The Stacks project

Proof. Represent $K$ by a K-injective complex $\mathcal{J}^\bullet$ of sheaves of $\Lambda$-modules on $X'_{\acute{e}tale}$. Choose a quasi-isomorphism $g'_!\mathcal{J}^\bullet \to \mathcal{I}^\bullet$ to a K-injective complex $\mathcal{I}^\bullet$ of sheaves of $\Lambda$-modules on $X_{\acute{e}tale}$. Then we can consider the map

where the first and third equality come from Lemma 63.3.4 and the second equality comes from Lemma 63.3.13 which tells us that both $g_! \circ f'_!$ and $f_! \circ g'_!$ are equal to $(g \circ f')_! = (f \circ g')_!$ as subsheaves of $(g \circ f')_* = (f \circ g')_*$.

Assume $\Lambda$ is torsion, i.e., we are in case (b). With notation as above, it suffices to show that $f_*g'_!\mathcal{J}^\bullet \to f_*\mathcal{I}^\bullet$ is an isomorphism. The question is local on $Y$. Hence we may assume that the dimension of fibres of $f$ is bounded, see Morphisms, Lemma 29.28.5. Then we see that $Rf_*$ has finite cohomological dimension, see Étale Cohomology, Lemma 59.92.2. Hence by Derived Categories, Lemma 13.32.2, if we show that $R^ qf_*(g'_!\mathcal{J}) = 0$ for $q > 0$ and any injective sheaf of $\Lambda$-modules $\mathcal{J}$ on $X'_{\acute{e}tale}$, then the result follows.

The stalk of $R^ qf_*(g'_!\mathcal{J})$ at a geometric point $\overline{y}$ is equal to $H^ q(X_{\overline{y}}, (g'_!\mathcal{J})|_{X_{\overline{y}}})$ by Étale Cohomology, Lemma 59.91.13. Since formation of $g'_!$ commutes with base change (Lemma 63.3.12) this is equal to

where $g'_{\overline{y}} : X'_{\overline{y}} \to X_{\overline{y}}$ is the induced morphism between geometric fibres. Since $Y' \to Y$ is locally quasi-finite, we see that $X'_{\overline{y}}$ is a disjoint union of the fibres $X'_{\overline{y}'}$ at geometric points $\overline{y}'$ of $Y'$ lying over $\overline{y}$. Denote $g'_{\overline{y}'} : X'_{\overline{y}'} \to X_{\overline{y}}$ the restriction of $g'_{\overline{y}}$ to $X'_{\overline{y}'}$. Thus the previous cohomology group is equal to

for example by Lemma 63.3.15 (but it is also obvious from the definition of $g'_{\overline{y}, !}$ in Section 63.3). Since taking étale cohomology over $X_{\overline{y}}$ commutes with direct sums (Étale Cohomology, Theorem 59.51.3) we conclude it suffices to show that

is zero. Observe that $g_{\overline{y}'} : X'_{\overline{y}'} \to X_{\overline{y}}$ is a morphism between proper scheme over $\overline{y}$ and hence is proper itself. As it is locally quasi-finite as well we conclude that $g_{\overline{y}'}$ is finite. Thus we see that $g'_{\overline{y}', !} = g'_{\overline{y}', *} = Rg'_{\overline{y}', *}$. By Leray we conlude that we have to show

is zero. As $\Lambda$ is torsion, this follows from proper base change (Étale Cohomology, Lemma 59.91.13) as the higher direct images of $\mathcal{J}$ under $f'$ are zero.

Proof in case (a). We will deduce this from case (b) by standard arguments. We will show that the induced map $g_! R^ pf'_* K \to R^ pf_*(g'_!K)$ is an isomorphism for all $p \in \mathbf{Z}$. Fix an integer $p_0 \in \mathbf{Z}$. Let $a$ be an integer such that $H^ j(K) = 0$ for $j < a$. We will prove $g_! R^ pf'_* K \to R^ pf_*(g'_!K)$ is an isomorphism for $p \leq p_0$ by descending induction on $a$. If $a > p_0$, then we see that the left and right hand side of the map are zero for $p \leq p_0$ by trivial vanishing, see Derived Categories, Lemma 13.16.1 (and use that $g_!$ and $g'_!$ are exact functors). Assume $a \leq p_0$. Consider the distinguished triangle

By induction we have the result for $\tau _{\geq a + 1}K$. In the next paragraph, we will prove the result for $H^ a(K)[-a]$. Then five lemma applied to the map between long exact sequence of cohomology sheaves associated to the map of distinguished triangles

gives the result for $K$. Some details omitted.

Let $\mathcal{F}$ be a torsion abelian sheaf on $X'_{\acute{e}tale}$. To finish the proof we show that $g_! Rf'_*\mathcal{F} \to R^ pf_*(g'_!\mathcal{F})$ is an isomorphism for all $p$. We can write $\mathcal{F} = \bigcup \mathcal{F}[n]$ where $\mathcal{F}[n] = \mathop{\mathrm{Ker}}(n : \mathcal{F} \to \mathcal{F})$. We have the isomorphism for $\mathcal{F}[n]$ by case (b). Since the functors $g_!$, $g'_!$, $R^ pf_*$, $R^ pf'_*$ commute with filtered colimits (follows from Lemma 63.3.17 and Étale Cohomology, Lemma 59.51.8) the proof is complete. $\square$

Proof. The statement means that if we write $R(g \circ f)_* = Rg_* \circ Rf_*$ and $R(g' \circ f')_* = Rg'_* \circ Rf'_*$, then the isomorphism $m_! \circ Rg'_* \circ Rf'_* \to Rg_* \circ Rf_* \circ k_!$ of the outer rectangle is equal to the composition

of the two maps of the squares in the diagram. To prove this choose a K-injective complex $\mathcal{J}^\bullet$ of $\Lambda$-modules on $X'_{\acute{e}tale}$ and a quasi-isomorphism $k_!\mathcal{J}^\bullet \to \mathcal{I}^\bullet$ to a K-injective complex $\mathcal{I}^\bullet$ of $\Lambda$-modules on $X_{\acute{e}tale}$. The proof of Lemma 63.8.1 shows that the canonical map

is a quasi-isomorphism and this quasi-isomorphism produces the second arrow on applying $Rg_*$. By Cohomology on Sites, Lemma 21.20.10 the complex $f_*\mathcal{I}^\bullet$, resp. $f'_*\mathcal{J}^\bullet$ is a K-injective complex of $\Lambda$-modules on $Y_{\acute{e}tale}$, resp. $Y'_{\acute{e}tale}$. (Using this is cheating and could be avoided.) In particular, the same reasoning gives that the canonical map

is a quasi-isomorphism and this quasi-isomorphism represents the first arrow. Finally, the proof of Lemma 63.8.1 show that $g_*l_!f'_!\mathcal{J}^\bullet$ represents $Rg_*(l_!f'_*\mathcal{J}^\bullet )$ because $f'_*\mathcal{J}^\bullet$ is K-injective. Hence $Rg_*(a) = g_*(a)$ and the composition $g_*(a) \circ b$ is the arrow of Lemma 63.8.1 for the rectangle. $\square$

Proof. The statement means that if we write $(h \circ h')_! = h_! \circ h'_!$ and $(g \circ g')_! = g_! \circ g'_!$ using the equalities of Lemma 63.3.13, then the isomorphism $h_! \circ h'_! \circ Rf''_* \to Rf_* \circ g_! \circ g'_!$ of the outer rectangle is equal to the composition

of the two maps of the squares in the diagram. To prove this choose a K-injective complex $\mathcal{I}^\bullet$ of $\Lambda$-modules on $X''_{\acute{e}tale}$ and a quasi-isomorphism $g'_!\mathcal{I}^\bullet \to \mathcal{J}^\bullet$ to a K-injective complex $\mathcal{J}^\bullet$ of $\Lambda$-modules on $X'_{\acute{e}tale}$. Next, choose a quasi-isomorphism $g_!\mathcal{J}^\bullet \to \mathcal{K}^\bullet$ to a K-injective complex $\mathcal{K}^\bullet$ of $\Lambda$-modules on $X_{\acute{e}tale}$. The proof of Lemma 63.8.1 shows that the canonical maps

are quasi-isomorphisms and these quasi-isomorphisms define the first and second arrow above. Since $g_!$ is an exact functor (Lemma 63.3.17) we find that $g_!g'_!\mathcal{I}^\bullet \to \mathcal{K}^\bullet$ is a quasi-ismorphism and hence the canonical map

is a quasi-isomorphism and represents the map for the outer rectangle in the derived category. Clearly this map is the composition of the other two and the proof is complete. $\square$

Proof. Represent $K$ by a K-injective complex $\mathcal{J}^\bullet$ of sheaves of $\Lambda$-modules on $X'_{\acute{e}tale}$. Choose a quasi-isomorphism $g'_!\mathcal{J}^\bullet \to \mathcal{I}^\bullet$ to a K-injective complex $\mathcal{I}^\bullet$ of sheaves of $\Lambda$-modules on $X_{\acute{e}tale}$. The proof of Lemma 63.8.1 constructs $g_!Rf'_*K \to Rf_*g'_!K$ as

Choose a quasi-isomorphism $(a')^{-1}\mathcal{J}^\bullet \to \mathcal{J}_1^\bullet$ to a K-injective complex $\mathcal{J}_1^\bullet$ of sheaves of $\Lambda$-modules on $X'_{1, {\acute{e}tale}}$. Then we can pick a diagram of complexes

commuting up to homotopy where all arrows are quasi-isomorphisms, the equality comes from Lemma 63.3.4, and $\mathcal{I}_1^\bullet$ is a K-injective complex of sheaves of $\Lambda$-modules on $X_{1, {\acute{e}tale}}$. The map $g_{1, !}Rf'_{1, *}(a')^{-1}K \to Rf_{1, *}g'_{1, !}(a')^{-1}K$ is given by

The identifications across the $3$ equal signs in both arrows are compatible with pullback maps, i.e., the diagram

of complexes of abelian sheaves commutes. To show this it is enough to show the diagram commutes with $g_!, g_{1, !}, g'_!, g'_{1, !}$ replaced by $g_*, g_{1, *}, g'_*, g'_{1, *}$ (because the shriek functors are defined as subfunctors of the $*$ functors and the base change maps are defined in a manner compatible with this, see proof of Lemma 63.3.12). For this new diagram the commutativity follows from the compatibility of pullback maps with horizontal and vertical stacking of diagrams, see Sites, Remarks 7.45.3 and 7.45.4 so that going around the diagram in either direction is the pullback map for the base change of $f \circ g' = g \circ f'$ by $b$. Since of course

commutes, to finish the proof it suffices to show that

commutes in the derived category, which holds by our choice of maps earlier. $\square$

Proof. This is a special case of Lemma 63.8.1 if $f$ and $g$ are separated. We urge the reader to skip the proof in the general case as we'll mainly use the case where $f$ and $g$ are separated.

Represent $K$ by a complex $\mathcal{K}^\bullet$ of sheaves of $\Lambda$-modules on $X_{\acute{e}tale}$. Choose a quasi-isomorphism $f_!\mathcal{K}^\bullet \to \mathcal{I}^\bullet$ into a K-injective complex $\mathcal{I}^\bullet$ of sheaves of $\Lambda$-modules on $Y_{\acute{e}tale}$. Consider the map

where the equalities are Lemmas 63.4.11 and 63.3.4. This map of complexes determines the map $g_!K \to Rh_*(f_!K)$ of the statement of the lemma.

Assume $\Lambda$ is torsion, i.e., we are in case (b). To check the map is an isomorphism we may work locally on $Z$. Hence we may assume that the dimension of fibres of $h$ is bounded, see Morphisms, Lemma 29.28.5. Then we see that $Rh_*$ has finite cohomological dimension, see Étale Cohomology, Lemma 59.92.2. Hence by Derived Categories, Lemma 13.32.2, if we show that $R^ qh_*(f_!\mathcal{F}) = 0$ for $q > 0$ and any sheaf $\mathcal{F}$ of $\Lambda$-modules on $X_{\acute{e}tale}$, then $h_*f_!\mathcal{K}^\bullet \to h_*\mathcal{I}^\bullet$ is a quasi-isomorphism.

Observe that $\mathcal{G} = f_!\mathcal{F}$ is a sheaf of $\Lambda$-modules on $Y$ whose stalks are nonzero only at points $y \in Y$ such that $\kappa (y)/\kappa (h(y))$ is a finite extension. This follows from the description of stalks of $f_!\mathcal{F}$ in Lemma 63.4.5 and the fact that both $f$ and $g$ are locally quasi-finite. Hence by the proper base change theorem (Étale Cohomology, Lemma 59.91.13) it suffices to show that $H^ q(Y_{\overline{z}}, \mathcal{H}) = 0$ where $\mathcal{H}$ is a sheaf on the proper scheme $Y_{\overline{z}}$ over $\kappa (\overline{z})$ whose support is contained in the set of closed points. Thus the required vanishing by Étale Cohomology, Lemma 59.97.3.

Case (a) follows from case (b) by the exact same argument as used in the proof of Lemma 63.8.1 (using Lemma 63.4.5 instead of Lemma 63.3.17). $\square$