The Stacks project

Proof. By Lemma 39.20.4 the quotient sheaves have a simple description which we will use below without further mention. We first show that

is injective. Namely, assume $a, b \in U(T)$ map to the same element on the right hand side. Then $f(a) = f(b)$. After replacing $T$ by the members of a $\tau$-covering we may assume that there exists an $r \in R(T)$ such that $a = s(r)$ and $b = t(r)$. Then $r' = f(r)$ is a $T$-valued point of $R'$ with $s'(r') = t'(r')$. Hence $r' = e'(f(a))$ (where $e'$ is the identity of the groupoid scheme associated to $j'$, see Lemma 39.13.3). Because the first diagram of the lemma is cartesian this implies that $r$ has to equal $e(a)$. Thus $a = b$.

Finally, we show that the displayed arrow is surjective. Let $T$ be a scheme over $S$ and let $(a', \overline{b})$ be a section of the sheaf $U' \times _{U'/R'} U/R$ over $T$. After replacing $T$ by the members of a $\tau$-covering we may assume that $\overline{b}$ is the class of an element $b \in U(T)$. After replacing $T$ by the members of a $\tau$-covering we may assume that there exists an $r' \in R'(T)$ such that $a' = t(r')$ and $s'(r') = f(b)$. Because the first diagram of the lemma is cartesian we can find $r \in R(T)$ such that $s(r) = b$ and $f(r) = r'$. Then it is clear that $a = t(r) \in U(T)$ is a section which maps to $(a', \overline{b})$. $\square$