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Define a map from $F$ the circular sector $0<r<\infty$, $0<\theta<2 \pi \sin \alpha$ where $ 0<2 \alpha<\pi$ to the cone $z = \cot \alpha \sqrt {x^2+y^2}$ as $F(r, \theta) = (r \sin \alpha \cos(\frac{\theta}{\sin \alpha}),r \sin \alpha \sin(\frac{\theta}{\sin \alpha}), r\cos\alpha)$ where $\theta \in (0,2\pi)$.

By construction, this map should be bijective, then I think it should be an isometry rather only a local isometry. But Do Carmo claims that it is a local isometry. Why?

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  • $\begingroup$ This is indeed a bijection so it is an isometry. He isn't wrong to say it's a local isometry as isometry implies local isometry. Did he write that it isn't an isometry? $\endgroup$ – Siddhant Apr 21 '17 at 12:44
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    $\begingroup$ This is NOT bijective. The range of $\theta$ is $0< \theta <2\pi \sin \alpha$ But not $0<\theta \le 2\pi \sin \alpha$. . $\endgroup$ – user99914 Apr 21 '17 at 12:49
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    $\begingroup$ Oh, right it is not onto $\endgroup$ – z.z Apr 21 '17 at 12:51
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    $\begingroup$ It is an isometry onto it's image. You are right $\endgroup$ – Siddhant Apr 21 '17 at 12:54
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    $\begingroup$ Note that certain points are closer on the cone than are the corresponding points on the unrolled pacman. $\endgroup$ – Ted Shifrin Apr 21 '17 at 16:52
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I commented earlier, but I want to emphasize the most important point. Take the points on the cone given by $F(1,\epsilon)$ and $F(1,2\pi-\epsilon)$. On the cone their distance is less than $2\epsilon$, but the distance in the circular sector is quite large.

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