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Let $ \displaystyle{ S^2 = \{(x,y,z) \in \mathbb R ^3 : x^2 + y^2 + z^2 =1 \} }$ and $ \displaystyle { U= \{ (x,y) \in \mathbb R ^2 : x^2 + y^2 < 1 \}}$. Consider the functions

$ \displaystyle{ f_1,f_2: U \to S^2 }$ where $ \displaystyle{ f_1 = (u,v, \sqrt{1-u^2 -v^2}) }$ and $ \displaystyle{ f_2 = (u,v, -\sqrt{1-u^2 -v^2}) }$.

Prove that $ f_2 ^{-1} \circ f_1$ is differentiable.

Do I have to find $ f_2 ^{-1} \circ f_1$ or I can prove that is differentiable without finding the function ?

Can you give some help?

Thanks in advance!

Edit: I still haven't made any progress. Any ideas? Thank you!

Sorry I made a mistake. It is $ f(U_1) \cup f(U_2) = S^2 - \{ (x,y,z) \in \mathbb R ^3 : z=0 \} $. So $ f_2 ^{-1} \circ f_1 $ is differentiable on this set.

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Well, I posted an answer which happened to miss a rather important point. I tried to correct things but somehow I erased both the answer and the comment that brought that tho my attention. Sorry, didn't mean to...in fact, I didn't even know I can erase others' comments! –  DonAntonio May 23 '12 at 12:08
At least like this, the second function gives only the south cap of the sphere, so that it doesn`t have a inverse. It this well written? –  Juan Simões May 23 '12 at 17:20
@JuanSimões: Yes it is well written. I came across with this in a proof that $ S^2$ is a diffential manifold. –  passenger May 23 '12 at 17:24
@JuanSimões: Sorry you are right! I have edit the question. –  passenger May 23 '12 at 17:31
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1 Answer

The original formulation of your question was quite ok. After the edit the answer (to the new question) is easy, it is just the composition of two functions which are differentiable (you need some basic knowledge about functions between submanifolds).

The map you defined in the edit extends, however, quite obviously, to all of $S^2$, by defining $f(x,y,0)= (x,y,0)$, and it makes sense to ask whether this extended map is continuous or even differentiable as a map $S^2\rightarrow S^2$.

Since I'm suspicious that this is homework I won't answer this question, but in order to figure this out you may want to ask yourself what this map is doing geometrically and whether it might extend even to $\mathbb{R}^3$ to a continuous or even differentiable map.

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$ f_2 $ is continious, 1-1, onto. Since $ S^2 $ is compact and $ \mathbb R ^3 $ is Hausdorff we get that $ f_2^{-1}$ is continious. But I can't see why is differentiable. Some hint? –  passenger May 23 '12 at 17:55
Don't look at $f_2$, look at $f$. $f$ is the restriction of a rather trivial map to the sphere. –  user20266 May 23 '12 at 18:35
Namely: $f$ is the restriction of $F(x,y,z)=(x,y,-z)$ to the sphere. –  ˈjuː.zɚ79365 Jun 18 '13 at 14:52
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