Lemma 78.22.2. Assume $B \to S$, $(U, R, s, t, c)$, and $\pi : \mathcal{S}_ U \to [U/R]$ are as in Lemma 78.20.2. The $2$-commutative square

of Lemma 78.20.3 is a $2$-fibre product of stacks in groupoids of $(\mathit{Sch}/S)_{fppf}$.

Lemma 78.22.2. Assume $B \to S$, $(U, R, s, t, c)$, and $\pi : \mathcal{S}_ U \to [U/R]$ are as in Lemma 78.20.2. The $2$-commutative square

\[ \xymatrix{ \mathcal{S}_ R \ar[r]_ s \ar[d]_ t & \mathcal{S}_ U \ar[d]^\pi \\ \mathcal{S}_ U \ar[r]^-\pi & [U/R] } \]

of Lemma 78.20.3 is a $2$-fibre product of stacks in groupoids of $(\mathit{Sch}/S)_{fppf}$.

**Proof.**
According to Stacks, Lemma 8.5.6 the lemma makes sense. It also tells us that we have to show that the functor

\[ \mathcal{S}_ R \longrightarrow \mathcal{S}_ U \times _{[U/R]} \mathcal{S}_ U \]

which maps $r : T \to R$ to $(T, t(r), s(r), \alpha (r))$ is an equivalence, where the right hand side is the $2$-fibre product as described in Categories, Lemma 4.32.3. This is, after spelling out the definitions, exactly the content of Lemma 78.22.1. (Alternative proof: Work out the meaning of Lemma 78.21.2 in this situation will give you the result also.) $\square$

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