Lemma 39.13.5. Let $S$ be a scheme. Let $(U, R, s, t, c, e, i)$ be a groupoid over $S$. The diagram

39.13.5.1
$$\label{groupoids-equation-pull} \xymatrix{ R \times _{t, U, t} R \ar@<1ex>[r]^-{\text{pr}_1} \ar@<-1ex>[r]_-{\text{pr}_0} \ar[d]_{(\text{pr}_0, c \circ (i, 1))} & R \ar[r]^ t \ar[d]^{\text{id}_ R} & U \ar[d]^{\text{id}_ U} \\ R \times _{s, U, t} R \ar@<1ex>[r]^-c \ar@<-1ex>[r]_-{\text{pr}_0} \ar[d]_{\text{pr}_1} & R \ar[r]^ t \ar[d]^ s & U \\ R \ar@<1ex>[r]^ s \ar@<-1ex>[r]_ t & U }$$

is commutative. The two top rows are isomorphic via the vertical maps given. The two lower left squares are cartesian.

Proof. The commutativity of the diagram follows from the axioms of a groupoid. Note that, in terms of groupoids, the top left vertical arrow assigns to a pair of morphisms $(\alpha , \beta )$ with the same target, the pair of morphisms $(\alpha , \alpha ^{-1} \circ \beta )$. In any groupoid this defines a bijection between $\text{Arrows} \times _{t, \text{Ob}, t} \text{Arrows}$ and $\text{Arrows} \times _{s, \text{Ob}, t} \text{Arrows}$. Hence the second assertion of the lemma. The last assertion follows from Lemma 39.13.4. $\square$

Comment #1869 by Laurent Moret-Bailly on

I find the notation $\mathrm{pr}_0\times c\circ(i,1)$ in the diagram confusing (fortunately, this map is clearly defined in the proof). I would have denoted the map by $(\mathrm{pr}_0, c\circ(i\times 1))$.

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