I am just starting to read Lee's "Riemannian Manifolds" and one of the first exercises in the text (2.7) is the following: given a smooth map $\phi:(M,g)\to (\bar{M},\bar{g})$, prove that for $\dim M=\dim\bar{M}$ we have that $\phi$ is a local isometry iff. $\phi^*\bar{g}=g$. Right to left is obvious. The converse statement, on the other hand, I'm not sure how to solve. I think I need to show that $\phi$ is a bijection, but how would I do this, only knowing that it is a local isometry? I feel like I'm making this more difficult in my head than it needs to be, and would appreciate if someone could offer a hint, or solution.
Here, the definition given for a local isometry is: for each $p\in M$ there is a neighourhood $U$ of $p$ such that $\phi |_U$ is an isometry.
An isometry is defined to be a diffeomorphism such that $\phi^*\bar{g}=g$.
Smooth map between Riemannian manifolds of same dimension is local isometry iff. metric is preserved
differential-geometrymetric-spacesriemannian-geometry
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Best Answer
A local isometry is not always a bijection, consider for example $\mathbb{R}\rightarrow S^1$ where $\mathbb{R}$ is endowed with the Eucledean metric and $S^1$ is the quotient of $\mathbb{R}$ by the translation $t(x)=x+1$.