Let $S_1 : [0, 2\pi r]\times [0, h]$

$S_2: x^2+y^2=r^2$

Let $f: S_1 \to S_2$

$(u,v)=(r\cos (\frac{u}{r}), r\sin (\frac{u}{r}), v)$ for $v\in [0,h]$ and $u\in [0, 2\pi r)$

How do I prove that $f$ is a diffeomorphism and $f$ is an isometry?

enter image description here

I know that the following theorem;

$f$ isometry $\iff$ for $S_1$ and $S_2$, the surface patches have the same first fundamental form.

But I don't know how to show these. Please can you help me. Thank you.

The definition of isomorphism: let $S_1$ and $S_2$ be two regular surfaces. Let $f: S_1 \to S_2$ be diffeomorphism. $f$ is isometry if the length of $\gamma$ in $S_1$ must be equal to the length of $f\circ \gamma$ in $S_2$ for any curve $\gamma$ in $S_1$. i.e $L_{S_1}(\gamma )= L_{S_2}(f \circ \gamma )$

The definition of difeomorphsim: $f:S_1\to S_2$ is a diffeomorphism


(1) $f$ is smooth.

(2) $f$ is one to ne and onto.

(3) $f^{-1} :S_2\to S_1$ is smooth.

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    $\begingroup$ $f$ is not a diffeomorphism because it is not a homeomorphism. It is then also not an isometry. $\endgroup$ – Dan Rust Nov 27 '13 at 14:48
  • $\begingroup$ But Dear @DanielRust , there should exist an diffeomorphism and isomorphism :( since my instructor said so. This is not homework or else. This is an example to explain the topic. $\endgroup$ – user315 Nov 27 '13 at 14:53
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    $\begingroup$ Is it possible they meant local diffeomorphism and local isometry? $\endgroup$ – Dan Rust Nov 27 '13 at 14:54
  • $\begingroup$ These were not stated. I took and posted the photo of my notebook. @DanielRust $\endgroup$ – user315 Nov 27 '13 at 14:55
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    $\begingroup$ Could you put the definitions for 'diffeomorphism' and 'isometry' in to your post please? They have fairly standard definitions, but it's possible you're not using the standard ones (in fact for diffeomorphism you are definitely not using the usual definition) $\endgroup$ – Dan Rust Nov 27 '13 at 15:11

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