Lemma 40.9.3. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Let $g : U' \to U$ and $g' : U'' \to U'$ be morphisms of schemes. Set $g'' = g \circ g'$. Let $(U', R', s', t', c')$ be the restriction of $R$ to $U'$. Let $h = s \circ \text{pr}_1 : U' \times _{g, U, t} R \to U$, let $h' = s' \circ \text{pr}_1 : U'' \times _{g', U', t} R \to U'$, and let $h'' = s \circ \text{pr}_1 : U'' \times _{g'', U, t} R \to U$. The following diagram is commutative

\[ \xymatrix{ U'' \times _{g', U', t} R' \ar[d]^{h'} & (U' \times _{g, U, t} R) \times _ U (U'' \times _{g'', U, t} R) \ar[l] \ar[r] \ar[d] & U'' \times _{g'', U, t} R \ar[d]_{h''} \\ U' & U' \times _{g, U, t} R \ar[l]_{\text{pr}_0} \ar[r]^ h & U } \]

with both squares cartesian where the left upper horizontal arrow is given by the rule

\[ \begin{matrix} (U' \times _{g, U, t} R) \times _ U (U'' \times _{g'', U, t} R)
& \longrightarrow
& U'' \times _{g', U', t} R'
\\ ((u', r_0), (u'', r_1))
& \longmapsto
& (u'', (c(r_1, i(r_0)), (g'(u''), u')))
\end{matrix} \]

with notation as explained in the proof.

**Proof.**
We work this out by exploiting the functorial point of view and reducing the lemma to a statement on arrows in restrictions of a groupoid category. In the last formula of the lemma the notation $((u', r_0), (u'', r_1))$ indicates a $T$-valued point of $(U' \times _{g, U, t} R) \times _ U (U'' \times _{g'', U, t} R)$. This means that $u', u'', r_0, r_1$ are $T$-valued points of $U', U'', R, R$ and that $g(u') = t(r_0)$, $g(g'(u'')) = g''(u'') = t(r_1)$, and $s(r_0) = s(r_1)$. It would be more correct here to write $g \circ u' = t \circ r_0$ and so on but this makes the notation even more unreadable. If we think of $r_1$ and $r_0$ as arrows in a groupoid category then we can represent this by the picture

\[ \xymatrix{ t(r_0) = g(u') & s(r_0) = s(r_1) \ar[l]_{r_0} \ar[r]^-{r_1} & t(r_1) = g(g'(u'')) } \]

This diagram in particular demonstrates that the composition $c(r_1, i(r_0))$ makes sense. Recall that

\[ R' = R \times _{(t, s), U \times _ S U, g \times g} U' \times _ S U' \]

hence a $T$-valued point of $R'$ looks like $(r, (u'_0, u'_1))$ with $t(r) = g(u'_0)$ and $s(r) = g(u'_1)$. In particular given $((u', r_0), (u'', r_1))$ as above we get the $T$-valued point $(c(r_1, i(r_0)), (g'(u''), u'))$ of $R'$ because we have $t(c(r_1, i(r_0))) = t(r_1) = g(g'(u''))$ and $s(c(r_1, i(r_0))) = s(i(r_0)) = t(r_0) = g(u')$. We leave it to the reader to show that the left square commutes with this definition.

To show that the left square is cartesian, suppose we are given $(v'', p')$ and $(v', p)$ which are $T$-valued points of $U'' \times _{g', U', t} R'$ and $U' \times _{g, U, t} R$ with $v' = s'(p')$. This also means that $g'(v'') = t'(p')$ and $g(v') = t(p)$. By the discussion above we know that we can write $p' = (r, (u_0', u_1'))$ with $t(r) = g(u'_0)$ and $s(r) = g(u'_1)$. Using this notation we see that $v' = s'(p') = u_1'$ and $g'(v'') = t'(p') = u_0'$. Here is a picture

\[ \xymatrix{ s(p) \ar[r]^-p & g(v') = g(u'_1) \ar[r]^-r & g(u'_0) = g(g'(v'')) } \]

What we have to show is that there exists a unique $T$-valued point $((u', r_0), (u'', r_1))$ as above such that $v' = u'$, $p = r_0$, $v'' = u''$ and $p' = (c(r_1, i(r_0)), (g'(u''), u'))$. Comparing the two diagrams above it is clear that we have no choice but to take

\[ ((u', r_0), (u'', r_1)) = ((v', p), (v'', c(r, p)) \]

Some details omitted.
$\square$

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