fixed important typo
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@ -179,7 +179,7 @@ since $X^X$ has these properties.
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\begin{proof}
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Let $G \coloneqq E(X,T)$ and let $d$ be a metric on $X$.
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\gist{
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For all $g \in G$ we need to show that $x \mapsto gx$ is bijective.
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For all $g \in G$ we need to show that $x \mapsto gx$ is injective.
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If we had $gx = gy$, then $d(gx,gy) = 0$.
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Then $\inf_{t \in T} d(tx,ty) = 0$, but the flow is distal,
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hence $x = y$.
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@ -195,7 +195,7 @@ since $X^X$ has these properties.
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It is $g' = g'gg'$,
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so $\forall x .~g'(x) = g'(g g'(x))$.
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Hence $g'$ is bijective
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Hence $g'$ is injective
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and $x = gg'(x)$,
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i.e.~$g g' = \id$.
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}{
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