The Stacks project

Lemma 39.21.2. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. The category of groupoid schemes cartesian over $(U, R, s, t, c)$ is equivalent to the category of pairs $(V, \varphi )$ where $V$ is a scheme over $U$ and

\[ \varphi : V \times _{U, t} R \longrightarrow R \times _{s, U} V \]

is an isomorphism over $R$ such that $e^*\varphi = \text{id}_ V$ and such that

\[ c^*\varphi = \text{pr}_1^*\varphi \circ \text{pr}_0^*\varphi \]

as morphisms of schemes over $R \times _{s, U, t} R$.

Proof. The pullback notation in the lemma signifies base change. The displayed formula makes sense because

\[ (R \times _{s, U, t} R) \times _{\text{pr}_1, R, \text{pr}_1} (V \times _{U, t} R) = (R \times _{s, U, t} R) \times _{\text{pr}_0, R, \text{pr}_0} (R \times _{s, U} V) \]

as schemes over $R \times _{s, U, t} R$.

Given $(V, \varphi )$ we set $U' = V$ and $R' = V \times _{U, t} R$. We set $t' : R' \to U'$ equal to the projection $V \times _{U, t} R \to V$. We set $s'$ equal to $\varphi $ followed by the projection $R \times _{s, U} V \to V$. We set $c'$ equal to the composition

\begin{align*} R' \times _{s', U', t'} R' & \xrightarrow {\varphi , 1} (R \times _{s, U} V) \times _ V (V \times _{U, t} R) \\ & \xrightarrow {} R \times _{s, U} V \times _{U, t} R \\ & \xrightarrow {\varphi ^{-1}, 1} V \times _{U, t} (R \times _{s, U, t} R) \\ & \xrightarrow {1, c} V \times _{U, t} R = R' \end{align*}

A computation, which we omit shows that we obtain a groupoid scheme over $(U, R, s, t, c)$. It is clear that this groupoid scheme is cartesian over $(U, R, s, t, c)$.

Conversely, given $f : (U', R', s', t', c') \to (U, R, s, t, c)$ cartesian then the morphisms

\[ U' \times _{U, t} R \xleftarrow {t', f} R' \xrightarrow {f, s'} R \times _{s, U} U' \]

are isomorphisms and we can set $V = U'$ and $\varphi $ equal to the composition $(f, s') \circ (t', f)^{-1}$. We omit the proof that $\varphi $ satisfies the conditions in the lemma. We omit the proof that these constructions are mutually inverse. $\square$

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