Let $S$ be a scheme. Let $(U, R, s, t, c, e, i)$ be a groupoid in algebraic spaces over $S$. Let $\overline{u}$ be a geometric point of $U$. In this section we explain what kind of structure we obtain on the local rings (Properties of Spaces, Definition 66.22.2)
The convention we will use is to denote the local ring homomorphisms induced by the morphisms $s, t, c, e, i$ by the corresponding letters. In particular we have a commutative diagram
of local rings. Thus if $I \subset B$ denotes the kernel of $e : B \to A$, then $B = s(A) \oplus I = t(A) \oplus I$. Let us denote
Then we have
because the localization $(B \otimes _{s, A, t} B)_{\mathfrak m_ B \otimes B + B \otimes \mathfrak m_ B}$ has separably closed residue field. Let $J \subset C$ be the ideal of $C$ generated by $I \otimes B + B \otimes I$. Then $J$ is also the kernel of the local ring homomorphism
The composition law $c : R \times _{s, U, t} R \to R$ corresponds to a ring map
sending $I$ into $J$.
Lemma 79.4.1. The map $I/I^2 \to J/J^2$ induced by $c$ is the composition where the second arrow comes from the equality $J = (I \otimes B + B \otimes I)C$. The map $i : B \to B$ induces the map $-1 : I/I^2 \to I/I^2$.
\[ I/I^2 \xrightarrow {(1, 1)} I/I^2 \oplus I/I^2 \to J/J^2 \]
Proof. To describe a local homomorphism from $C$ to another henselian local ring it is enough to say what happens to elements of the form $b_1 \otimes b_2$ by Algebra, Lemma 10.155.6 for example. Keeping this in mind we have the two canonical maps
corresponding to the embeddings $R \to R \times _{s, U, t} R$ given by $r \mapsto (r, e(s(r)))$ and $r \mapsto (e(t(r)), r)$. These maps define maps $J/J^2 \to I/I^2$ which jointly give an inverse to the map $I/I^2 \oplus I/I^2 \to J/J^2$ of the lemma. Thus to prove statement we only have to show that $e_1 \circ c : B \to B$ and $e_2 \circ c : B \to B$ are the identity maps. This follows from the fact that both compositions $R \to R \times _{s, U, t} R \to R$ are identities.
The statement on $i$ follows from the statement on $c$ and the fact that $c \circ (1, i) = e \circ t$. Some details omitted. $\square$
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