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
\delta _1 \circ \delta _2 = c(\delta _1 \circ t \circ \delta _2, \delta _2)
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
\vcenter { \xymatrix{ & \bullet \ar[dl] \\ \bullet \ar@(ul, dl)[] & \bullet \ar[d] \\ & \bullet \ar[lu] } } \quad \circ \quad \vcenter { \xymatrix{ & \bullet \ar[d] \\ \bullet \ar[ru] & \bullet \ar[d] \\ & \bullet \ar[lu] } } \quad = \quad \quad \vcenter { \xymatrix{ & \bullet \ar@/^/[dd] \\ \bullet \ar@(ul, dl)[] & \bullet \ar[l] \\ & \bullet \ar[lu] } }
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
\Gamma = \{ \delta : U \to R \mid s \circ \delta = \text{id}_ U\}
and composition \circ : \Gamma \times \Gamma \to \Gamma is given by the rule above
79.5.0.1
\begin{equation} \label{spaces-more-groupoids-equation-composition} \delta _1 \circ \delta _2 = c(\delta _1 \circ t \circ \delta _2, \delta _2) \end{equation}
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
\Gamma _0 = \{ \delta \in \Gamma \mid \delta |_{U_0} = e_0\}
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
\xymatrix{ Y \ar[r] \ar[d] & 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 \\ U \ar[r]_\delta & R \ar@<1ex>[r]^ s \ar@<-1ex>[r]_ t & U }
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
\xymatrix{ a_0^*\mathcal{C}_{R_0/R} \ar[rr] & & \mathcal{C}_{Y_0/Y} \\ a_0^*t_0^*\mathcal{C}_{U_0/U} \ar[u] \ar@{=}[rr] & & (t_0 \circ a_0)^*\mathcal{C}_{U_0/U} \ar[u] }
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
\xymatrix{ (U_0 \subset U) \ar@{..>}[rr]_{(e_0, \delta )} \ar[rd]_{(\text{id}_{U_0}, \text{id}_ U)} & & (R_0 \subset R) \ar[ld]^{(s_0, s)} \\ & (U_0 \subset U) }
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
79.5.1.1
\begin{equation} \label{spaces-more-groupoids-equation-isomorphism} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{U_0}}(e_0^*\Omega _{R_0/U_0}, \mathcal{C}_{U_0/U}) \longrightarrow \Gamma _0 \end{equation}
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
B \to C = (B \otimes _{s, A, t} B)^ h_{\mathfrak m_ B \otimes B + B \otimes \mathfrak m_ B} \to A
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
D(\zeta ) = (\delta _2 \circ t)\left(D_1(\zeta )\right) + D_2(\zeta )
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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