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Given a path connected metric space $X$ and a cover $\tilde{X}$ which is also a path connected metric space with covering map $E$, then is $E$ a local isometry?

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Not true: For example take $X= S^1$ with subspace topology of $\mathbb R^2$.

and $\tilde{X}= \mathbb R$. Define map $p:\mathbb R\to S^1$ by $p(t)= (\cos 2\pi t, \sin2\pi t)$ . Then this is covering map but not local isometry.

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