The Stacks project

Lemma 83.8.5. In Lemma 83.8.4 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 83.5.6 part (1); we will instead deduce the result from this. As $\epsilon _{Y, *}$ is the identity functor on underlying presheaves, it reflects isomorphisms. Lemma 83.8.1 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 83.5.6.

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 83.8.4 and Cohomology on Sites, Lemma 21.19.2. Then second equality is Lemma 83.8.2. The third is Lemma 83.5.6 part (2). The fourth is Lemma 83.8.2 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 83.8.1 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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