Let $(N, h)$ and $(M, g)$ riemannian manifolds with $dim M = dim N$.
We say that M is locally isometric to N if there is a smoother application $F: M\rightarrow N$ such that $$g(v,w)=h(dF_p(v), dF_p(w)),\ \forall v,w \in T_pM,\ \forall p \in M .$$
Another definition is: M is locally isometric to N if for each $p \in M $ exists in an open $U\subset M$ (with $ p \in U $), an open $S\subset N$ and a diffeomorphism $ f : U \rightarrow S $ such that $$g(v,w)=h(df_p(v), df_p(w)),\ \forall v,w \in T_pU,\ \forall p \in U. $$
Of course the first definition implies the second. But these definitions are equivalent? If it is true, how to prove the other implication?
Thanks
Here is an example showing that both things are not equivalent. Take $M = S^1$ the unit circle with its Riemannian metric (induced by $\mathbb{R}^2$) and let $N=\mathbb{R}$ with its stantard metric. Then $\dim (M) = \dim (N) = 1$ and $M,N$ are locally isometric because of the arclength parameter. But there are no map $F : M \to N$ satisfying $$g(v,w) = h(dF_p(v),dF_p(w))$$ for all $p, v,w \in T_pM$. Indeed, such a map $F$ should be a submersion, but there are no submersion from $S^1$ to $\mathbb{R}$ due to the existence of a critical point e.g. a maximum of $F$.