Lemma 85.29.1. Let $(U, R, s, t, c, e, i)$ be a groupoid scheme over $S$. There exists a simplicial scheme $X$ over $S$ with the following properties

1. $X_0 = U$, $X_1 = R$, $X_2 = R \times _{s, U, t} R$,

2. $s_0^0 = e : X_0 \to X_1$,

3. $d^1_0 = s : X_1 \to X_0$, $d^1_1 = t : X_1 \to X_0$,

4. $s_0^1 = (e \circ t, 1) : X_1 \to X_2$, $s_1^1 = (1, e \circ t) : X_1 \to X_2$,

5. $d^2_0 = \text{pr}_1 : X_2 \to X_1$, $d^2_1 = c : X_2 \to X_1$, $d^2_2 = \text{pr}_0$, and

6. $X = \text{cosk}_2 \text{sk}_2 X$.

For all $n$ we have $X_ n = R \times _{s, U, t} \ldots \times _{s, U, t} R$ with $n$ factors. The map $d^ n_ j : X_ n \to X_{n - 1}$ is given on functors of points by

$(r_1, \ldots , r_ n) \longmapsto (r_1, \ldots , c(r_ j, r_{j + 1}), \ldots , r_ n)$

for $1 \leq j \leq n - 1$ whereas $d^ n_0(r_1, \ldots , r_ n) = (r_2, \ldots , r_ n)$ and $d^ n_ n(r_1, \ldots , r_ n) = (r_1, \ldots , r_{n - 1})$.

Proof. We only have to verify that the rules prescribed in (1), (2), (3), (4), (5) define a $2$-truncated simplicial scheme $U'$ over $S$, since then (6) allows us to set $X = \text{cosk}_2 U'$, see Simplicial, Lemma 14.19.2. Using the functor of points approach, all we have to verify is that if $(\text{Ob}, \text{Arrows}, s, t, c, e, i)$ is a groupoid, then

$\xymatrix{ \text{Arrows} \times _{s, \text{Ob}, t} \text{Arrows} \ar@<8ex>[d]^{\text{pr}_0} \ar@<0ex>[d]_ c \ar@<-8ex>[d]_{\text{pr}_1} \\ \text{Arrows} \ar@<4ex>[d]^ t \ar@<-4ex>[d]_ s \ar@<4ex>[u]^{1, e} \ar@<-4ex>[u]_{e, 1} \\ \text{Ob} \ar@<0ex>[u]_ e }$

is a $2$-truncated simplicial set. We omit the details.

Finally, the description of $X_ n$ for $n > 2$ follows by induction from the description of $X_0$, $X_1$, $X_2$, and Simplicial, Remark 14.19.9 and Lemma 14.19.6. Alternately, one shows that $\text{cosk}_2$ applied to the $2$-truncated simplicial set displayed above gives a simplicial set whose $n$th term equals $\text{Arrows} \times _{s, \text{Ob}, t} \ldots \times _{s, \text{Ob}, t} \text{Arrows}$ with $n$ factors and degeneracy maps as given in the lemma. Some details omitted. $\square$

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