fixed important typo

inputs/lecture_17.tex

@@ -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.~$g g' = \id$.
 }{