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Let $$ \varphi:\mathbb{R}^2 \longrightarrow \mathbb{S}^2-{N} \subset \mathbb{R}^3$$ be the (inverse) stereographic projection from the North pole on the unit sphere centred at the origin.

$$\varphi(x,y) = \left( \frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1} {x^2+y^2+1} \right) $$ Let $$\hat{f}(x,y) = (\alpha x, \alpha y) $$ where $\alpha >0. $ $$f: \mathbb{S}^2 \longrightarrow \mathbb{S}^2 = \left\{ \begin{array}{ll} \varphi \circ \hat{f} \circ \varphi^{-1} & on \quad \mathbb{R}^2 \longrightarrow \mathbb{S}^2-{N} \\ f(N)=N & \\ \end{array} \right. $$

How would I show that the map is smooth at $N$ and compute $Df|_N$ at the north pole?

In the Jacobian representation, $Df|_N$ goes to infinity.

I tried taking $$\lim_{\epsilon=(a,b,c)\rightarrow (0,0,0)} \frac{f(N+\epsilon)-f(N)}{\epsilon} $$ But the same problem happens.

Is there a reformulation of $f$, say $h$ which is equal to $f$ but expressed entirely differently (for example in polar coordinates?)

Any hints or clues are appreciated!

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    $\begingroup$ Your $f$ is the identity, so the derivative is the identity at any point. You need a chart around the north pole, e.g. stereographic projection from the south pole, to compute the derivative in the chart representation. $\endgroup$ – Daniel Fischer Nov 11 '14 at 19:05
  • $\begingroup$ Sorry I've missed a piece out - just editing it in now $\endgroup$ – ZahaMan Nov 11 '14 at 19:14
  • $\begingroup$ Okay. The second part of my comment still applies. You need a chart around $N$ and express $f$ in that chart. $\endgroup$ – Daniel Fischer Nov 11 '14 at 19:22
  • $\begingroup$ So something like -(φS o fhat o φS^-1) around and including N $\endgroup$ – ZahaMan Nov 11 '14 at 19:29
  • $\begingroup$ More something like $\varphi_S^{-1}\circ f \circ \varphi_S$. You want a function $\mathbb{R}^2 \to \mathbb{R}^2$ to compute the derivative. $\endgroup$ – Daniel Fischer Nov 11 '14 at 19:39

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