Lemma 39.23.2. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $U = \mathop{\mathrm{Spec}}(A)$ and $R = \mathop{\mathrm{Spec}}(B)$ are affine and $s, t : R \to U$ finite locally free. Let $C$ be as in (39.23.0.1). Let $f \in A$. Then $\text{Norm}_{s^\sharp }(t^\sharp (f)) \in C$.
Proof. Consider the commutative diagram
\[ \xymatrix{ & U & \\ R \ar[d]_ s \ar[ru]^ t & R \times _{s, U, t} R \ar[l]^-{\text{pr}_0} \ar[d]^{\text{pr}_1} \ar[r]_-c & R \ar[d]^ s \ar[lu]_ t \\ U & R \ar[l]_ t \ar[r]^ s & U } \]
of Lemma 39.13.4. Think of $f \in \Gamma (U, \mathcal{O}_ U)$. The commutativity of the top part of the diagram shows that $\text{pr}_0^\sharp (t^\sharp (f)) = c^\sharp (t^\sharp (f))$ as elements of $\Gamma (R \times _{S, U, t} R, \mathcal{O})$. Looking at the right lower cartesian square the compatibility of the norm construction with base change shows that $s^\sharp (\text{Norm}_{s^\sharp }(t^\sharp (f))) = \text{Norm}_{\text{pr}_1^\sharp }(c^\sharp (t^\sharp (f)))$. Similarly we get $t^\sharp (\text{Norm}_{s^\sharp }(t^\sharp (f))) = \text{Norm}_{\text{pr}_1^\sharp }(\text{pr}_0^\sharp (t^\sharp (f)))$. Hence by the first equality of this proof we see that $s^\sharp (\text{Norm}_{s^\sharp }(t^\sharp (f))) = t^\sharp (\text{Norm}_{s^\sharp }(t^\sharp (f)))$ as desired. $\square$
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