3
$\begingroup$

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

$\endgroup$
7
  • 1
    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Mar 21, 2021 at 14:09
  • $\begingroup$ Just that I've asked, this is: how to prove that $\phi=f:S\rightarrow S^2$ is bijective (whithout using $h$)? $\endgroup$
    – Silvinha
    Mar 21, 2021 at 14:12
  • 1
    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Mar 21, 2021 at 15:34
  • 2
    $\begingroup$ Quite a coincidence that this question showed up yesterday. $\endgroup$ Mar 21, 2021 at 18:49
  • 3
    $\begingroup$ "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. $\endgroup$ Mar 21, 2021 at 19:12

1 Answer 1

2
+100
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$
    – Silvinha
    Mar 21, 2021 at 19:56
  • $\begingroup$ 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). $\endgroup$
    – Silvinha
    Mar 21, 2021 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.