Lemma 77.15.1. Let $B \to S$ as in Section 77.3. Let $(G, m)$ be a group algebraic space over $B$ with identity $e_ G$ and inverse $i_ G$. Let $X$ be an algebraic space over $B$ and let $a : G \times _ B X \to X$ be an action of $G$ on $X$ over $B$. Then we get a groupoid in algebraic spaces $(U, R, s, t, c, e, i)$ over $B$ in the following manner:

1. We set $U = X$, and $R = G \times _ B X$.

2. We set $s : R \to U$ equal to $(g, x) \mapsto x$.

3. We set $t : R \to U$ equal to $(g, x) \mapsto a(g, x)$.

4. We set $c : R \times _{s, U, t} R \to R$ equal to $((g, x), (g', x')) \mapsto (m(g, g'), x')$.

5. We set $e : U \to R$ equal to $x \mapsto (e_ G(x), x)$.

6. We set $i : R \to R$ equal to $(g, x) \mapsto (i_ G(g), a(g, x))$.

Proof. Omitted. Hint: It is enough to show that this works on the set level. For this use the description above the lemma describing $g$ as an arrow from $v$ to $a(g, v)$. $\square$

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