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.
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. $\square$
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