# Why is a Riemannian submersion a submetry

I have encountered a conclusion in the book but I do not know how to prove it.

If $F$ is a Riemannian submersion from $(M,g)$ to $(N,\tilde g)$, then it is a submetry. (A mapping $F$ is a submetry if for any $p \in M$, we can find $r>0$ such that for each $\epsilon \leq r$ we have $F(B(p,\epsilon))=B(F(p),\epsilon)$ where $B$ is the open ball in the metric space with metric induced by distance)

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It is clear that one has $F(B(p,\epsilon)) \subseteq B(F(p),\epsilon))$ since it is easy to see that the length of a curve decreases under a submersion (that is $\mathcal L (F \circ c) \leq \mathcal L (c)$, where $\mathcal L$ denotes both the length on $M$ with respect to $g$ and on $N$ with respect to $\tilde g$).
So one needs to show that $F(B(p,\epsilon)) \supseteq B(F(p),\epsilon)$. To do so let $\tilde q \in B(F(p),\epsilon))$ and $\tilde \gamma$ be a minimal geodesic connecting $F(p)$ and $\tilde q$ parametrized on $[0,1]$. Since $F$ is a submersion there exists a smooth horizontal lift $\gamma$ of $\tilde \gamma$ starting at $p$, that is $\gamma(0) = p$ and $\tilde \gamma = F \circ \gamma$. Consequently $\dot {\tilde \gamma}(t) = dF_{\gamma (t)}(\dot \gamma (t))$. Now, since the submersion is also riemannian, one concludes that $\mathcal L (\gamma) = \mathcal L (\tilde \gamma) < \epsilon$ , i.e. $\gamma \subset B(p,\epsilon)$ so $\tilde q = \tilde \gamma(1) = F(\gamma(1)) \in F(B(p,\epsilon)).$
Thanks for the reference. It seems to me that $M$ should be complete in order to guarantee the lifting on closed interval. –  Hezudao Dec 4 '12 at 4:31