**Proof.**
By Lemma 56.94.5 the lemmas in Cohomology on Sites, Section 21.29 all apply to our current setting. To translate the results observe that the category $\mathcal{A}_ X$ of Cohomology on Sites, Lemma 21.29.2 is the essential image of $a_ X^{-1} : \textit{Ab}(X_{\acute{e}tale}) \to \textit{Ab}((\mathit{Sch}/X)_{fppf})$.

Part (1) is equivalent to $(V_ n)$ for all $n$ which holds by Cohomology on Sites, Lemma 21.29.8.

Part (2) follows by applying $\epsilon _ Y^{-1}$ to the conclusion of Cohomology on Sites, Lemma 21.29.3.

Part (3) follows from Cohomology on Sites, Lemma 21.29.8 part (1) because $\pi _ X^{-1}K$ is in $D^+_{\mathcal{A}'_ X}((\mathit{Sch}/X)_{\acute{e}tale})$ and $a_ X^{-1} = \epsilon _ X^{-1} \circ a_ X^{-1}$.

Part (4) follows from Cohomology on Sites, Lemma 21.29.8 part (2) for the same reason.

Part (5). We use that

\begin{align*} R\epsilon _{Y, *}Rf_{big, fppf, *}a_ X^{-1}K & = Rf_{big, {\acute{e}tale}, *}R\epsilon _{X, *}a_ X^{-1}K \\ & = Rf_{big, {\acute{e}tale}, *}\pi _ X^{-1}K \\ & = \pi _ Y^{-1}Rf_{small, *}K \\ & = R\epsilon _{Y, *} a_ Y^{-1}Rf_{small, *}K \end{align*}

The first equality by the commutative diagram in Lemma 56.94.2 and Cohomology on Sites, Lemma 21.19.2. The second equality is (3). The third is Lemma 56.93.5 part (2). The fourth is (3) again. Thus the base change map $a_ Y^{-1}(Rf_{small, *}K) \to Rf_{big, fppf, *}(a_ X^{-1}K)$ induces an isomorphism

\[ R\epsilon _{Y, *}a_ Y^{-1}Rf_{small, *}K \to R\epsilon _{Y, *}Rf_{big, fppf, *}a_ X^{-1}K \]

The proof is finished by the following remark: a map $\alpha : a_ Y^{-1}L \to M$ with $L$ in $D^+(Y_{\acute{e}tale})$ and $M$ in $D^+((\mathit{Sch}/Y)_{fppf})$ such that $R\epsilon _{Y, *}\alpha $ is an isomorphism, is an isomorphism. Namely, we show by induction on $i$ that $H^ i(\alpha )$ is an isomorphism. This is true for all sufficiently small $i$. If it holds for $i \leq i_0$, then we see that $R^ j\epsilon _{Y, *}H^ i(M) = 0$ for $j > 0$ and $i \leq i_0$ by (1) because $H^ i(M) = a_ Y^{-1}H^ i(L)$ in this range. Hence $\epsilon _{Y, *}H^{i_0 + 1}(M) = H^{i_0 + 1}(R\epsilon _{Y, *}M)$ by a spectral sequence argument. Thus $\epsilon _{Y, *}H^{i_0 + 1}(M) = \pi _ Y^{-1}H^{i_0 + 1}(L) = \epsilon _{Y, *}a_ Y^{-1}H^{i_0 + 1}(L)$. This implies $H^{i_0 + 1}(\alpha )$ is an isomorphism (because $\epsilon _{Y, *}$ reflects isomorphisms as it is the identity on underlying presheaves) as desired.
$\square$

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