Lemma 84.8.5. In Lemma 84.8.4 if $f$ is proper, then we have

1. $a_ Y^{-1} \circ f_{small, *} = f_{big, ph, *} \circ a_ X^{-1}$, and

2. $a_ Y^{-1}(Rf_{small, *}K) = Rf_{big, ph, *}(a_ X^{-1}K)$ for $K$ in $D^+(X_{\acute{e}tale})$ with torsion cohomology sheaves.

Proof. Proof of (1). You can prove this by repeating the proof of Lemma 84.5.6 part (1); we will instead deduce the result from this. As $\epsilon _{Y, *}$ is the identity functor on underlying presheaves, it reflects isomorphisms. Lemma 84.8.1 shows that $\epsilon _{Y, *} \circ a_ Y^{-1} = \pi _ Y^{-1}$ and similarly for $X$. To show that the canonical map $a_ Y^{-1}f_{small, *}\mathcal{F} \to f_{big, ph, *}a_ X^{-1}\mathcal{F}$ is an isomorphism, it suffices to show that

\begin{align*} \pi _ Y^{-1}f_{small, *}\mathcal{F} & = \epsilon _{Y, *}a_ Y^{-1}f_{small, *}\mathcal{F} \\ & \to \epsilon _{Y, *}f_{big, ph, *}a_ X^{-1}\mathcal{F} \\ & = f_{big, {\acute{e}tale}, *} \epsilon _{X, *}a_ X^{-1}\mathcal{F} \\ & = f_{big, {\acute{e}tale}, *}\pi _ X^{-1}\mathcal{F} \end{align*}

is an isomorphism. This is part (1) of Lemma 84.5.6.

To see (2) we use that

\begin{align*} R\epsilon _{Y, *}Rf_{big, ph, *}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 84.8.4 and Cohomology on Sites, Lemma 21.19.2. Then second equality is Lemma 84.8.2. The third is Lemma 84.5.6 part (2). The fourth is Lemma 84.8.2 again. Thus the base change map $a_ Y^{-1}(Rf_{small, *}K) \to Rf_{big, ph, *}(a_ X^{-1}K)$ induces an isomorphism

$R\epsilon _{Y, *}a_ Y^{-1}Rf_{small, *}K \to R\epsilon _{Y, *}Rf_{big, ph, *}a_ X^{-1}K$

The proof is finished by the following remark: consider a map $\alpha : a_ Y^{-1}L \to M$ with $L$ in $D^+(Y_{\acute{e}tale})$ having torsion cohomology sheaves and $M$ in $D^+((\textit{Spaces}/Y)_{ph})$. If $R\epsilon _{Y, *}\alpha$ is an isomorphism, then $\alpha$ 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 Lemma 84.8.1 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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