78.11 Groupoids in algebraic spaces
Please refer to Groupoids, Section 39.13 for notation.
Definition 78.11.1. Let B \to S as in Section 78.3.
A groupoid in algebraic spaces over B is a quintuple (U, R, s, t, c) where U and R are algebraic spaces over B, and s, t : R \to U and c : R \times _{s, U, t} R \to R are morphisms of algebraic spaces over B with the following property: For any scheme T over B the quintuple
is a groupoid category.
A morphism f : (U, R, s, t, c) \to (U', R', s', t', c') of groupoids in algebraic spaces over B is given by morphisms of algebraic spaces f : U \to U' and f : R \to R' over B with the following property: For any scheme T over B 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 in algebraic spaces over B. Note that there are unique morphisms of algebraic spaces e : U \to R and i : R \to R over B such that for every scheme T over B 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) of Groupoids, Section 39.13. Conversely given a septuple with this property the quintuple (U, R, s, t, c) is a groupoid in algebraic spaces over B. Note that i is an isomorphism, and e is a section of both s and t. Moreover, given a groupoid in algebraic spaces over B we denote
j = (t, s) : R \longrightarrow U \times _ B U
which is compatible with our conventions in Section 78.4 above. We sometimes say “let (U, R, s, t, c, e, i) be a groupoid in algebraic spaces over B” to stress the existence of identity and inverse.
Lemma 78.11.2. Let B \to S as in Section 78.3. Given a groupoid in algebraic spaces (U, R, s, t, c) over B the morphism j : R \to U \times _ B U is a pre-equivalence relation.
Proof.
Omitted. This is a nice exercise in the definitions.
\square
Lemma 78.11.3. Let B \to S as in Section 78.3. Given an equivalence relation j : R \to U \times _ B U over B there is a unique way to extend it to a groupoid in algebraic spaces (U, R, s, t, c) over B.
Proof.
Omitted. This is a nice exercise in the definitions.
\square
Lemma 78.11.4. Let B \to S as in Section 78.3. Let (U, R, s, t, c) be a groupoid in algebraic spaces over B. 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 78.11.5. Let B \to S be as in Section 78.3. Let (U, R, s, t, c, e, i) be a groupoid in algebraic spaces over B. The diagram
78.11.5.1
\begin{equation} \label{spaces-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 \times 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 78.11.4.
\square
Lemma 78.11.6. Let B \to S be as in Section 78.3. Let (U, R, s, t, c) be a groupoid in algebraic spaces over B. Let B' \to B be a morphism of algebraic spaces. Then the base changes U' = B' \times _ B U, R' = B' \times _ B R endowed with the base changes s', t', c' of the morphisms s, t, c form a groupoid in algebraic spaces (U', R', s', t', c') over B' and the projections determine a morphism (U', R', s', t', c') \to (U, R, s, t, c) of groupoids in algebraic spaces over B.
Proof.
Omitted. Hint: R' \times _{s', U', t'} R' = B' \times _ B (R \times _{s, U, t} R).
\square
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