Q:

Show that the map on spheres is smooth

For each of the following maps between spheres, compute sufficiently many coordinate representations to prove that it is smooth.
$(a):$ $p_{n}:\mathbb{S}^{1}\rightarrow \mathbb{S}^{1}$ is the $n$th power map for $n\in\mathbb{Z}$, given in complex notation by $p_{n}(z)=z^{n}$.
$(b):$ $\alpha:\mathbb{S}^{n}\rightarrow \mathbb{S}^{n}$ is the antipodal map $\alpha(x)=-x.$
$(c):$ $F:\mathbb{S}^{3}\rightarrow \mathbb{S}^{2}$ is given by $F(w,z)=(z\overline{w}+w\overline{z},iw\overline{z}-iz\overline{w},z\overline{z}-w\overline{w})$, where we think of $\mathbb{S}^{3}$ as the subset $\{(w,z):|w|^{2}+|z|^{2}=1\}$ of $\mathbb{C}^{2}$.
Now my questions is only for part $(a)$ but if you want to share for the others thats ok. Since I have the stereographic projection $\sigma:\mathbb{S}^{n}\setminus \{N\}\rightarrow \mathbb{R}^{n}$ by $$\sigma(x^{1},\dots ,x^{n+1})=\frac{(x^{1},\dots ,x^{n})}{1-x^{n+1}}.$$  I somehow want to make use of this function and the function $\widetilde{\sigma}(x)=-\sigma(-x)$ for $x\in \mathbb{S}^{n}\setminus \{S\}$. But I don't see any nice way to go about it since to write $z^{n}$ in $\mathbb{R}^{2}$ is kinda messy any suggestions? Or I'm I just doing it wrong?

A:

Since every point in $\mathbb S^1 \subseteq \mathbb C$ can be written as $\cos \theta + i \sin \theta$ for $\theta \in [0, 2 \pi)$, the $n$-th power map, in this notation, can be rewritten as 
$$
 p_n( \cos \theta + i \sin \theta ) = p_n( e^{i n \theta } ) = e^{i n \theta} = \cos (n\theta) + i \sin (n \theta)
$$
Also, $\sigma, \widetilde{\sigma}, \sigma^{-1}, \widetilde{\sigma}^{-1}$ will be expressed as
$$
 \sigma ( e^{i \theta } ) = \frac{ \cos \theta }{ 1 - \sin \theta } \; \; \; \; \;  \widetilde{\sigma}( e^{i \theta } ) = \frac{ \cos \theta }{ 1+ \sin \theta} \; \; \; \;   \\
  \sigma^{-1}(u) = \frac{ 2u + i( u^2 - 1 ) }{u^2 + 1 } \; \; \; \;  \widetilde{\sigma}^{-1}(u) = \frac{2 u + i (1 - u ^2 ) }{ 1 + u^2 }
$$
We know $\widehat{p_n}$, the coordinate representation of $p_n$, is of the form $\sigma_2 \circ p_n \circ \sigma_1$ where $\sigma_1$ is either $\sigma^{-1}$ or $\widetilde{\sigma}^{-1}$ and $\sigma_2$ is either $\sigma$ or $\widetilde{\sigma}$. We proceed to show $\widehat{p_n}: \mathbb R \rightarrow \mathbb R$ is a smooth function (in the sense of $\mathbb R$). We know for $u \in \mathbb R$,
$$
 \sigma_1 = \frac{ 2u  \pm i (1 - u^2) }{ 1 + u^2 }
$$
the plus minus sign depends on whether $\sigma_1$ is $\sigma^{-1}$ or $\widetilde{\sigma}^{-1}$. Next, we write 
$$
  \cos \phi = \frac{ 2u }{ 1 + u^2  }\text{ and } \sin \phi = \pm \frac{ 1 - u^2 }{ 1 + u^2 }
$$
Choose appropriate domains for the inverse trigonometric function, we have $u = \arccos \phi$. Therefore
\begin{align*}
 \widehat{p_n}( u ) &= (\sigma_2 \circ p_n \circ \sigma_1) (u ) = \sigma_2 ( p_n( \sigma_1( u ) ) ) = \sigma_2( p_n( \cos \phi + i \sin \phi) ) \\
   &= \sigma_2( \cos (n \phi ) + i \sin ( n \phi ) ) = \frac{ \cos ( n \phi ) }{ 1 \pm \sin (n \phi ) } = \frac{ \cos (n \arccos u ) }{ 1 \pm \sin (n \arccos u ) }
\end{align*}
Since $f(x) = \sin x, \cos x, \arccos x$ are smooth functions, it follows that $\widehat{ p_n }$ is also a smooth function.

