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Correct paragraph between 4.5.7 and 4.5.8. It now avoids an inmplicit…
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… use of delooping. The former footnote has made its way to the text. A forward reference to delooping has been made in a new footnote.
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pierrecagne committed Aug 8, 2024
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Expand Up @@ -1828,7 +1828,7 @@ \section{The sign homomorphism}
\emph{can} be identified with $\set{\pm1}$,\footnote{%
In this section, we identify $\USG_2$ with the set $\set{\pm1}$,
which has a compatible abstract group structure given by multiplication.}
according to whether it transposes the elements of the $T$, or not.
according to whether it transposes the elements of $T$, or not.
Hence, we can define the sign of any permutation of a finite set:
\begin{definition}\label{def:sgn-permutation}
Let $A$ be a finite set, and let $\sigma$ be a permutation of $A$.
Expand All @@ -1839,17 +1839,20 @@ \section{The sign homomorphism}
set $\Bsgn_\div(A)$, or not.
We write $\sgn(\sigma):\set{\pm1}$ for the sign of $\sigma$.
\end{definition}
For permutations of the standard $n$-element set,
this is the same as the value $\Usgn(\sigma) : \USG_2$.
Note that $\sgn$ defines an abstract homomorphism from $\Aut(A)$ to $\SG_2$
for each $A$, since it does so for $A \jdeq \sh_{\SG_n}$.
Hence we in fact have homomorphisms $\sgn^A : \Hom(\Aut(A),\SG_2)$
for all finite sets $A$.\footnote{%
We need to add the decoration signifying which finite set $A$
is considered as reference, since the classifying map depends on it:
We can take $\Bsgn^A_\div(B) \defeq (\Bsgn_\div(A) \eqto \Bsgn_\div(B))$
with the pointing given by the above identification of $\Bsgn^A_\div(A)$
with $\set{\pm1}$.}
For permutations of the standard $n$-element set, this is the same as the value
$\Usgn(\sigma) : \USG_2$. Note that $\sgn$ defines an abstract homomorphism from
$\Aut(A)$ to $\SG_2$ for each $A$, since it does so for $A \jdeq
\sh_{\SG_n}$. Even better, this abstract homomorphism comes from a concrete one
$\sgn^A : \Hom(\Aut(A),\SG_2)$ for each finite set $A$. Indeed, since
$T \eqto U$ is a 2-element set for any 2-element sets $T$ and $U$, we can
consider the map $\Bsgn^A_\div : \BAut(A) \to \BSG_2$ that associates
$B : \BAut(A)$ with $(\Bsgn_\div(A) \eqto \Bsgn_\div(B))$. The identification of
$\Bsgn^A_\div(A)$ with $\{\pm1\}$ mentioned above makes $\Bsgn^A_\div$ into a
pointed map $\Bsgn^A : \BAut(A) \ptdto \BSG_2$, i.e., it defines an homomorphism
$\sgn^A : \Hom(\Aut(A),\SG_2)$, as announced.%
\footnote{This is an instance of a more general construction, called {\em
delooping} (see \cref{sec:delooping}). The formula for $\Bsgn^A_\div$ here is
very simple since $\SG_2$ is a fairly simple group.} %

\begin{lemma}\label{lem:sign-properties}
\begin{enumerate}
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