The Stacks project

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

  1. (associativity) $c \circ (1, c) = c \circ (c, 1)$ as maps $\text{Arrows} \times _{s, \text{Ob}, t} \text{Arrows} \times _{s, \text{Ob}, t} \text{Arrows} \to \text{Arrows}$,

  2. (identity) there exists a map $e : \text{Ob} \to \text{Arrows}$ such that

    1. $s \circ e = t \circ e = \text{id}$ as maps $\text{Ob} \to \text{Ob}$,

    2. $c \circ (1, e \circ s) = c \circ (e \circ t, 1) = 1$ as maps $\text{Arrows} \to \text{Arrows}$,

  3. (inverse) there exists a map $i : \text{Arrows} \to \text{Arrows}$ such that

    1. $s \circ i = t$, $t \circ i = s$ as maps $\text{Arrows} \to \text{Ob}$, and

    2. $c \circ (1, i) = e \circ t$ 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.

  1. 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.

  2. 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
\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: $R' \times _{s', U', t'} R' = S' \times _ S (R \times _{s, U, t} R)$. $\square$

Comments (4)

Comment #5917 by Souparna Purohit on

Typo: in the statement of Lemma 39.13.3, I think should say

Comment #8371 by Shubhankar Sahai on

It might be useful to add that giving a groupoid object of a category is the same as giving a simplicial object with certain additional compatibilities. At least one direction is explained in tag 07TN, but it might be useful to mention this here as well, since this seems to be a more natural place for people interested in more stacky applications to come looking.

Comment #8976 by on

This would have to be done somewhere in the chapter on simplicial methods I think and then this section could refer back to that. Going to leave as is, but I would welcome a small set of changes implementing this.

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