There is one exercise I proved involving that a smooth map $\varphi$ from $M$ to $\widetilde{M}$ is a local isometry if and only if (pullback) $(\varphi^* \tilde{g})=g.$
So I am wondering if the proof is correct? Below is both the exercise and my proof:
Exercise:
- Prove that if $(M,g)$ and $(\widetilde{M}, \tilde{g})$ are Riemannian manifolds with $\dim M = \dim \widetilde{M}$, then a smooth map $\varphi : M \to \widetilde{M}$ is a local isometry if and only if $\varphi^* \tilde{g} = g$.
My proof:
Recall that an isometry from $(M,g)$ to $(\widetilde{M}, \tilde{g})$ is given by a diffeomorphism $\varphi : M \to \widetilde{M}$ with the pullback $\varphi^* \tilde{g} = g$, i.e.
\begin{equation*}
\forall v, w \in T_pM, \forall p \in M : \, g_p(v,w) = \tilde{g}_{\varphi(p)}(d\varphi_p(v), d\varphi_p(w)).
\end{equation*}
Now, let $(M,g)$ and $(\widetilde{M}, \tilde{g})$ be Riemannian manifolds with the same dimension, and define a smooth map as
$$
C^\infty(M)\ni \varphi : M \to \widetilde{M}.
$$
($\implies$) Assume $\varphi$ is a local isometry. The pullback $\varphi^* \tilde{g} = g$ is then
$$
\left(\varphi^* \tilde{g}\right)_p(v,w) = \tilde{g}_{\varphi(p)}(d\varphi_p(v), d\varphi_p(w)),
$$
since there exists an isometry between $T_pM$ and $T_{\varphi(p)} \widetilde{M}$, then we get
$$
\left(\varphi^* \tilde{g}\right)_p(v,w) = \tilde{g}_{\varphi(p)}(d\varphi_p(v), d\varphi_p(w)) = g_p(v,w).
$$
Hence it follows that $\varphi^*\tilde{g} = g$.
($\impliedby$) Suppose $\varphi^*\tilde{g} = g$. Then for all $p \in M$ and arbitrary chosen tangent vectors $v,w \in T_pM$ we have from before
$$
\left(\varphi^* \tilde{g}\right)_p(v,w) = \tilde{g}_{\varphi(p)}(d\varphi_p(v), d\varphi_p(w)) = g_p(v,w).
$$
Since the differential $d\varphi_p : T_pM \to T_{\varphi(p)}\widetilde{M}$ preserves the inner product and it is a linear isometry, therefore $\varphi$ is a local isometry and $\varphi^*\tilde{g} = g$.
Best Answer
The first statement is a tautology, given a local isometry $\phi$, we must have that $\phi^*\tilde{g}=g$. Now let $\phi$ be a smooth map, we need to show that given $\phi^*\tilde{g}=g$, $\phi$ is a local diffeomorphism.
By the inverse function theorem, it suffices to check that $D_p\phi$ is an isomorphism for all $p\in M$, and since dimensions are equal it suffices to check that $D_p\phi$ is injective. We proceed by contradiction, suppose that $D_p\phi$ is not injective, and let $v\in \ker D_p\phi$. Then for all $w\in T_pM$: $$g_p(v,w)=(\phi^*\tilde{g})_p(v,w)=g_{\phi(p)}(D_p\phi v,D_p\phi w)=0$$ It follows that $g_p$ is degenerate, a contradiction. So $D_p\phi$ is injective at all $p\in M$, and $\phi$ is a local diffeomorphism.