# Why is this a covering map of Riemann manifolds?

Can you tell me, wh the following is true: If $f: M \rightarrow N$ is a smooth map between complete and connected Riemannian manifolds which fulfills $f^{*}(g_n)=g_M$ for the metrics $g_m$ and $g_N$ on $M$ and $N$, then it is a covering map?

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Since the metric pulls-back, the Levi-Cevita connection pulls back as well, so the image of a geodesic is a geodesic. Use exponential coordinates in $N$, by design it lifts to exponential coordinates based at the fibres of points $f^{-1}(p)$ for $p \in N$. That's the idea. Basically the unique path-lifting property of a covering space is guaranteed via the exponential map. –  Ryan Budney Feb 11 '12 at 0:47
I think you need an additional hypothesis or 2. For example, if $M =\mathbb{R}$ and $N = \mathbb{R}^2$ with their standard metrics and $f$ is the natural inclusion, then $f$ pulls back the metrics, but is not a covering. Perhaps some kind of surjectivity condition is all that is needed. My copy of Do Carmo is at school - I know he proves something similar to this in the spirit of Ryan's comment, but I don't recall all the hypothesis. –  Jason DeVito Feb 11 '12 at 1:41
@Jason: I think you just need the manifolds have the same dimension and the map is a submersion. –  Ryan Budney Feb 11 '12 at 3:33
If the dimensions are equal and the metric pulls back to the metric, don't we have automagically a submersion? –  Mariano Suárez-Alvarez Feb 11 '12 at 3:56
The passage I was thinking of in Do Carmo is not quite as I remembered. They hypothesis is simply that $\|d_p f v\| \geq \|v\|$ for all vectors $v$ (much weaker than $f^*(g_n) = g_m$), but he also assumes $f$ is a local diffeomorphism, which is quite strong. On the other hand, being a local diffeomorphism follows from $f^*(g_n) = g_m$ together with the fact that, say, $M$ and $N$ have the same dimension (as Mariano suggested). –  Jason DeVito Feb 13 '12 at 20:45