smoothness of hopf fibration projection with respect to standard differentiable structure on unit sphere We know that steographic projection defines a differentiable structure on $S^n$ by sending points on $S^n$ to hyperplane $\{x^{n+1}\}=0$. In fact, stereographic projection $\sigma_P: S^n- P\to R^n $ is a map which sends a point $X\in S^n- P$ to the unique point where PX intersect $R^n$, where P is N or S respectively.
Now , it can be calculated that $\sigma_N(x^1,x^2,...,x^{n+1})= \frac{1}{1-x^{n+1}}(x^1,x^2,...,x^n)$ and $\sigma_S(x^1,x^2,...,x^{n+1})= \frac{1}{1+x^{n+1}}(x^1,x^2,...,x^n)$ . Also , it is easy to show that $\{S^n- N,\sigma_N\}$,$\{S^n- S,\sigma_S\}$ defines a differentiable structure on $S^n$.
I was further asked to compute the coordinate representation for each of these maps below in stereographic coordinates and use this to show each map is smooth.
a) antipodal map $ a:S^n \to S^n $ that sends x to -x.
b) Hopf fibration projection $\pi:S^3\to S^2 $,defined by $\pi(z_1,z_2)=(2Re(z_1z_2),2Im(z_1\bar{z_2}),|z_1|^2-|z_2|^2)$ ,where $S^3$ is the unit sphere, $\{|z_1|^2+|z_2|^2=1\}$,in $C^2$.
I have no idea how to do this part . Can anyone help me with this part?
 A: I'll sketch the Hopf fibration case, leaving you the antipodal map case.
The point of charts is that, locally, they give a way of thinking about a map $f:M\rightarrow N$ between two manifolds as a map between Euclidean spaces.  More specifically, suppose $M$ is $m$-dimensional and $N$ is $n$-dimensional.  Let $p\in M$ and choose a chart $(U,\varphi)$ around $p$ and a chart $(V,\rho)$ around $f(p)$.  In other words, $\varphi:U\rightarrow \mathbb{R}^m$ is a homeomorphism onto an open subset of $\mathbb{R}^m$ and likewise for $\rho:V\rightarrow \mathbb{R}^n$.
Then the composition $\rho \circ f \circ \varphi^{-1}$ makes sense and actually defines a map from an open subset of $\mathbb{R}^m$ to $\mathbb{R}^n$.  In this setting, we know how to make sense of the statement $\rho \circ f\circ \varphi^{-1}$ is smooth (or $C^1$, differentiable, etc) and we define the analytic properties of $f$ to be those of $\rho \circ f \circ \varphi^{-1}$.
Thus, in order to prove that $f:M\rightarrow N$ is $C^\infty$, we really have to show that for every point $p\in M$, $\rho \circ f \circ \varphi^{-1}$ is smooth.
Now, to your particular example, we must pick a point $p$ in $S^3$, pick charts around both $p$ and $\pi(p)$, and then compute the corresponding map from $\mathbb{R}^3$ to $\mathbb{R}^2$.
First, using the map from wikipedia, we have $$\pi(z_1, z_2) = \left(2\Re(z_1 \overline{z_2}),\, 2\Im(z_1\overline{z_2}),\, |z_1|^2 - |z_2|^2 \right).$$
First note that $\pi(z_1,z_2)$ really is an element of $S^2$ with this definition.  This follows because \begin{align} |\pi(z_1,z_2)| &= 4\Re(z_1\overline{z_2})^2 + 4\Im(z_1\overline{z_2})^2 + \left(|z_1|^2 - |z_2|^2\right)^2\\ &= 4|z_1\overline{z_2}|^2 + |z_1|^4 -2|z_1|^2|z_2|^2 + |z_2|^4\\ &= |z_1|^4 + 2|z_1|^2|z_2|^2 + |z_2|^4\\ &= (|z_1|^2 + |z_2|^2)^2 \\ &= 1.\end{align}
Now, to check smoothness, instead of using complex number "coordinates" on $S^3$, I'm going to use regular good old fashioned $(x_1, x_2, x_3, x_4)$ where $z_1 = x_1 + i x_2$ and $z_2 = x_3 + i x_4$ and, of course, $\sum x_i^2 = 1$.
With these "coordinates", $\pi$ looks like $$\pi(x_1,x_2,x_3, x_4) = (2x_1x_3 +2x_2x_4, \,2x_2 x_3 - 2x_1 x_4,\, x_1^2 + x_2^2 - x_3^3 - x_4^2).$$
Now, choose any point $p = (x_1, x_2, x_3, x_4)\in S^3$ where we'll assume that $p$ is not the north pole of $S^3$, meaning, say, $x_4\neq 1$.  We will also assume that $\pi(p)$ is not the north pole of $S^2$.  With these assumptions, we can use $\varphi = \sigma^3_N$ on $S^3$ and $\rho = \sigma^2_N$ on $S^2$.  (The superscripts on the $\sigma$s refer to which sphere they're defined on.)
Then (according to wikipedia) $$\varphi^{-1}(x_1, x_2, x_3) = \frac{1}{1+x_1^2 + x_2^2 + x_3^2}(2x_1,\, 2x_2,\, 2x_3,\, x_1^2 + x_2^2 + x_3^2 - 1).$$
Now, I'll leave it to you to actually compute $\rho \circ \pi \circ \varphi^{-1}$.  This ends up being a map from $\mathbb{R}^3$ to $\mathbb{R}^2$ which is a rational function in each coordinate.  I will also leave it to you to show that, perhaps after shrinking the chart around $p$, the denominator of each rational function is always nonzero, so that $\rho \circ \pi circ \varphi^{-1}$ is actually smooth.  Then, by definition, $\pi$ is smooth at the point $p$.
To finish up this example, one should remove the assumptions that both $p$ and $f(p)$ are not their respective north poles.  To remove these assumptions, use the other charts $\sigma_S^3$ and $\sigma_S^2$.
