Lemma 54.55.5. Assumptions and notations as in Theorem 54.55.3. There is a functorial bijection

**Proof.**
We can prove this using formal arguments and the result of Theorem 54.55.3 as follows. Given a sheaf $\mathcal{F}$ corresponding to the $G$-set $M = \mathcal{F}_{\overline{s}}$ we have

Here the first identification is explained in Sites, Sections 7.2 and 7.12, the second results from Theorem 54.55.3 and the third is clear. We will also give a direct proof^{1}.

Suppose that $t \in \Gamma (S, \mathcal{F})$ is a global section. Then the triple $(S, \overline{s}, t)$ defines an element of $\mathcal{F}_{\overline{s}}$ which is clearly invariant under the action of $G$. Conversely, suppose that $(U, \overline{u}, t)$ defines an element of $\mathcal{F}_{\overline{s}}$ which is invariant. Then we may shrink $U$ and assume $U = \mathop{\mathrm{Spec}}(L)$ for some finite separable field extension of $K$, see Proposition 54.26.2. In this case the map $\mathcal{F}(U) \to \mathcal{F}_{\overline{s}}$ is injective, because for any morphism of étale neighbourhoods $(U', \overline{u}') \to (U, \overline{u})$ the restriction map $\mathcal{F}(U) \to \mathcal{F}(U')$ is injective since $U' \to U$ is a covering of $S_{\acute{e}tale}$. After enlarging $L$ a bit we may assume $K \subset L$ is a finite Galois extension. At this point we use that

where the maps $\mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(L \otimes _ K L)$ come from the ring maps $a \otimes b \mapsto a\sigma (b)$. Hence we see that the condition that $(U, \overline{u}, t)$ is invariant under all of $G$ implies that $t \in \mathcal{F}(\mathop{\mathrm{Spec}}(L))$ maps to the same element of $\mathcal{F}(\mathop{\mathrm{Spec}}(L) \times _{\mathop{\mathrm{Spec}}(K)} \mathop{\mathrm{Spec}}(L))$ via restriction by either projection (this uses the injectivity mentioned above; details omitted). Hence the sheaf condition of $\mathcal{F}$ for the étale covering $\{ \mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(K)\} $ kicks in and we conclude that $t$ comes from a unique section of $\mathcal{F}$ over $\mathop{\mathrm{Spec}}(K)$. $\square$

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## Comments (2)

Comment #3198 by Dario Weißmann on

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