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

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? ($$S=\{(a, b, c)\in \mathbb{R}^3: f(a, b, c)=a\}\; \text{with}\; a>3$$)

• This question is very unfocussed. There are lots of steps on this page and I would not wish to guess which of these many steps is dis-satisfactory to you. Mar 21, 2021 at 14:09
• Just that I've asked, this is: how to prove that $\phi=f:S\rightarrow S^2$ is bijective (whithout using $h$)? Mar 21, 2021 at 14:12
• Then you should rewrite your question to state, right at the beginning, exactly what you want. As currently written, we have to scroll through a long attachment, and then we have to figure out that what you are really asking is buried in what looks like a parenthetical comment. Mar 21, 2021 at 15:34
• Quite a coincidence that this question showed up yesterday. Mar 21, 2021 at 18:49
• "I need your help in 12 hours." It sure sounds like this is an exam question that you've found in a book. I didn't downvote, but I would normally vote to close. Mar 21, 2021 at 19:12

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.
• Thank you very much for your answer. This solved my doubt related to understand $h$. Congratulations for your gift in have an easy geometric view of these kind of problems. Sure, I will vote your answer! =D Mar 21, 2021 at 19:56
• PS. If you have any idea in how to prove this bijection without defining that $h$, please let me know. Because I tried in another way,but I failed. (Thanks for everything). Mar 21, 2021 at 19:58