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

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?

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? – guaraqe 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

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.
$f_2$ 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