xca:fequiviffHt=Hu added

circle.tex

@@ -2266,6 +2266,14 @@ \section{Connected \coverings over the circle}
   $\zs(z) \sim_H \zs(z')$.
   It is clear that $(X,t)$ is a cycle with $H_t = H$.
 \end{proof}
+
+\begin{xca}\label{xca:fequiviffHt=Hu}
+Let $(X,t)$ and $(Y,u)$ be cycles, and $f:X\to Y$ a map such that $uf=ft$.
+Show that $f$ is an equivalence if and only if $H_t =_{\Sub_\zet} H_u$.
+Hint: prove first that any such $f$ is surjective. Then prove that
+$H_u \subseteq H_t$ implies that $f$ is injective.
+\end{xca}
+
 The components of $\Cyc$ will pop up many times from now on,
 so we make the following definitions to make it easier to talk about them.
 \begin{definition}\label{def:Order}
@@ -2283,7 +2291,8 @@ \section{Connected \coverings over the circle}
   We write $\ord\defeq\settrunc{\blank}:\Cyc\to\Order$ for the map
   from cycles to their orders, and we write $\ord(t)\defeq\ord(X,t)$ for short.
 
-  We say that the order $d\defeq\ord(X,t)$ \emph{divides} the order $k\defeq\ord(Y,u)$, written $d | k$,
+  We say that the order $d\defeq\ord(X,t)$ \emph{divides} the 
+  order $k\defeq\ord(Y,u)$, written $d | k$,
   for cycles $(X,t),(Y,u)$, if $H_u \subseteq H_t$.
 \end{definition}
 We have a canonical injection $\NN \mono \Order$,
@@ -2988,11 +2997,8 @@ \section{The \texorpdfstring{$m$\th}{mᵗʰ} root:
   We also need an identification
   $(X,t)\eqto\cdg{m}(V,t^m)\jdeq(\bn m\times V,\sqrt[\uproot{2}m]{t^m})$.
   This we define via a map $e : \bn m\times V \to X$, $e(k,x) \defeq t^k(x)$.
-  This is an equivalence as long as the orders match.\footnote{\MB{?
-  Surjectivity: Let $x:X$. We may assume an $x_0: V$. Then also $x_0:X$,
-  and we may assume an $n:\zet$ such that $t^n(x_0)=x$.
-  Write $n=mq+r$ with $r:\bn m$, then $e(r,t^{mq}(x_0))=x$.
-  Injectivity: ...}}
+  This is an equivalence as long as the orders match,
+  see \cref{xca:fequiviffHt=Hu}.
   So let $n:\zet$, and assume first that $t^n=\id_X$.
   Then $P$ implies that we may write $n=qm$ for some $q:\zet$,
   so