Prove cylinder local isometric to the plane Let $C = \{(x,y,z)\mid x^2+y^2 = 1\}\subset \Bbb{R}^3$ be the cylinder and $P = \{(x,y,z)\mid z = 0\}$.
Prove if we gives the induced metric to $C$ and $P$ from $\Bbb{R}^3$.they are local isometry.
Here local isometry means a smooth map $\varphi:P\to C$ such that $\varphi^*(g_c) = g_p$.
I have proved it clear,but seems trapped into some detail.Is there some clear proof of this exercise?
I constructed as follows:$$\varphi:P\to C \\(x,y,0) \to (\cos x,\sin x ,y)$$
The hard part is to clearly show that it's local isometry. First pick local parametrization for $C$ as $X(u,v) = (\cos u,\sin u,v)$.hence we see that :let $i_C :C\to \Bbb{R}^3$,then the induced metric on $C$ under the coordinate chart is $du^2 + dv^2$.If we compute $\varphi^*(du^2 +dv^2) = d(u\circ\varphi)^2 + d(v\circ \varphi)^2$ where $u \circ \varphi (x,y,0) = x + C$ and $v \circ \varphi (x,y,0) = y + C$ ,for some constant $C$ , hence we have $\varphi^*(du^2 + dv^2) = dx^2 + dy^2$ on $P$ coinside with metric on $P$
 A: If $A \subset \mathbb{R}^3$ is a submanifold, let us denote by $\iota_A$ the inclusion map and ${\iota_A}^*g$ the induced Riemannian metric.
The map $\varphi \colon P \to C$ is a local isometry iff $\varphi^* \left({\iota_C}^*g\right) = {\iota_P}^*g$, and thanks to the property of pull-back maps, $\varphi$ is a local isometry iff
$$
{\iota_P}^*g = \left(\iota_C\circ \varphi\right)^*g.
$$
Let $u$ an $v$ be two tangent vectors at $(x,y,0) \in P$. Then
$$
 \left(\iota_C\circ \varphi\right)^*g(u,v) = g\left({\iota_C}_* \circ \varphi_*u,  {\iota_C}_*\circ \varphi_* v\right).
$$
If $u = (u^1,u^2,0)$ and $v = (v^1,v^2,0)$, then
\begin{align}
\left({\iota_C}_*\circ\varphi_*\right)_{(x,y,0)} u &=  u^1\left(-\sin x,\cos x,0 \right) + u^2\left(0,0,1\right), \\
\left({\iota_C}_*\circ\varphi_*\right)_{(x,y,0)} v &=  v^1\left(-\sin x,\cos x,0 \right) + v^2\left(0,0,1\right),
\end{align}
(${\iota_C}_*$ is just the inclusion of the tangent spaces into $\mathbb{R}^3$, thus does not affect the computations in coordinates). It follows that
\begin{align}
\left( {\iota_C} \circ \varphi\right)^*g_{(x,y,0)}(u,v) &= \left\langle \left(-u^1\sin x, u^1 \cos x, u^2\right) , \left(-v^1\sin x, v^1 \cos x, v^2\right)\right\rangle \\
&= \left(-u^1\sin x\right)\left(-v^1\sin x\right) + \left(u^1 \cos x\right) \left(v^1\cos x\right) + u^2 v^2\\
&= u^1v^1\left(\cos^2 x + \sin^2 x\right) + u^2v^2\\
&= u^1v^1 + u^2v^2\\
&= {{\iota_P}^*g}_{(x,y,0)}(u,v) 
\end{align}
which is what we wanted to show.
