done with 4.4.26

group.tex

@@ -876,7 +876,7 @@ \section{Abstract groups}
 
 \begin{xca}
   \label{xca:conj}
-  Let $\mathcal G\jdeq(S,e,\mu,\iota)$ be an abstract group and let $g:S$.  If $s:S$, let $\conj^g(s)\defequi g\cdot s\cdot g^{-1}$.  Show that the resulting function $c^g:S\to S$ preserves the group structure (for instance $g\cdot(s\cdot s')\cdot g^{-1}=(g\cdot s\cdot g^{-1} )\cdot(g\cdot s\cdot g^{-1})$) and is an equivalence.  The resulting identification $\conj^g:\mathcal G\eqto\mathcal G$ is called \emph{conjugation} by $g$.\index{conjugation}
+  Let $\mathcal G\jdeq(S,e,\mu,\iota)$ be an abstract group and let $g:S$.  If $s:S$, let $\conj^g(s)\defequi g\cdot s\cdot g^{-1}$.  Show that the resulting function $\conj^g:S\to S$ preserves the group structure (for instance $g\cdot(s\cdot s')\cdot g^{-1}=(g\cdot s\cdot g^{-1} )\cdot(g\cdot s\cdot g^{-1})$) and is an equivalence.  The resulting identification $\conj^g:\mathcal G\eqto\mathcal G$ is called \emph{conjugation} by $g$.\index{conjugation}
 \end{xca}
 
 \begin{remark}
@@ -1407,40 +1407,35 @@ \section{Homomorphisms}
 
 \begin{example}
 \label{exa:conj-concrete}
- In terms of concrete groups, one can think of conjugation as moving
+ In terms of concrete groups one can think of conjugation as moving
  the shape of the concrete group along a path in the classifying type. 
  This path can in particular be a loop at the shape. More precisely,
- let $G$ be a group, $y$ an element of $\BG$, and $p$ a path of type $\shape_G\eqto y$.
- Then $(\id_\BG,\inv p)$ is a pointed equivalence of type $\BG \equivto_* (\BG_\div , y)$
- and hence induces an isomorphism from $G$ to $\mkgroup(\BG_\div , y)$.
- By \cref{remark:groupsasunivalenttype} we then get an identification of these groups.
+ let $G$ be a group, $y$ an element of $\BG$, and $p$ a path of 
+ type $\shape_G\eqto y$.
+ Then $(\id_\BG,\inv p)$ is a pointed equivalence of type 
+ $\BG \equivto_* (\BG_\div , y)$ and hence induces an isomorphism 
+ from $G$ to $\mkgroup(\BG_\div , y)$.\footnote{%
+ One may wonder why $\inv p$ in $(\id_\BG,\inv p)$.
+ The reason is our convention for the direction of the
+ pointing path of a pointed map in combination with the
+ convention that $\conj^p$ is transport along $p$.}
+ By \cref{remark:groupsasunivalenttype} we then get an 
+ identification of these groups.
  Moreover, by path induction on $p$, the equivalence
- $\USym(\mkgroup(\id_\BG,\inv p))\jdeq\loops(\id_\BG,\inv p)$ of type $(\sh_G\eqto\sh_G)\equivto(y\eqto y)$
- can be identified with the map $g \mapsto p g \inv p$. In the special case
+ $\USym(\mkgroup(\id_\BG,\inv p))\jdeq\loops(\id_\BG,\inv p)$ 
+ of type $(\sh_G\eqto\sh_G)\equivto(y\eqto y)$\footnote{%
+ Note that $\USym(\mkgroup(\BG_\div,y)) \jdeq 
+ \loops(BG_\div,y) \jdeq (y\eqto y)$.}
+ can be identified with the map $g \mapsto p g \inv p$. 
+ In the special case
  of $y\jdeq\sh_G$ this map is precisely $\conj^p$ from \cref{xca:conj}.
- The more general form introduced here is also called conjugation.\index{conjugation}
-
-  \MB{OLD OBSOLTE?}In terms of concrete groups, it is not so easy to define the
-  classifying map of conjugation by $g$, as homomorphism (indeed, automorphism)
-  $\conj^g : \Hom(G,G)$, for any particular given $g:\USymG$.\footnote{%
-    We shall develop a general method of doing this in~\cref{sec:homabsisconcr}
-    below.}
-  However, conjugation defines uniformly a homomorphism
-  $\conj : \Hom(G,\Aut(G))$.
-  And this has a very pretty classifying map,
-  $\Bconj : \BG \ptdto \BAut(G)$, defined by $t \mapsto \mkgroup(\BG_\div,t)$,
-  that is, if we have a shape $t$ for $G$,
-  then we get a new group in the component of $G$
-  by taking $t$ as the new designed shape.
-
-  To see that this really captures conjugation as defined above on underlying
-  symmetries, we generalize, and prove for all $t:\BG$ and all $g : \shape_G \eqto t$,
-  that $\ap{\Bconj}(g)$ applied to any $s : \USymG$ equals
-  $gsg^{-1} : t \eqto t$.\footnote{%
-    Note that $\USym(\Bconj(t)) \jdeq \USym(\mkgroup(\BG_\div,t)) \jdeq \loops(BG_\div,t)
-    \jdeq (t\eqto t)$.}
-  And this follows by induction on $g$.
-\MB{ENDOLD}
+ \footnote{%
+ One can also start from $\conj^p$ and ask for a pointed map 
+ $f: \BG_\div \ptdto \BG_\div$ such that $\loops(f) = \conj^p$.
+ We shall develop a general method for constructing
+ such a map $f$  in~\cref{sec:homabsisconcr} below.}
+ The more general form introduced here is also called conjugation.
+ \index{conjugation}
 \end{example}
 
 The above example motivates and justifies the following definition of
@@ -3520,17 +3515,17 @@ \section{$G$-sets vs $\abstr(G)$-sets}
 What abstract $\abstr(G)$-set does this correspond to?
 In particular, under the equivalence $\abstr:\Hom(H,G)\to\Hom^\abstr(\abstr(H),\abstr(G))$, what is the corresponding action of $\abstr(G)$ on the abstract homomorphisms?
 
-The answer is that $g:\USym G$ acts on $\Hom^\abstr(\abstr(H),\abstr(G))$ by postcomposing with conjugation $c^g$ by $g$ as defined in \cref{ex:conjhomo}.
+The answer is that $g:\USym G$ acts on $\Hom^\abstr(\abstr(H),\abstr(G))$ by postcomposing with conjugation $\conj^g$ by $g$ as defined in \cref{ex:conjhomo}.
 
 Let us spell this out in some detail:
 If $(F,p):\Hom(H,G)(\shape_G)\defequi
- \sum_{F:\BH_\div\to \BG_\div}(\shape_G=F(\shape_H))$ and $g:\USym G$, then $g\cdot(F,p)\defequi(F,p\,g^{-1})$.  If we show that the action of $g$ sends $\abstr(F,p)$ to $c^g\circ\abstr(F,p)$ we are done.
+ \sum_{F:\BH_\div\to \BG_\div}(\shape_G=F(\shape_H))$ and $g:\USym G$, then $g\cdot(F,p)\defequi(F,p\,g^{-1})$.  If we show that the action of $g$ sends $\abstr(F,p)$ to $\conj^g\circ\abstr(F,p)$ we are done.
 
 Recall that $\abstr(F,p)$ consists of the composite
 $$\xymatrix{\USym H\ar[r]^-{F^=}&(F(\shape_H)=F(\shape_G))\ar[rr]^-{t\mapsto p^{-1}t\,p}&&\USym G},$$
 (\ie $\abstr(F,p)$ applied to $q:\USym H $ is  $p^{-1}F^=(q)\,p$)  together with the proof that this is an abstract group homomorphism.
 We see that $\abstr(F,p\,g^{-1})$ is given by conjugation:
-$q\mapsto(p\,g^{-1})^{-1}F^=(q)\,(p\,g^{-1})=g\,(p^{-1}F^=(q)\,p)\,g^{-1}$, or in other words $c^g\circ\abstr(F,p)$.
+$q\mapsto(p\,g^{-1})^{-1}F^=(q)\,(p\,g^{-1})=g\,(p^{-1}F^=(q)\,p)\,g^{-1}$, or in other words $\conj^g\circ\abstr(F,p)$.
 \end{example}
 For reference we list the conclusion of this example as a lemma:
 \begin{lemma}\label{lem:abstrandconj}

macros.tex

@@ -627,7 +627,7 @@
 \newBcommand{Hom}
 \newBcommand{Iso}
 \newBcommand{sgn} % sign homomorphism
-\newBcommand[c]{conj} % conjugation homomorphism
+\newBcommand{conj} % conjugation homomorphism MB c := conj 040124
 \newBcommand{Z} % center
 \newBcommand[F]{FG} % free group
 \newBcommand[V]{VG} % Vierer group