## 78.14 The finite part of a groupoid

In this section we are going to use the idea explained in Section 78.13 to take the finite part of a groupoid in algebraic spaces.

Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $(U, R, s, t, c, e, i)$ be a groupoid in algebraic spaces over $B$. Assumption: The morphisms $s, t$ are separated and locally of finite type. This notation and assumption will we be fixed throughout this section.

Denote $R_ s$ the algebraic space $R$ seen as an algebraic space over $U$ via $s$. Let $U' = (R_ s/U, e)_{fin}$. Since $s$ is separated and locally of finite type, by Proposition 78.12.11 and Lemma 78.12.15, we see that $U'$ is an algebraic space endowed with an étale morphism $g : U' \to U$. Moreover, by Lemma 78.12.1 there exists a universal open subspace $Z_{univ} \subset R \times _{s, U, g} U'$ which is finite over $U'$ and such that $(1_{U'}, e \circ g) : U' \to R \times _{s, U, g} U'$ factors through $Z_{univ}$. Moreover, by Lemma 78.12.4 the open subspace $Z_{univ}$ is also closed in $R \times _{s, U', g} U$. Picture so far:

$\xymatrix{ Z_{univ} \ar[d] \ar[rd] & \\ R \times _{s, U, g} U' \ar[d] \ar[r] & U' \ar[d]^ g \\ R \ar[r]^ s & U }$

Let $T$ be a scheme over $B$. We see that a $T$-valued point of $Z_{univ}$ may be viewed as a triple $(u, Z, z)$ where

1. $u : T \to U$ is a $T$-valued point of $U$,

2. $Z \subset R \times _{s, U, u} T$ is an open and closed subspace finite over $T$ such that $(e \circ u, 1_ T)$ factors through it, and

3. $z : T \to R$ is a $T$-valued point of $R$ with $s \circ z = u$ and such that $(z, 1_ T)$ factors through $Z$.

Having said this, it is morally clear from the discussion in Section 78.13 that we can turn $(Z_{univ}, U')$ into a groupoid in algebraic spaces over $B$. To make sure will define the morphisms $s', t', c', e', i'$ one by one using the functorial point of view. (Please don't read this before reading and understanding the simple construction in Section 78.13.)

The morphism $s' : Z_{univ} \to U'$ corresponds to the rule

$s' : (u, Z, z) \mapsto (u, Z).$

The morphism $t' : Z_{univ} \to U'$ is given by the rule

$t' : (u, Z, z) \mapsto (t \circ z, c(Z, i \circ z)).$

The entry $c(Z, i \circ z)$ makes sense as the map $c(-, i \circ z) : R \times _{s, U, u} T \to R \times _{s, U, t \circ z} T$ is an isomorphism with inverse $c(-, z)$. The morphism $e' : U' \to Z_{univ}$ is given by the rule

$e' : (u, Z) \mapsto (u, Z, (e \circ u, 1_ T)).$

Note that this makes sense by the requirement that $(e \circ u, 1_ T)$ factors through $Z$. The morphism $i' : Z_{univ} \to Z_{univ}$ is given by the rule

$i' : (u, Z, z) \mapsto (t \circ z, c(Z, i \circ z), i \circ z).$

Finally, composition is defined by the rule

$c' : ((u_1, Z_1, z_1), (u_2, Z_2, z_2)) \mapsto (u_2, Z_2, z_1 \circ z_2).$

We omit the verification that the axioms of a groupoid in algebraic spaces hold for $(U', Z_{univ}, s', t', c', e', i')$.

A final piece of information is that there is a canonical morphism of groupoids

$(U', Z_{univ}, s', t', c', e', i') \longrightarrow (U, R, s, t, c, e, i)$

Namely, the morphism $U' \to U$ is the morphism $g : U' \to U$ which is defined by the rule $(u, Z) \mapsto u$. The morphism $Z_{univ} \to R$ is defined by the rule $(u, Z, z) \mapsto z$. This finishes the construction. Let us summarize our findings as follows.

Lemma 78.14.1. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $(U, R, s, t, c, e, i)$ be a groupoid in algebraic spaces over $B$. Assume the morphisms $s, t$ are separated and locally of finite type. There exists a canonical morphism

$(U', Z_{univ}, s', t', c', e', i') \longrightarrow (U, R, s, t, c, e, i)$

of groupoids in algebraic spaces over $B$ where

1. $g : U' \to U$ is identified with $(R_ s/U, e)_{fin} \to U$, and

2. $Z_{univ} \subset R \times _{s, U, g} U'$ is the universal open (and closed) subspace finite over $U'$ which contains the base change of the unit $e$.

Proof. See discussion above. $\square$

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