better xca:map-of-cycles, polish

circle.tex

@@ -2267,11 +2267,10 @@ \section{Connected \coverings over the circle}
   It is clear that $(X,t)$ is a cycle with $H_t = H$.
 \end{proof}
 
-\begin{xca}\label{xca:fequiviffHt=Hu}
+\begin{xca}\label{xca:map-of-cycles}
 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.
+Show: (i) $H_t \subseteq H_u$; (ii) $f$ is surjective; (iii) 
+if $H_u \subseteq H_t$ then $f$ is injective.
 \end{xca}
 
 The components of $\Cyc$ will pop up many times from now on,
@@ -2996,10 +2995,10 @@ \section{The \texorpdfstring{$m$\th}{mᵗʰ} root:
   then so does $t^m(x)$.}
   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,
-  see \cref{xca:fequiviffHt=Hu}.
-  So let $n:\zet$, and assume first that $t^n=\id_X$.
+  This we define via a map $e' : \bn m\times V \to X$, $e'(k,v) \defeq t^k(v)$.
+  This is an equivalence if $H_t \subseteq H_{\mthroot{m}{t^m}}$,
+  by \cref{xca:map-of-cycles}.
+  So let $n:\zet$, and assume that $t^n=\id_X$.
   Then $P$ implies that we may write $n=qm$ for some $q:\zet$,
   so
   \[
@@ -3008,11 +3007,11 @@ \section{The \texorpdfstring{$m$\th}{mᵗʰ} root:
     = (\id_{\bn m} \times t^m)^q = (\id_{\bn m} \times t^{mq}) 
     = (\id_{\bn m} \times \id_V)= \id_{\bn m \times V}.
   \]
-  Conversely, we know from \cref{lem:m-root-id} again,
-  that if $(\sqrt[\uproot{2}m]{t^m})^n=\id_{\bn m \times V}$,
-  then we may write $n=qm$ for some $q:\zet$,
-  and $(t^m)^q=\id_V$, which by \cref{lem:cycle-order-point-ap}
-  implies that $t^n = \id_X$.
+%  Conversely, we know from \cref{lem:m-root-id} again,
+ % that if $(\sqrt[\uproot{2}m]{t^m})^n=\id_{\bn m \times V}$,
+  %then we may write $n=qm$ for some $q:\zet$,
+   %and $(t^m)^q=\id_V$, which by \cref{lem:cycle-order-point-ap}
+    %implies that $t^n = \id_X$.
 
   Straight from these definitions, we see that $g\circ h=\id$.
   We leave to the reader to check that $h\circ g=\id$.