$3.3$ Lemma. Let be $M$ a Riemannian complete manifold and let $f: M \longrightarrow N$ be a local diffeomorphism onto a Riemannian manifold $N$ which has the following property: for all $p \in M$ and for all $v \in T_pM$, we have $|df_p(v)| \geq |v|$. Then $f$ is a covering map.

My doubt is concerning the first three lines of the proof:

By a general covering spaces (Cf. M. do Carmo [dC $2$], p. $383$), it suffices to show that $f$ has the path lifting property for curves in $N$

The reference cited on the proof is

[dC $2$] CARMO, M. do, Differentiable Curves and surfaces, Prentice-Hall, New Jersey, 1976.

and on the page $383$ of this book just have this proposition:

$\textbf{PROPOSITION}$ $6.$ Let $\pi: \tilde{B} \longrightarrow B$ a local homeomorphism with the property of lifting arcs. Assume $B$ is locally simply connected and that $\tilde{B}$ is locally arcwise connected. Then $\pi$ is a covering map.

I think the codomain of $f$ is locally simply connected because, given $p \in M$ arbitrary, we can have a convex neighborhood $V$ of $p$ and $f$ send convex neighborhood in convex neighborhood (this is an application of the intermediate value theorem in paths like $[x,y]$ where $x$ and $y$ are in the convex neighborhood of $p$), then the convex neighborhood $\pi(V)$ will be a star-shaped region and, by this topic, the neighborhood will be simply connected.

I don't sure how to show the condition on $\tilde{B}$. I would appreciate if someone can help me in understand how $f$ on lemma $3.3$ satisfy the conditions of the proposition $6$ and if what I tried to do is correct.

Thanks in advance!

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    $\begingroup$ A manifold is locally arcwise connected, because open balls in $\mathbb{R}^n$ are arcwise connected. $\endgroup$ – user10354138 May 21 at 4:21
  • $\begingroup$ Thanks! Is my argument to justify that $B$ is locally simply connected right? $\endgroup$ – George May 21 at 10:38
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    $\begingroup$ You don't need that either --- $N$ is a manifold, so again, locally $N$ can be thought of as open ball in $\mathbb{R}^n$, which is simply connected. $\endgroup$ – user10354138 May 21 at 10:40

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