# Lemma $3.3$ - Chapter $7$ - Do Carmo's Riemannian Geometry

$$3.3$$ 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.

• A manifold is locally arcwise connected, because open balls in $\mathbb{R}^n$ are arcwise connected. – user10354138 May 21 at 4:21
• Thanks! Is my argument to justify that $B$ is locally simply connected right? – George May 21 at 10:38
• 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. – user10354138 May 21 at 10:40