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Let $S^2$ be the unit sphere in $\Bbb{R}^3$. It is a regular surface.

Let $\phi:S^2\rightarrow S^2$ be the reflection function, $p=(0,0,1)$ the north pole on the sphere, and $v=(x,y,0)^T$ a vector in the tangent plane $T_p S^2$ (which is the $XY$ plane).

I need to show that $d\phi |_p (v)=-v$. For that, I use the definition: $d\phi|_p (v)=d(\phi \circ X)|_{X^{-1}(p)}(w)$ when $w\in\Bbb{R}^2$ is a vector fulfilling $dX|_p(w)=v$ (there exists one, from the definition on a tangent plane) and $X:U\subset\mathbb{R}^2\rightarrow S^2$ a parameterization of $S^2$.

My problem was in finding $w$. If I take $X(\theta,\phi)=(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)$, then $X^{-1}(p)=(\theta=0,\phi)$. $\phi$ is general, because it can be anything, right? Now, I want to find a $w$ such that $dX|_{(\theta=0,\phi)}w=(x,y,0)^T$. But $$dX|_{(\theta=0,\phi)}=\begin{pmatrix} \cos\phi & 0 \\ \sin\phi & 0 \\ 0 & 0 \\ \end{pmatrix}$$

So $dX|_{(\theta=0,\phi)}\begin{pmatrix} w_1 \\ w_2 \\ \end{pmatrix}=\begin{pmatrix} w_1 \cos\phi\\ w_1 \sin\phi\\ 0\\ \end{pmatrix}$ which can't be equal to $\begin{pmatrix} x\\ y\\ 0\\ \end{pmatrix}$ with general $x,y\in\Bbb{R}$ unless we add $\phi$ as a variable, which is wierd because it shouldn't influence the answer.

Note: if I ignore the fact I can't find $w$ and treat it as just a vector fulfilling $dX|_{(\theta=0,\phi)}w=(x,y,0)^T$, I can prove the statement easily. However, it really bothered me that I couldn't find $w$, even though it is not crucial for this very question.

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    How is the reflection function defined? Is there a formula for it? If so I would suggest working with a curve on the sphere having your tangent vector as a derivative. – 2017-02-11
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    Note that $\phi$ is minus the identity map. No need to use the atlas. +1 for the effort. – 2017-02-11
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    @Test123 It's true. As I noted at the end, the statement is easily proven without finding $w$. Still, I want to find it, because by theory it should exist and it bothers me. That's why the title is finding $w$ and not proving the statement. – 2017-02-11
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    Note that $v\in T_pS^2$ corresponds to a point $q$ on $S^2$ using the exponential map. Of course the length of $v$ determines the position of $q$. Whether $q$ is on the upper or lower hemisphere you may use one of the two charts. If $|v|< 1$ you can find such $w$ (the exponential map is non singular). – 2017-02-11
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    @Test123 I'm sorry, I don't know what the exponential map is. Thanks anyway! Math536's answer was what I was looking for. – 2017-02-12
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    @Whyka It's an analogous argument I used. It's worth learning the exponential map. – 2017-02-12

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If you want to use $X$ to compute $d\phi|_p$, you need $X$ to be a regular parameterization of a neighborhood of $p$ in $S^2$. In particular, this means that

  1. $X$ should be injective: if $x_1 , x_2 \in U$ and $X(x_1) = X(x_2)$, then $x_1 = x_2$;
  2. the derivative $dX|_x$ should have rank $2$ for all $x \in U$.

However, if you read your question again, you will see that you have shown that $X(\theta,\phi)=(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)$ satisfies neither of these conditions!

Of course, a parameterization doesn't need to be defined on all of $\mathbb{R}^2$. If we restrict the domain of $X$ to the open set $U = (0,\pi) \times (0,2\pi)$, say, then we do obtain a regular parameterization of the sphere. However, the north pole is then no longer in the image of $X$, so this can't be used to compute $d\phi_p$. I suggest that you try using the following parameterization instead: $$X(u,v) = (u,v,\sqrt{1 - u^2 - v^2}).$$

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    So the parameterization I wrote is a local parameterization for any $x\in S^2$ but the poles... Now I get it. I think this will prevent many confusions in the future regarding the sphere. Thanks a lot! – 2017-02-12