## 62.6 Upper shriek for locally quasi-finite morphisms

For a locally quasi-finite morphism $f : X \to Y$ of schemes, the functor $f_! : \textit{Ab}(X_{\acute{e}tale}) \to \textit{Ab}(Y_{\acute{e}tale})$ commutes with direct sums and is exact, see Lemma 62.4.5. This suggests that it has a right adjoint which we will denote $f^!$.

Warning: This functor is the non-derived version!

Lemma 62.6.1. Let $f : X \to Y$ be a locally quasi-finite morphism of schemes.

The functor $f_! : \textit{Ab}(X_{\acute{e}tale}) \to \textit{Ab}(Y_{\acute{e}tale})$ has a right adjoint $f^! : \textit{Ab}(Y_{\acute{e}tale}) \to \textit{Ab}(X_{\acute{e}tale})$.

We have $f^!(\overline{y}_*A) = \prod _{f(\overline{x}) = \overline{y}} \overline{x}_*A$.

If $\Lambda $ is a ring, then the functor $f_! : \textit{Mod}(X_{\acute{e}tale}, \Lambda ) \to \textit{Mod}(Y_{\acute{e}tale}, \Lambda )$ has a right adjoint $f^! : \textit{Mod}(Y_{\acute{e}tale}, \Lambda ) \to \textit{Mod}(X_{\acute{e}tale}, \Lambda )$ which agrees with $f^!$ on underlying abelian sheaves.

**Proof.**
Proof of (1). Let $E \subset \mathop{\mathrm{Ob}}\nolimits (\textit{Ab}(Y_{\acute{e}tale}))$ be the class consisting of products of skyscraper sheaves. We claim that

every $\mathcal{G}$ in $\textit{Ab}(Y_{\acute{e}tale})$ is a subsheaf of an element of $E$, and

for every $\mathcal{G} \in E$ there exists an object $\mathcal{H}$ of $\textit{Ab}(X_{\acute{e}tale})$ such that $\mathop{\mathrm{Hom}}\nolimits (f_!\mathcal{F}, \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits (\mathcal{F}, \mathcal{H})$ functorially in $\mathcal{F}$.

Once the claim has been verified, the dual of Homology, Lemma 12.29.6 produces the adjoint functor $f^!$.

Part (a) is true because we can map $\mathcal{G}$ to the sheaf $\prod \overline{y}_*\mathcal{G}_{\overline{y}}$ where the product is over all geometric points of $Y$. This is an injection by Étale Cohomology, Theorem 59.29.10. (This is the first step in the Godement resolution when done in the setting of abelian sheaves on topological spaces.)

Part (b) and part (2) of the lemma can be seen as follows. Suppose that $\mathcal{G} = \prod \overline{y}_*A_{\overline{y}}$ for some abelian groups $A_{\overline{y}}$. Then

\[ \mathop{\mathrm{Hom}}\nolimits (f_!\mathcal{F}, \mathcal{G}) = \prod \mathop{\mathrm{Hom}}\nolimits (f_!\mathcal{F}, \overline{y}_*A_{\overline{y}}) \]

Thus it suffices to find abelian sheaves $\mathcal{H}_{\overline{y}}$ on $X_{\acute{e}tale}$ representing the functors $\mathcal{F} \mapsto \mathop{\mathrm{Hom}}\nolimits (f_!\mathcal{F}, \overline{y}_*A_{\overline{y}})$ and to take $\mathcal{H} = \prod \mathcal{H}_{\overline{y}}$. This reduces us to the case $\mathcal{H} = \overline{y}_*A$ for some fixed geometric point $\overline{y} : \mathop{\mathrm{Spec}}(k) \to Y$ and some fixed abelian group $A$. We claim that in this case $\mathcal{H} = \prod _{f(\overline{x}) = \overline{y}} \overline{x}_*A$ works. This will finish the proof of parts (1) and (2) of the lemma. Namely, we have

\[ \mathop{\mathrm{Hom}}\nolimits (f_!\mathcal{F}, \overline{y}_*A) = \mathop{\mathrm{Hom}}\nolimits _{\textit{Ab}}((f_!\mathcal{F})_{\overline{y}}, A) = \mathop{\mathrm{Hom}}\nolimits _{\textit{Ab}}(\bigoplus \nolimits _{f(\overline{x}) = \overline{y}} \mathcal{F}_{\overline{x}}, A) \]

by the description of stalks in Lemma 62.4.5 on the one hand and on the other hand we have

\[ \mathop{\mathrm{Hom}}\nolimits (\mathcal{F}, \mathcal{H}) = \prod \nolimits _{f(\overline{x}) = \overline{y}} \mathop{\mathrm{Hom}}\nolimits (\mathcal{F}, \overline{x}_*A) = \prod \nolimits _{f(\overline{x}) = \overline{y}} \mathop{\mathrm{Hom}}\nolimits _{\textit{Ab}}(\mathcal{F}_{\overline{x}}, A) \]

We leave it to the reader to identify these as functors of $\mathcal{F}$.

Proof of part (3). Observe that an object $\textit{Mod}(X_{\acute{e}tale}, \Lambda )$ is the same thing as an object $\mathcal{F}$ of $\textit{Ab}(X_{\acute{e}tale})$ together with a map $\Lambda \to \text{End}(\mathcal{F})$. Hence the functors $f_!$ and $f^!$ in (1) define functors $f_!$ and $f^!$ as in (3). A straightforward computation shows that they are adjoints.
$\square$

Lemma 62.6.2. Let $j : U \to X$ be an étale morphism. Then $j^! = j^{-1}$.

**Proof.**
This is true because $j_!$ as defined in Section 62.4 agrees with $j_!$ as defined in Étale Cohomology, Section 59.70, see Lemma 62.4.3. Finally, in Étale Cohomology, Section 59.70 the functor $j_!$ is defined as the left adjoint of $j^{-1}$ and hence we conclude by uniqueness of adjoint functors.
$\square$

Lemma 62.6.3. Let $f : X \to Y$ and $g : Y \to Z$ be separated and locally quasi-finite morphisms. There is a canonical isomorphism $(g \circ f)^! \to f^! \circ g^!$. Given a third locally quasi-finite morphism $h : Z \to T$ the diagram

\[ \xymatrix{ (h \circ g \circ f)^! \ar[r] \ar[d] & f^! \circ (h \circ g)^! \ar[d] \\ (g \circ f)^! \circ h^! \ar[r] & f^! \circ g^! \circ h^! } \]

commutes.

**Proof.**
By uniqueness of adjoint functors, this immediately translates into the corresponding (dual) statement for the functors $f_!$. See Lemma 62.4.12.
$\square$

Lemma 62.6.4. Let $j : U \to X$ and $j' : V \to U$ be étale morphisms. The isomorphism $(j \circ j')^{-1} = (j')^{-1} \circ j^{-1}$ and the isomorphism $(j \circ j')^! = (j')^! \circ j^!$ of Lemma 62.6.3 agree via the isomorphism of Lemma 62.6.2.

**Proof.**
Omitted.
$\square$

Lemma 62.6.5. Consider a cartesian square

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

of schemes with $f$ locally quasi-finite. For any abelian sheaf $\mathcal{F}$ on $Y'_{\acute{e}tale}$ we have $(g')_*(f')^!\mathcal{F} = f^!g_*\mathcal{F}$.

**Proof.**
By uniqueness of adjoint functors, this follows from the corresponding (dual) statement for the functors $f_!$. See Lemma 62.4.10.
$\square$

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