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I was asked to prove that the inverse function of the function $f$ isn't continuous.

$f: [0,2\pi)\rightarrow S^1$; $f(t)=(\cos(t),\sin(t))$

$S^1$ denotes the circle of radius one centered in the origin.

I don't understand what the inverse function is.... The natural construction for me would be $f^{-1}(x,y)=\arccos(x)$ for $-1\leq x\leq 1$, $y\geq 0$ and $f^{-1}(x,y)=\arccos(x)+\pi$ for $-1\leq x\leq 1$, $y< 0$. Is this right?

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You don't need to find a formula for the inverse function to show that it isn't continuous. Think about compactness... –  David Mitra Jan 15 '13 at 1:20
    
But, I believe you want $f^{-1}(x,y)=2\pi-\arccos x$ for $y<0$. –  David Mitra Jan 15 '13 at 1:32
    
Alternatively, think about sequential continuity of the inverse function. –  Matemáticos Chibchas Jan 15 '13 at 2:06

1 Answer 1

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Not quite. You want $f^{-1}(x,y)=2\pi-\arccos x$ for $y<0$.

To show that $f^{-1}$ is not continuous, show that it isn't continuous at the point $(1,0)$. Note that in any neighborhood of $(1,0)$ in $S^1$, there are points as close together as you like that get mapped to points that are not close to each other (one to a point near $t=0$ and one to a point near $t=2\pi$).

Alternatively, you could use the fact that $S^1$ is compact, while $[0,2\pi)$ isn't.

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