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I am trying to prove the following:

enter image description here

So I think I must prove first that $A$ is differentiable. I am looking at the following definition:

enter image description here

I don't understand where the $\Bbb{R}^2$'s (or $\Bbb{R}^3$) appear here. I thought about the following maps: $X(x,y,z)=(-x,-y,-z)$ from $\Bbb{R}^3 \to \Bbb{R}^3$ and then a map $Y$ from $\Bbb{R}^3\to S^2$ now we have: $A=Y\circ X$ but I'm not sure how to proceed from this.

Red Banana
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  • I have to believe that "$R^2$" is a mistake and it should be "$R^3$". – George Ivey Aug 31 '22 at 00:27
  • @GeorgeIvey: There is no mistake, it is completely correct as is. – Michael Albanese Aug 31 '22 at 00:44
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    Do you know what a parameterization is? If not, you need to understand what that is before you can understand the definition in the post. If you do know what a parameterization is, do you know what the standard parameterizations of $S^2$ are? Presumably this would have been discussed earlier in whatever book you're using. – Michael Albanese Aug 31 '22 at 00:47
  • @MichaelAlbanese Yes. The standard parametrizations are $(x,y,\pm \sqrt{1-x^2-y^2})$. And permutations of the arguments and variables. – Red Banana Aug 31 '22 at 00:50
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    So these are the parameterizations $\mathbf{x}_1$ and $\mathbf{x}_2$ you need to use in the definition to verify that $A$ is differentiable. – Michael Albanese Aug 31 '22 at 00:55
  • @MichaelAlbanese Is $A$ defined as I wrote? – Red Banana Aug 31 '22 at 00:59
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    Yes, that would be the map taking the place of $\varphi$ in the definition. – Michael Albanese Aug 31 '22 at 01:10

1 Answers1

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In this case, there is a result that makes life easier. Example $3$ is this section states the following: Let $S_1$ and $S_2$ be regular surfaces. Assume that $S_1 \subset V \subset \mathbb R^3$, where $V$ is an open set of $\mathbb R^3,$ and that $\varphi: V \to \mathbb R^3$ is a differentiable map such that $\varphi(S_1) \subset S_2$. Then the restriction $\varphi |_{S_1}: S_1 \to S_2$ is a differentiable map.

So basically, if we take $V = \mathbb R^3$, then we just have to prove that $A: \mathbb R^3 \to \mathbb R^3$ is a differentiable map, and this should present no great difficulty.

Now to show that $A$ is a diffeomorphism, we also need to show that $A^{-1}$ is differentiable. Hint: $A^2 = \text{id}$, so what does this tell us about $A^{-1}$?

Elliot Herrington
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