Lemma 39.13.4. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid over $S$. In the commutative diagram

the two lower squares are fibre product squares. Moreover, the triangle on top (which is really a square) is also cartesian.

Lemma 39.13.4. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid over $S$. In the commutative diagram

\[ \xymatrix{ & U & \\ R \ar[d]_ s \ar[ru]^ t & R \times _{s, U, t} R \ar[l]^-{\text{pr}_0} \ar[d]^{\text{pr}_1} \ar[r]_-c & R \ar[d]^ s \ar[lu]_ t \\ U & R \ar[l]_ t \ar[r]^ s & U } \]

the two lower squares are fibre product squares. Moreover, the triangle on top (which is really a square) is also cartesian.

**Proof.**
Omitted. Exercise in the definitions and the functorial point of view in algebraic geometry.
$\square$

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