fixed important typo
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@ 179,7 +179,7 @@ since $X^X$ has these properties.


\begin{proof}


Let $G \coloneqq E(X,T)$ and let $d$ be a metric on $X$.


\gist{


For all $g \in G$ we need to show that $x \mapsto gx$ is bijective.


For all $g \in G$ we need to show that $x \mapsto gx$ is injective.


If we had $gx = gy$, then $d(gx,gy) = 0$.


Then $\inf_{t \in T} d(tx,ty) = 0$, but the flow is distal,


hence $x = y$.



@ 195,7 +195,7 @@ since $X^X$ has these properties.




It is $g' = g'gg'$,


so $\forall x .~g'(x) = g'(g g'(x))$.


Hence $g'$ is bijective


Hence $g'$ is injective


and $x = gg'(x)$,


i.e.~$g g' = \id$.


}{




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