow \mathbb{S}^2$. Would someone explain this bijection please? And also help me to find another way to prove this? ($S=\{(a, b, c)\in \mathbb{R}^3: f(a, b, c)=a\}\; \text{with}\; a>3$)

Answer 1

It's a slightly confusingly-worded proof, but the essential idea is to show that for every $(x,y,z)\ne 0$, the half line $\{ t(x,y,z) \mid t\ge 0\}$ intersects the surface $S$ at precisely one point. This is what the argument with $h$ shows: since $h(t)$ ranges from 3 to $+\infty$ as $t$ ranges from $0$ to $+\infty$, it must equal $a$ for at least one value of $t$. Since $h'(t)>0$, $h$ is monotonically increasing, and so it equals $a$ for at most one value of $t$. Therefore $h(t)$ equals $a$ for $precisely$ one value of $t$. For this value $t^*$ of $t$, $t^*(x,y,z)$ is an element of $S$.

Now the half line $\{ t(x,y,z) \mid t\ge 0\}$ also intersects the sphere $\mathbb{S}^2$ at precisely one point (obviously). The mapping $f:S\to \mathbb{S}^2$ given by $f(p) = \frac{p}{|p|}$ maps the point in the half line intersecting $S$ to the point intersecting $\mathbb{S}^2$, and so is injective. It's also surjective - take any $(x,y,z)\in \mathbb{S}^2$ and apply the above argument to find a $t^*(x,y,z)\in S$. Then $f(t^*(x,y,z)) = (x,y,z)$. Therefore $f$ is bijective.