diff --git a/group.tex b/group.tex
index 1dab4c7..d1ed5fe 100644
--- a/group.tex
+++ b/group.tex
@@ -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 $A$.
@@ -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}