Lemma 79.5.1. In the situation discussed in this section, let $\delta \in \Gamma _0$ and $f = t \circ \delta : U \to U$. If $s, t$ are flat, then the canonical map $\mathcal{C}_{U_0/U} \to \mathcal{C}_{U_0/U}$ induced by $f$ (More on Morphisms of Spaces, Lemma 76.5.3) is the identity map.

## 79.5 Groupoid of sections

Suppose we have a groupoid $(\text{Ob}, \text{Arrows}, s, t, c, e, i)$. Then we can construct a monoid $\Gamma $ whose elements are maps $\delta : \text{Ob} \to \text{Arrows}$ with $s \circ \delta = \text{id}_{\text{Ob}}$ and composition given by

In other words, an element of $\Gamma $ is a rule $\delta $ which prescribes an arrow emanating from every object and composition is the natural thing. For example

with obvious notation

The same procedure can be applied to a groupoid in algebraic spaces $(U, R, s, t, c, e, i)$ over a scheme $S$. Namely, as elements of $\Gamma $ we take the set

and composition $\circ : \Gamma \times \Gamma \to \Gamma $ is given by the rule above

The identity is given by $e \in \Gamma $. The groupoid $\Gamma $ is not a group in general because there may be elements $\delta \in \Gamma $ which do not have an inverse. Namely, it is clear that $\delta \in \Gamma $ will have an inverse if and only if $t \circ \delta $ is an automorphism of $U$ and in this case $\delta ^{-1} = i \circ \delta \circ (t \circ \delta )^{-1}$.

For later use we discuss what happens with the subgroupoid $\Gamma _0$ of $\Gamma $ of sections which are infinitesimally close to the identity $e$. More precisely, suppose given an $R$-invariant closed subspace $U_0 \subset U$ such that $U$ is a first order thickening of $U_0$. Denote $R_0 = s^{-1}(U_0) = t^{-1}(U_0)$ and let $(U_0, R_0, s_0, t_0, c_0, e_0, i_0)$ be the corresponding groupoid in algebraic spaces. Set

If $s$ and $t$ are flat, then every element in $\Gamma _0$ is invertible. This follows because $t \circ \delta $ will be a morphism $U \to U$ inducing the identity on $\mathcal{O}_{U_0}$ and on $\mathcal{C}_{U_0/U}$ (Lemma 79.5.1) and we conclude because we have a short exact sequence $0 \to \mathcal{C}_{U_0/U} \to \mathcal{O}_ U \to \mathcal{O}_{U_0} \to 0$.

**Proof.**
To see this we extend the bottom of the diagram (79.3.0.2) as follows

where the left square is cartesian and this is our definition of $Y$; we will not need to know more about $Y$. There is a similar diagram with similar properties obtained by base change to $U_0$ everywhere. We are trying to show that $\text{id}_ U = s \circ \delta $ and $f = t \circ \delta $ induce the same maps on conormal sheaves. Since $s$ is flat and surjective, it suffices to prove the same thing for the two compositions $a, b : Y \to R$ along the top row. Observe that $a_0 = b_0$ and that one of $a$ and $b$ is an isomorphism as we know that $s \circ \delta $ is an isomorphism. Therefore the two morphisms $a, b : Y \to R$ are morphisms between algebraic spaces flat over $U$ (via the morphism $t : R \to U$ and the morphism $t \circ a = t \circ b : Y \to U$). This implies what we want. Namely, by the compatibility with compositions in More on Morphisms of Spaces, Lemma 76.5.4 we conclude that both maps $a_0^*\mathcal{C}_{R_0/R} \to \mathcal{C}_{Y_0/Y}$ fit into a commutative diagram

whose vertical arrows are isomorphisms by More on Morphisms of Spaces, Lemma 76.18.1. Thus the lemma holds. $\square$

Let us identify the group $\Gamma _0$. Applying the discussion in More on Morphisms of Spaces, Remarks 76.17.3 and 76.17.7 to the diagram

we see that $\delta = \theta \cdot e$ for a unique $\mathcal{O}_{U_0}$-linear map $\theta : e_0^*\Omega _{R_0/U_0} \to \mathcal{C}_{U_0/U}$. Thus we get a bijection

by applying More on Morphisms of Spaces, Lemma 76.17.5.

Lemma 79.5.2. The bijection (79.5.1.1) is an isomorphism of groups.

**Proof.**
Let $\delta _1, \delta _2 \in \Gamma _0$ correspond to $\theta _1, \theta _2$ as above and the composition $\delta = \delta _1 \circ \delta _2$ in $\Gamma _0$ correspond to $\theta $. We have to show that $\theta = \theta _1 + \theta _2$. Recall (More on Morphisms of Spaces, Lemma 76.17.2) that $\theta _1, \theta _2, \theta $ correspond to derivations $D_1, D_2, D : e_0^{-1}\mathcal{O}_{R_0} \to \mathcal{C}_{U_0/U}$ given by $D_1 = \theta _1 \circ \text{d}_{R_0/U_0}$ and so on. It suffices to check that $D = D_1 + D_2$.

We may check equality on stalks. Let $\overline{u}$ be a geometric point of $U$ and let us use the local rings $A, B, C$ introduced in Section 79.4. The morphisms $\delta _ i$ correspond to ring maps $\delta _ i : B \to A$. Let $K \subset A$ be the ideal of square zero such that $A/K = \mathcal{O}_{U_0, \overline{u}}$. In other words, $K$ is the stalk of $\mathcal{C}_{U_0/U}$ at $\overline{u}$. The fact that $\delta _ i \in \Gamma _0$ means exactly that $\delta _ i(I) \subset K$. The derivation $D_ i$ is just the map $\delta _ i - e : B \to A$. Since $B = s(A) \oplus I$ we see that $D_ i$ is determined by its restriction to $I$ and that this is just given by $\delta _ i|_ I$. Moreover $D_ i$ and hence $\delta _ i$ annihilates $I^2$ because $I = \mathop{\mathrm{Ker}}(I)$.

To finish the proof we observe that $\delta $ corresponds to the composition

where the first arrow is $c$ and the second arrow is determined by the rule $b_1 \otimes b_2 \mapsto \delta _2(t(\delta _1(b_1))) \delta _2(b_2)$ as follows from (79.5.0.1). By Lemma 79.4.1 we see that an element $\zeta $ of $I$ maps to $\zeta \otimes 1 + 1 \otimes \zeta $ plus higher order terms. Hence we conclude that

However, by Lemma 79.5.1 the action of $\delta _2 \circ t$ on $K = \mathcal{C}_{U_0/U, \overline{u}}$ is the identity and we win. $\square$

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