## 62.8 Preliminaries to derived lower shriek via compactifications

In this section we prove some lemmas on the existence of certain natural isomorphisms of functors which follow immediately from proper base change.

Lemma 62.8.1. Consider a commutative diagram of schemes

$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }$

with $f$ and $f'$ proper and $g$ and $g'$ separated and locally quasi-finite. Let $\Lambda$ be a ring. Functorially in $K \in D(X'_{\acute{e}tale}, \Lambda )$ there is a canonical map

$g_!Rf'_*K \longrightarrow Rf_*(g'_!K)$

in $D(Y_{\acute{e}tale}, \Lambda )$. This map is an isomorphism if (a) $K$ is bounded below and has torsion cohomology sheaves, or (b) $\Lambda$ is a torsion ring.

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

$g_!f'_*\mathcal{J}^\bullet = g_!f'_!\mathcal{J}^\bullet = f_!g'_!\mathcal{J}^\bullet = f_*g'_!\mathcal{J}^\bullet \to f_*\mathcal{I}^\bullet$

where the first and third equality come from Lemma 62.3.4 and the second equality comes from Lemma 62.3.13 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 62.3.12) this is equal to

$H^ q(X_{\overline{y}}, g'_{\overline{y}, !}(\mathcal{J}|_{X'_{\overline{y}}}))$

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

$H^ q(X_{\overline{y}}, \bigoplus \nolimits _{\overline{y}'/\overline{y}} g'_{\overline{y}', !}(\mathcal{J}|_{X'_{\overline{y}'}}))$

for example by Lemma 62.3.15 (but it is also obvious from the definition of $g'_{\overline{y}, !}$ in Section 62.3). 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

$H^ q(X_{\overline{y}}, g'_{\overline{y}', !}(\mathcal{J}|_{X'_{\overline{y}'}}))$

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

$H^ q(X'_{\overline{y}'}, \mathcal{J}|_{X'_{\overline{y}'}})$

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

$H^ a(K)[-a] \to K \to \tau _{\geq a + 1}K$

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

$\xymatrix{ g_! Rf'_*(H^ a(K)[-a]) \ar[d] \ar[r] & g_! Rf'_* K \ar[r] \ar[d] & g_! Rf'_* \tau _{\geq a + 1} K \ar[d] \\ Rf_*(g'_!(H^ a(K)[-a])) \ar[r] & Rf_*(g'_!K) \ar[r] & Rf_*(g'_!\tau _{|geq a + 1}K) }$

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 62.3.17 and Étale Cohomology, Lemma 59.51.8) the proof is complete. $\square$

Lemma 62.8.2. Consider a commutative diagram of schemes

$\xymatrix{ X' \ar[r]_ k \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]_ l \ar[d]_{g'} & Y \ar[d]^ g \\ Z' \ar[r]^ m & Z }$

with $f$, $f'$, $g$ and $g'$ proper and $k$, $l$, and $m$ separated and locally quasi-finite. Then the isomorphisms of Lemma 62.8.1 for the two squares compose to give the isomorphism for the outer rectangle (see proof for a precise statement).

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

$m_! \circ Rg'_* \circ Rf'_* \to Rg_* \circ l_! \circ Rf'_* \to Rg_* \circ Rf_* \circ k_!$

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 62.8.1 shows that the canonical map

$a : l_!f'_*\mathcal{J}^\bullet \to f_*\mathcal{I}^\bullet$

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

$b : m_!g'_*f'_*\mathcal{J}^\bullet \to g_*f_*\mathcal{I}^\bullet$

is a quasi-isomorphism and this quasi-isomorphism represents the first arrow. Finally, the proof of Lemma 62.8.1 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 62.8.1 for the rectangle. $\square$

Lemma 62.8.3. Consider a commutative diagram of schemes

$\xymatrix{ X'' \ar[r]_{g'} \ar[d]_{f''} & X' \ar[r]_ g \ar[d]_{f'} & X \ar[d]^ f \\ Y'' \ar[r]^{h'} & Y' \ar[r]^ h & Y }$

with $f$, $f'$, and $f''$ proper and $g$, $g'$, $h$, and $h'$ separated and locally quasi-finite. Then the isomorphisms of Lemma 62.8.1 for the two squares compose to give the isomorphism for the outer rectangle (see proof for a precise statement).

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 62.3.13, then the isomorphism $h_! \circ h'_! \circ Rf''_* \to Rf_* \circ g_! \circ g'_!$ of the outer rectangle is equal to the composition

$h_! \circ h'_! \circ Rf''_* \to h_! \circ Rf'_* \circ g'_! \to Rf_* \circ g_! \circ g'_!$

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 62.8.1 shows that the canonical maps

$h'_!f''_*\mathcal{I}^\bullet \to f'_*\mathcal{J}^\bullet \quad \text{and}\quad h_!f'_*\mathcal{J}^\bullet \to f_*\mathcal{K}^\bullet$

are quasi-isomorphisms and these quasi-isomorphisms define the first and second arrow above. Since $g_!$ is an exact functor (Lemma 62.3.17) we find that $g_!g'_!\mathcal{I}^\bullet \to \mathcal{K}^\bullet$ is a quasi-ismorphism and hence the canonical map

$h_!h'_!f''_*\mathcal{I}^\bullet \to f_*\mathcal{K}^\bullet$

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. $\square$

Remark 62.8.4. Consider a commutative diagram

$\xymatrix{ X'' \ar[r]_{k'} \ar[d]_{f''} & X' \ar[r]_ k \ar[d]_{f'} & X \ar[d]^ f \\ Y'' \ar[r]^{l'} \ar[d]_{g''} & Y' \ar[r]^ l \ar[d]_{g'} & Y \ar[d]^ g \\ Z'' \ar[r]^{m'} & Z' \ar[r]^ m & Z }$

of schemes whose vertical arrows are proper and whose horizontal arrows are separated and locally quasi-finite. Let us label the squares of the diagram $A$, $B$, $C$, $D$ as follows

$\begin{matrix} A & B \\ C & D \end{matrix}$

Then the maps of Lemma 62.8.1 for the squares are (where we use $Rf_* = f_*$, etc)

$\begin{matrix} \gamma _ A : l'_! \circ f''_* \to f'_* \circ k'_! & \gamma _ B : l_! \circ f'_* \to f_* \circ k_! \\ \gamma _ C : m'_! \circ g''_* \to g'_* \circ l'_! & \gamma _ D : m_! \circ g'_* \to g_* \circ l_! \end{matrix}$

For the $2 \times 1$ and $1 \times 2$ rectangles we have four further maps

$\begin{matrix} \gamma _{A + B} : (l \circ l')_! \circ f''_* \to f_* \circ (k \circ k')_* \\ \gamma _{C + D} : (m \circ m')_! \circ g''_* \to g_* \circ (l \circ l')_! \\ \gamma _{A + C} : m'_! \circ (g'' \circ f'')_* \to (g' \circ f')_* \circ k'_! \\ \gamma _{B + D} : m_! \circ (g' \circ f')_* \to (g \circ f)_* \circ k_! \end{matrix}$

By Lemma 62.8.3 we have

$\gamma _{A + B} = \gamma _ B \circ \gamma _ A, \quad \gamma _{C + D} = \gamma _ D \circ \gamma _ C$

and by Lemma 62.8.2 we have

$\gamma _{A + C} = \gamma _ A \circ \gamma _ C, \quad \gamma _{B + D} = \gamma _ B \circ \gamma _ D$

Here it would be more correct to write $\gamma _{A + B} = (\gamma _ B \star \text{id}_{k'_!}) \circ (\text{id}_{l_!} \star \gamma _ A)$ with notation as in Categories, Section 4.28 and similarly for the others. Having said all of this we find (a priori) two transformations

$m_! \circ m'_! \circ g''_* \circ f''_* \longrightarrow g_* \circ f_* \circ k_! \circ k'_!$

namely

$\gamma _ B \circ \gamma _ D \circ \gamma _ A \circ \gamma _ C = \gamma _{B + D} \circ \gamma _{A + C}$

and

$\gamma _ B \circ \gamma _ A \circ \gamma _ D \circ \gamma _ C = \gamma _{A + B} \circ \gamma _{C + D}$

The point of this remark is to point out that these transformations are equal. Namely, to see this it suffices to show that

$\xymatrix{ m_! \circ g'_* \circ l'_! \circ f''_* \ar[r]_{\gamma _ D} \ar[d]_{\gamma _ A} & g_* \circ l_! \circ l'_! \circ f''_* \ar[d]^{\gamma _ A} \\ m_! \circ g'_* \circ f'_* \circ k'_! \ar[r]^{\gamma _ D} & g_* \circ l_! \circ f'_* \circ k'_! }$

commutes. This is true because the squares $A$ and $D$ meet in only one point, more precisely by Categories, Lemma 4.28.2 or more simply the discussion preceding Categories, Definition 4.28.1.

Lemma 62.8.5. Let $b : Y_1 \to Y$ be a morphism of schemes. Consider a commutative diagram of schemes

$\vcenter { \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } } \quad \text{and let}\quad \vcenter { \xymatrix{ X'_1 \ar[r]_{g'_1} \ar[d]_{f'_1} & X_1 \ar[d]^{f_1} \\ Y'_1 \ar[r]^{g_1} & Y_1 } }$

be the base change by $b$. Assume $f$ and $f'$ proper and $g$ and $g'$ separated and locally quasi-finite. For a ring $\Lambda$ and $K$ in $D(X'_{\acute{e}tale}, \Lambda )$ there is commutative diagram

$\xymatrix{ b^{-1}g_!Rf'_*K \ar[d] \ar[r] & g_{1, !}(b')^{-1}Rf'_*K \ar[r] & g_{1, !}Rf'_{1, *}(a')^{-1}K \ar[d] \\ b^{-1}Rf_*g'_!K \ar[r] & Rf_{1, *}a^{-1}g'_!K \ar[r] & Rf_{1, *}g'_{1, !}(a')^{-1}K }$

in $D(Y_{1, {\acute{e}tale}}, \Lambda )$ where $a : X_1 \to X$, $a' : X'_1 \to X'$, $b' : Y'_1 \to Y'$ are the projections, the vertical maps are the arrows of Lemma 62.8.1 and the horizontal arrows are the base change map (from Étale Cohomology, Section 59.86) and the base change map of Lemma 62.3.12.

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 62.8.1 constructs $g_!Rf'_*K \to Rf_*g'_!K$ as

$g_!f'_*\mathcal{J}^\bullet = g_!f'_!\mathcal{J}^\bullet = f_!g'_!\mathcal{J}^\bullet = f_*g'_!\mathcal{J}^\bullet \to f_*\mathcal{I}^\bullet$

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

$\xymatrix{ g'_{1, !}\mathcal{J}_1^\bullet \ar[rr] & & \mathcal{I}_1^\bullet \\ g'_{1, !}(a')^{-1}\mathcal{J}^\bullet \ar[u] \ar@{=}[r] & a^{-1}g'_!\mathcal{J}^\bullet \ar[r] & a^{-1}\mathcal{I}^\bullet \ar[u] }$

commuting up to homotopy where all arrows are quasi-isomorphisms, the equality comes from Lemma 62.3.4, 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

$g_{1, !}f'_{1, *}\mathcal{J}_1^\bullet = g_{1, !}f'_{1, !}\mathcal{J}_1^\bullet = f_{1, !}g'_{1, !}\mathcal{J}_1^\bullet = f_{1, *}g'_{1, !}\mathcal{J}_1^\bullet \to f_{1, *}\mathcal{I}_1^\bullet$

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

$\xymatrix{ b^{-1}g_!f'_*\mathcal{J}^\bullet \ar@{=}[d] \ar[r] & g_{1, !}(b')^{-1}f'_*\mathcal{J}^\bullet \ar[r] & g_{1, !}f'_{1, *}(a')^{-1}\mathcal{J}^\bullet \ar@{=}[d] \\ b^{-1}f_*g'_!\mathcal{J}^\bullet \ar[r] & f_{1, *}a^{-1}g'_!\mathcal{J}^\bullet \ar[r] & f_{1, *}g'_{1, !}(a')^{-1}\mathcal{J}^\bullet }$

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 62.3.12). 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

$\xymatrix{ g_{1, !}f'_{1, *}(a')^{-1}\mathcal{J}^\bullet \ar@{=}[d] \ar[r] & g_{1, !}f'_{1, *}\mathcal{J}_1^\bullet \ar@{=}[d] \\ f_{1, *}g'_{1, !}(a')^{-1}\mathcal{J}^\bullet \ar[r] & f_{1, *}g'_{1, !}\mathcal{J}_1^\bullet }$

commutes, to finish the proof it suffices to show that

$\xymatrix{ b^{-1}f_*g'_!\mathcal{J}^\bullet \ar[r] \ar[d] & f_{1, *}a^{-1}g'_!\mathcal{J}^\bullet \ar[r] \ar[d] & f_{1, *}g'_{1, !}(a')^{-1}\mathcal{J}^\bullet \ar[r] & f_{1, *}g'_{1, !}\mathcal{J}_1^\bullet \ar[d] \\ b^{-1}f_*\mathcal{I}^\bullet \ar[r] & f_{1, *}a^{-1}\mathcal{I}^\bullet \ar[rr] & & f_{1, *}\mathcal{I}_1^\bullet }$

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

Lemma 62.8.6. Consider a commutative diagram of schemes

$\xymatrix{ X \ar[r]_ f \ar[rd]_ g & Y \ar[d]^ h \\ & Z }$

with $f$ and $g$ locally quasi-finite and $h$ proper. Let $\Lambda$ be a ring. Funtorially in $K \in D(X_{\acute{e}tale}, \Lambda )$ there is a canonical map

$g_!K \longrightarrow Rh_*(f_!K)$

in $D(Z_{\acute{e}tale}, \Lambda )$. This map is an isomorphism if (a) $K$ is bounded below and has torsion cohomology sheaves, or (b) $\Lambda$ is a torsion ring.

Proof. This is a special case of Lemma 62.8.1 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

$g_!\mathcal{K}^\bullet = h_!f_!\mathcal{K}^\bullet = h_*f_!\mathcal{K}^\bullet \longrightarrow h_*\mathcal{I}^\bullet$

where the equalities are Lemmas 62.4.11 and 62.3.4. 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 62.4.5 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 62.8.1 (using Lemma 62.4.5 instead of Lemma 62.3.17). $\square$

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