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Let $X$ be a locally compact complete inner metric space. $G$ is a subset of the isometry group (ie $\mathrm{iso}(X)$) of $X$. Then the projection $X \to X/G$ is a submetry? A map $\sigma :X \to Y$ between locally compact complete inner metric spaces is called a submetry if $\sigma \left( {{B_r}\left( p \right)} \right) = {B_r}\left( {\sigma \left( p \right)} \right)$ for all $r>0$ and $p \in X$.

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    $\begingroup$ What is an inner metric space? $\endgroup$ – tomasz Jun 23 '13 at 15:03
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    $\begingroup$ @tomasz: A metric space $(X,d)$ is an inner metric space if for all $x,y \in X$ $$d(x,y) = \inf\{\ell(\gamma) \mid \gamma\colon [0,1] \to X \text{ is a rectifiable curve with }\gamma(0) = x, \gamma(1) = y\},$$ that is, the distance between two points can be approximated arbitarily well by the length of a rectifiable curve. If a metric space is rectifiably connected, the right hand side always defines a metric $d_i$ called the inner distance. We always have $d_i \geq d$ on $X$ and the condition is that equality holds for all $x,y \in X$. $\endgroup$ – Martin Jun 24 '13 at 13:26
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Be $x\in B_r(p)$, that is, $d(x,p)<r$. Now $d(\sigma(x),\sigma(p)) = \min_{y = g(x): g\in G}d(y,p) \le d(x,p)$, thus $\sigma(x)\in B_r(\sigma(p))$. So $\sigma(B_r(p))\subset B_r(\sigma(p))$

Be $x\in B_r(\sigma(p))$. Then there exists a point $y\in X$ with $\sigma(y)=x$ so that $d(y,p)=d(x,\sigma(p))<r$. That is, $y\in B_r(p)$, and therefore $x\in \sigma(B_r(p))$. Thus $B_r(\sigma(p))\subset \sigma(B_r(p))$

Together we therefore get $\sigma(B_r(p)) = B_r(\sigma(p))$, that is, $\sigma$ is indeed a submetry.

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