In my lecture on differential geometry, we introduced the following definition:
Definition 1: Let $\phi : M \longrightarrow N$ be a smooth map between regular surfaces. Then the map $\phi$ is called an isometry if for all smooth curves $\gamma : I \longrightarrow M$ it holds that $\mathrm{len}(\gamma) = \mathrm{len}(\delta)$, where $\delta = \phi \circ \gamma$.
However, I have now read other lecture notes on differential geometry, where an isometry is defined as
Definition 2: Let $\alpha : U \longrightarrow \mathbb{R}^3$ and $\beta : U \longrightarrow \mathbb{R}^3$ be parameterisations of $M = \alpha(U)$ and $N = \beta(U)$. A map $\phi : M \longrightarrow N$ is called an isometry iff the fundamental forms $I_{\alpha}(p)$ and $I_{\beta}(\phi(p))$ coincide for all $p \in M$.
Now, I tried to prove the equivalence of these two definitions. By intuition this should be an easy task. Unfortunately, I fail to solve this easy task.
My approach was to show that
$$\big( I_{\alpha}\big)_{ij} = \big< \partial_i \alpha, \partial_j \alpha \big> \overset{!}{=} \big< \partial_i \beta, \partial_j \beta \big> = \big( I_{\beta} \big)_{ij}$$
So I calculated
\begin{align*} \big< \partial_i \beta, \partial_j \beta \big> &= \big< \partial_i (\phi \circ \alpha ), \partial_j (\phi \circ \alpha) \big> \\ &{} \\ &= (\partial_i \alpha_m) (\partial_j \alpha_n) \cdot \big< \partial_{\alpha_m} \phi, \partial_{\alpha_n} \phi \big> \\ &{} \\ &= ( \operatorname{Jac}\alpha )_{mi} \cdot ( \operatorname{Jac}\alpha )_{nj} \cdot \big< \partial_{\alpha_m} \phi, \partial_{\alpha_n} \phi \big> \end{align*}
This coincides with the fundamental form $I_{\alpha}$ if
$$ \big< \partial_{\alpha_m} \phi, \partial_{\alpha_n} \phi \big> = \delta_{mn} $$
Is this approach correct so far? If so, how do I proceed/how do I show that last step?