39.13 Groupoids
Recall that a groupoid is a category in which every morphism is an isomorphism, see Categories, Definition 4.2.5. Hence a groupoid has a set of objects $\text{Ob}$, a set of arrows $\text{Arrows}$, a source and target map $s, t : \text{Arrows} \to \text{Ob}$, and a composition law $c : \text{Arrows} \times _{s, \text{Ob}, t} \text{Arrows} \to \text{Arrows}$. These maps satisfy exactly the following axioms
(associativity) $c \circ (1, c) = c \circ (c, 1)$ as maps $\text{Arrows} \times _{s, \text{Ob}, t} \text{Arrows} \times _{s, \text{Ob}, t} \text{Arrows} \to \text{Arrows}$,
(identity) there exists a map $e : \text{Ob} \to \text{Arrows}$ such that
$s \circ e = t \circ e = \text{id}$ as maps $\text{Ob} \to \text{Ob}$,
$c \circ (1, e \circ s) = c \circ (e \circ t, 1) = 1$ as maps $\text{Arrows} \to \text{Arrows}$,
(inverse) there exists a map $i : \text{Arrows} \to \text{Arrows}$ such that
$s \circ i = t$, $t \circ i = s$ as maps $\text{Arrows} \to \text{Ob}$, and
$c \circ (1, i) = e \circ t$ and $c \circ (i, 1) = e \circ s$ as maps $\text{Arrows} \to \text{Arrows}$.
If this is the case the maps $e$ and $i$ are uniquely determined and $i$ is a bijection. Note that if $(\text{Ob}', \text{Arrows}', s', t', c')$ is a second groupoid category, then a functor $f : (\text{Ob}, \text{Arrows}, s, t, c) \to (\text{Ob}', \text{Arrows}', s', t', c')$ is given by a pair of set maps $f : \text{Ob} \to \text{Ob}'$ and $f : \text{Arrows} \to \text{Arrows}'$ such that $s' \circ f = f \circ s$, $t' \circ f = f \circ t$, and $c' \circ (f, f) = f \circ c$. The compatibility with identity and inverse is automatic. We will use this below. (Warning: The compatibility with identity has to be imposed in the case of general categories.)
Definition 39.13.1. Let $S$ be a scheme.
A groupoid scheme over $S$, or simply a groupoid over $S$ is a quintuple $(U, R, s, t, c)$ where $U$ and $R$ are schemes over $S$, and $s, t : R \to U$ and $c : R \times _{s, U, t} R \to R$ are morphisms of schemes over $S$ with the following property: For any scheme $T$ over $S$ the quintuple
\[ (U(T), R(T), s, t, c) \]
is a groupoid category in the sense described above.
A morphism $f : (U, R, s, t, c) \to (U', R', s', t', c')$ of groupoid schemes over $S$ is given by morphisms of schemes $f : U \to U'$ and $f : R \to R'$ with the following property: For any scheme $T$ over $S$ the maps $f$ define a functor from the groupoid category $(U(T), R(T), s, t, c)$ to the groupoid category $(U'(T), R'(T), s', t', c')$.
Let $(U, R, s, t, c)$ be a groupoid over $S$. Note that, by the remarks preceding the definition and the Yoneda lemma, there are unique morphisms of schemes $e : U \to R$ and $i : R \to R$ over $S$ such that for every scheme $T$ over $S$ the induced map $e : U(T) \to R(T)$ is the identity, and $i : R(T) \to R(T)$ is the inverse of the groupoid category. The septuple $(U, R, s, t, c, e, i)$ satisfies commutative diagrams corresponding to each of the axioms (1), (2)(a), (2)(b), (3)(a) and (3)(b) above, and conversely given a septuple with this property the quintuple $(U, R, s, t, c)$ is a groupoid scheme. Note that $i$ is an isomorphism, and $e$ is a section of both $s$ and $t$. Moreover, given a groupoid scheme over $S$ we denote
\[ j = (t, s) : R \longrightarrow U \times _ S U \]
which is compatible with our conventions in Section 39.3 above. We sometimes say “let $(U, R, s, t, c, e, i)$ be a groupoid over $S$” to stress the existence of identity and inverse.
Lemma 39.13.2. Given a groupoid scheme $(U, R, s, t, c)$ over $S$ the morphism $j : R \to U \times _ S U$ is a pre-equivalence relation.
Proof.
Omitted. This is a nice exercise in the definitions.
$\square$
Lemma 39.13.3. Given an equivalence relation $j : R \to U \times _ S U$ over $S$ there is a unique way to extend it to a groupoid $(U, R, s, t, c)$ over $S$.
Proof.
Omitted. This is a nice exercise in the definitions.
$\square$
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.
$\square$
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
\begin{equation} \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 } \end{equation}
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.
$\square$
Lemma 39.13.6. Let $(U, R, s, t, c)$ be a groupoid over a scheme $S$. Let $S' \to S$ be a morphism. Then the base changes $U' = S' \times _ S U$, $R' = S' \times _ S R$ endowed with the base changes $s'$, $t'$, $c'$ of the morphisms $s, t, c$ form a groupoid scheme $(U', R', s', t', c')$ over $S'$ and the projections determine a morphism $(U', R', s', t', c') \to (U, R, s, t, c)$ of groupoid schemes over $S$.
Proof.
Omitted. Hint: $R' \times _{s', U', t'} R' = S' \times _ S (R \times _{s, U, t} R)$.
$\square$
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