Lemma 77.25.3. Notation and assumption as in Lemma 77.21.1. Assume that

$\xymatrix{ R \ar[d]_ s \ar[r]_ f & R' \ar[d]^{s'} \\ U \ar[r]^ f & U' }$

is cartesian. Then

$\xymatrix{ \mathcal{S}_ U \ar[d] \ar[r] & [U/R] \ar[d]^{[f]} \\ \mathcal{S}_{U'} \ar[r] & [U'/R'] }$

is a $2$-fibre product square.

Proof. Applying the inverse isomorphisms $i : R \to R$ and $i' : R' \to R'$ to the (first) cartesian diagram of the statement of the lemma we see that

$\xymatrix{ R \ar[d]_ t \ar[r]_ f & R' \ar[d]^{t'} \\ U \ar[r]^ f & U' }$

is cartesian as well. By Lemma 77.21.2 we have a $2$-fibre square

$\xymatrix{ [U''/R''] \ar[d] \ar[r] & [U/R] \ar[d] \\ \mathcal{S}_{U'} \ar[r] & [U'/R'] }$

where $U'' = U \times _{f, U', t'} R'$ and $R'' = R \times _{f \circ s, U', t'} R'$. By the above we see that $(t, f) : R \to U''$ is an isomorphism, and that

$R'' = R \times _{f \circ s, U', t'} R' = R \times _{s, U} U \times _{f, U', t'} R' = R \times _{s, U, t} \times R.$

Explicitly the isomorphism $R \times _{s, U, t} R \to R''$ is given by the rule $(r_0, r_1) \mapsto (r_0, f(r_1))$. Moreover, $s'', t'', c''$ translate into the maps

$R \times _{s, U, t} R \to R, \quad s''(r_0, r_1) = r_1, \quad t''(r_0, r_1) = c(r_0, r_1)$

and

$\begin{matrix} c'' : & (R \times _{s, U, t} R) \times _{s'', R, t''} (R \times _{s, U, t} R) & \longrightarrow & R \times _{s, U, t} R, \\ & ((r_0, r_1), (r_2, r_3)) & \longmapsto & (c(r_0, r_2), r_3). \end{matrix}$

Precomposing with the isomorphism

$R \times _{s, U, s} R \longrightarrow R \times _{s, U, t} R, \quad (r_0, r_1) \longmapsto (c(r_0, i(r_1)), r_1)$

we see that $t''$ and $s''$ turn into $\text{pr}_0$ and $\text{pr}_1$ and that $c''$ turns into $\text{pr}_{02} : R \times _{s, U, s} R \times _{s, U, s} R \to R \times _{s, U, s} R$. Hence we see that there is an isomorphism $[U''/R''] \cong [R/R \times _{s, U, s} R]$ where as a groupoid in algebraic spaces $(R, R \times _{s, U, s} R, s'', t'', c'')$ is the restriction of the trivial groupoid $(U, U, \text{id}, \text{id}, \text{id})$ via $s : R \to U$. Since $s : R \to U$ is a surjection of fppf sheaves (as it has a right inverse) the morphism

$[U''/R''] \cong [R/R \times _{s, U, s} R] \longrightarrow [U/U] = \mathcal{S}_ U$

is an equivalence by Lemma 77.25.2. This proves the lemma. $\square$

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