Sign error in proof lemma in examples-stacks.tex

Thanks to Yogesh More
http://stacks.math.columbia.edu/tag/04UV#comment-1427

examples-stacks.tex

@@ -1317,33 +1317,33 @@ \section{Quotients by group actions}
 diagram
 $$
 \xymatrix{
-G \times_{B, \pi \circ x} U \ar[d] \ar[r]_{R_g} &
+G \times_{B, \pi \circ x} U \ar[d] \ar[r]_{R_{g^{-1}}} &
 G \times_{B, \pi \circ x'} U' \ar[d] \\
 U \ar[r]^f & U'
 }
 $$
-where $R_g$ on $T$-valued points is given by
+where $R_{g^{-1}}$ on $T$-valued points is given by
 $$
-R_g(g', u) = (m(g', g), f(u))
+R_{g^{-1}}(g', u) = (m(g', i(g)), f(u))
 $$
 To see that this works we have to verify that
 $$
 a \circ (\text{id}_G \times x)
 =
-a \circ (\text{id}_G \times x') \circ R_g
+a \circ (\text{id}_G \times x') \circ R_{g^{-1}}
 $$
 which is true because the right hand side applied to the $T$-valued point
-$(g', u)$ gives
+$(g', u)$ gives the desired equality
 \begin{align*}
-a((\text{id}_G \times x')(m(g', g), f(u)))
+a((\text{id}_G \times x')(m(g', i(g)), f(u)))
 & =
-a(m(g', g), x'(f(u))) \\
+a(m(g', i(g)), x'(f(u))) \\
 & =
-a(g', a(g, x'(f(u)))) \\
+a(g', a(i(g), x'(f(u)))) \\
 & =
 a(g', x(u))
 \end{align*}
-because $a(g, x) = x' \circ f$ as desired.
+because $a(g, x) = x' \circ f$ and hence $a(i(g), x' \circ f) = x$.
 
 \medskip\noindent
 By the universal property of stackification from
@@ -1391,8 +1391,8 @@ \section{Quotients by group actions}
 G \times_{B, \pi \circ x' \circ u} T
 $$
 of trivial $G_T$-torsors compatible with the given maps to $X$.
-Since the torsors are trivial we see that $R = R_g$ (right multiplication)
-by some $g \in G(T)$. Compatibility with the maps
+Since the torsors are trivial we see that $R = R_{g^{-1}}$
+(right multiplication) by some $g \in G(T)$. Compatibility with the maps
 $a \circ (1_G, x \circ u), a \circ (1_G, x' \circ u) : G \times_B T \to X$
 is equivalent to the condition that $a(g, x \circ u) = x' \circ u$.
 Hence we obtain the desired equality of $\mathit{Isom}$-sheaves.