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Need to know how to compute the transition map of $h = g^{-1}\circ f$

The unit circle has charts $f(s) = (\cos(s),\sin(s))$ is element of $\Bbb R^2$, for $-\pi < s < \pi$, and

$g(t) = \left(\cfrac{2t}{t^2 + 1}, \cfrac{t^2 - 1}{t^2 + 1}\right)$ for $t\in\Bbb R$. Compute the transition map of $h = g^{-1}\circ f$

Sorry for my poor structure of the question. Thanks in advance.

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Where exactly are you stuck? For example, can you compute $g^{-1}$? –  Jason DeVito Nov 13 '12 at 23:43
    
That's exactly where I am stuck! I think i should be able to figure it out if i get g^-1 –  Denis Nov 14 '12 at 9:53
    
Well, set $x = \frac{2t}{t^2+1}$ and solve for $t$ using the quadratic formula. In the quadratic formula, looking at the squareroot part (which ends up simplifying to $\pm\sqrt{1-x^2}$), notice that this is just another name for $y$. In the end, I got $t = \frac{1+y}{x}$ as my answer. –  Jason DeVito Nov 14 '12 at 18:05
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