minor Ch. 6

Author: Marc Bezem

absgroup.tex

@@ -769,7 +769,7 @@ \section{Homomorphisms, from abstract to concrete and back}
   \]
   to be the set quotient of $(\sh_H\eqto y)\times X(\sh_G)$ modulo
   the equivalence relation $(p,x)\sim(p\, \phi(\inv g),g\cdot_X x)$ for all 
-  $g:\sh_G\eqto\sh_G$.\footnote{\MB{Induction? $\exists y$?}}
+  $g:\sh_G\eqto\sh_G$.\footnote{\MB{Should be map to $\Prop$. Induction? $\exists g$?}}
 \end{definition}
 
 \begin{lemma}\label{lem:abshom_*}
@@ -798,26 +798,30 @@ \section{Homomorphisms, from abstract to concrete and back}
 $$\pathsp{}:\Hom(G,H)\we
 \bigl((\typetorsor_G,\princ G)\ptdto(\typetorsor_H,\princ H)\bigr)
 $$
-by mapping $\Bf$ to $\princ H\circ \Bf \circ \invprinc G$.
-Then $A\defeq ({\abstr}\circ\pathsp{}^{-1})$ is a map
+by mapping, for any $f:\Hom(G,H)$, $\Bf$ to 
+$\pathsp{\blank}^H\circ \Bf \circ (\pathsp{\blank}^G)^{-1}$.
+Then $A\defeq ({\abstr}\circ\pathsp{}^{-1})$ is a map\footnote{%
+\MB{To show that 
+$\abstr:\Hom(G,H)\to \absHom(\abstr(G),\abstr(H))$ is an equivalence,
+we factor it as $A\circ \pathsp{}$ and prove that both factors are
+equivalences.}}
 \[
 A:\bigl((\typetorsor_G,\princ G)\ptdto(\typetorsor_H,\princ H)\bigr)
     \to \absHom(\abstr(G),\abstr(H))
 \]
-satisfying ${\loops f}\circ \pathsp{\blank}^G=\pathsp{\blank}^H \circ A(f)$
-for all $f$ in the domain of $A$:
+satisfying, for all $g$ in the domain of $A$,
+ ${\loops g}\circ \pathsp{\blank}^G=\pathsp{\blank}^H \circ A(g)$:
 %see \cref{fig:mapA}
 \[%begin{marginfigure}
       \begin{tikzcd}[ampersand replacement=\&]
-        \USymG \ar[r,"A(f)"]\ar[d,eql,"{\pathsp{\blank}^G}"'] \& 
+        \USymG \ar[r,"A(g)"]\ar[d,eql,"{\pathsp{\blank}^G}"'] \& 
         \USymH \ar[d,eqr,"{\pathsp{\blank}^H}"] \\
-        \princ G\eqto\princ G \ar[r,"{\loops f}"'] \& \princ H\eqto\princ H
+        \princ G\eqto\princ G \ar[r,"{\loops g}"'] \& \princ H\eqto\princ H
       \end{tikzcd}
      % \caption{\label{fig:mapA}caption}
 \]%end{marginfigure}
-(together with the proof that $A(f)$ is an abstract group homomorphism). 
+(together with the proof that $A(g)$ is an abstract group homomorphism). 
 We are done if we show that $A$ is an equivalence.
-\marginnote{The reason to complicate $\abstr$ this way is that it gets easier to write out the inverse function.\MB{MB doesn't understand.}}
 
 For any $\phi:\absHom(\abstr(G),\abstr(H))$, recall the pointed map
 \[

actions.tex

@@ -2132,6 +2132,10 @@ \subsection{Homomorphisms and torsors}
     types $\Hom_H(f_*X,Y)$ and $\Hom_G(X,f^*\,Y)$, 
     for all $G$-sets $X$ and $H$-sets $Y$.
 \end{xca}
+% Solution 2nd part: elaborating $\Hom_H(f_*X,Y)$ you can leave out
+% the set truncation since $Y$ is an $H$-set. Now turn the sum
+% over $z:\BG$ into an upfront product. Then eliminate the product
+% over $w:\BH$ by using $f(z)\eqto w$.
 
 \begin{remark}\label{rem:^*-_*-as-(pre)image}\MB{New:}
 One may view $f_*X$ and $f^*Y$ as generalizations of

macros.tex

@@ -843,7 +843,7 @@
 \newcommand*{\absHom}{\Hom^{\abstr}}
 \newcommand*{\absGSetvar}[1][X]{\mathcal{#1}}
 \newcommand*{\pathsp}[1]{\constant{\mathbb P}_{\!#1}} % NB negative thin space
-\newcommand*{\invpathsp}[1]{\constant{P}_{\!#1}^{-1}} % NB negative thin space
+\newcommand*{\invpathsp}[1]{\constant{\mathbb P}_{\!#1}^{-1}} % NB negative thin space
 \newcommand*{\uc}[1]{{\pathsp{#1}}}%universal set bundle
 \newcommand*{\abstr}{\casop{\constant{abs}}}
 \newcommand*{\agp}[1]{\mathcal #1} %generic abstract group