The Stacks project

Lemma 21.31.7. Let $X$ be an object of $\textit{LC}_{Zar}$. Then

  1. for $\mathcal{F} \in \textit{Ab}(X)$ we have $H^ n_{Zar}(X, \pi _ X^{-1}\mathcal{F}) = H^ n(X, \mathcal{F})$,

  2. $\pi _{X, *} : \textit{Ab}(\textit{LC}_{Zar}/X) \to \textit{Ab}(X)$ is exact,

  3. the unit $\text{id} \to \pi _{X, *} \circ \pi _ X^{-1}$ of the adjunction is an isomorphism, and

  4. for $K \in D(X)$ the canonical map $K \to R\pi _{X, *} \pi _ X^{-1}K$ is an isomorphism.

Let $f : X \to Y$ be a morphism of $\textit{LC}_{Zar}$. Then

  1. there is a commutative diagram

    \[ \xymatrix{ \mathop{\mathit{Sh}}\nolimits (\textit{LC}_{Zar}/X) \ar[r]_{f_{Zar}} \ar[d]_{\pi _ X} & \mathop{\mathit{Sh}}\nolimits (\textit{LC}_{Zar}/Y) \ar[d]^{\pi _ Y} \\ \mathop{\mathit{Sh}}\nolimits (X_{Zar}) \ar[r]^ f & \mathop{\mathit{Sh}}\nolimits (Y_{Zar}) } \]

    of topoi,

  2. for $L \in D^+(Y)$ we have $H^ n_{Zar}(X, \pi _ Y^{-1}L) = H^ n(X, f^{-1}L)$,

  3. if $f$ is proper, then we have

    1. $\pi _ Y^{-1} \circ f_* = f_{Zar, *} \circ \pi _ X^{-1}$ as functors $\mathop{\mathit{Sh}}\nolimits (X) \to \mathop{\mathit{Sh}}\nolimits (\textit{LC}_{Zar}/Y)$,

    2. $\pi _ Y^{-1} \circ Rf_* = Rf_{Zar, *} \circ \pi _ X^{-1}$ as functors $D^+(X) \to D^+(\textit{LC}_{Zar}/Y)$.

Proof. Proof of (1). The equality $H^ n_{Zar}(X, \pi _ X^{-1}\mathcal{F}) = H^ n(X, \mathcal{F})$ is a general fact coming from the trivial observation that coverings of $X$ in $\textit{LC}_{Zar}$ are the same thing as open coverings of $X$. The reader who wishes to see a detailed proof should apply Lemma 21.8.2 to the functor $X_{Zar} \to \textit{LC}_{Zar}$.

Proof of (2). This is true because $\pi _{X, *} = \tau _ X^{-1}$ for some morphism of topoi $\tau _ X : \mathop{\mathit{Sh}}\nolimits (X_{Zar}) \to \mathop{\mathit{Sh}}\nolimits (\textit{LC}_{Zar})$ as follows from Sites, Lemma 7.21.8 applied to the functor $X_{Zar} \to \textit{LC}_{Zar}/X$ used to define $\pi _ X$.

Proof of (3). This is true because $\tau _ X^{-1} \circ \pi _ X^{-1}$ is the identity functor by Sites, Lemma 7.21.8. Or you can deduce it from the explicit description of $\pi _ X^{-1}$ in Lemma 21.31.6.

Proof of (4). Apply (3) to an complex of abelian sheaves representing $K$.

Proof of (5). The morphism of topoi $f_{Zar}$ comes from an application of Sites, Lemma 7.25.8 and in our case comes from the continuous functor $Z/Y \mapsto Z \times _ Y X/X$ by Sites, Lemma 7.27.3. The diagram commutes simply because the corresponding continuous functors compose correctly (see Sites, Lemma 7.14.4).

Proof of (6). We have $H^ n_{Zar}(X, \pi _ Y^{-1}\mathcal{G}) = H^ n_{Zar}(X, f_{Zar}^{-1}\pi _ Y^{-1}\mathcal{G})$ for $\mathcal{G}$ in $\textit{Ab}(Y)$, see Lemma 21.8.1. This is equal to $H^ n_{Zar}(X, \pi _ X^{-1}f^{-1}\mathcal{G})$ by the commutativity of the diagram in (5). Hence we conclude by (1) in the case $L$ consists of a single sheaf in degree $0$. The general case follows by representing $L$ by a bounded below complex of abelian sheaves.

Proof of (7a). Let $\mathcal{F}$ be a sheaf on $X$. Let $g : Z \to Y$ be an object of $\textit{LC}_{Zar}/Y$. Consider the fibre product

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

Then we have

\[ (f_{Zar, *}\pi _ X^{-1}\mathcal{F})(Z/Y) = (\pi _ X^{-1}\mathcal{F})(Z'/X) = \Gamma (Z', (g')^{-1}\mathcal{F}) = \Gamma (Z, f'_*(g')^{-1}\mathcal{F}) \]

the second equality by Lemma 21.31.6. On the other hand

\[ (\pi _ Y^{-1}f_*\mathcal{F})(Z/Y) = \Gamma (Z, g^{-1}f_*\mathcal{F}) \]

again by Lemma 21.31.6. Hence by proper base change for sheaves of sets (Cohomology, Lemma 20.19.3) we conclude the two sets are canonically isomorphic. The isomorphism is compatible with restriction mappings and defines an isomorphism $\pi _ Y^{-1}f_*\mathcal{F} = f_{Zar, *}\pi _ X^{-1}\mathcal{F}$. Thus an isomorphism of functors $\pi _ Y^{-1} \circ f_* = f_{Zar, *} \circ \pi _ X^{-1}$.

Proof of (7b). Let $K \in D^+(X)$. By Lemma 21.21.6 the $n$th cohomology sheaf of $Rf_{Zar, *}\pi _ X^{-1}K$ is the sheaf associated to the presheaf

\[ (g : Z \to Y) \longmapsto H^ n_{Zar}(Z', \pi _ X^{-1}K) \]

with notation as above. Observe that

\begin{align*} H^ n_{Zar}(Z', \pi _ X^{-1}K) & = H^ n(Z', (g')^{-1}K) \\ & = H^ n(Z, Rf'_*(g')^{-1}K) \\ & = H^ n(Z, g^{-1}Rf_*K) \\ & = H^ n_{Zar}(Z, \pi _ Y^{-1}Rf_*K) \end{align*}

The first equality is (6) applied to $K$ and $g' : Z' \to X$. The second equality is Leray for $f' : Z' \to Z$ (Cohomology, Lemma 20.14.1). The third equality is the proper base change theorem (Cohomology, Theorem 20.19.2). The fourth equality is (6) applied to $g : Z \to Y$ and $Rf_*K$. Thus $Rf_{Zar, *}\pi _ X^{-1}K$ and $\pi _ Y^{-1}Rf_*K$ have the same cohomology sheaves. We omit the verification that the canonical base change map $\pi _ Y^{-1}Rf_*K \to Rf_{Zar, *}\pi _ X^{-1}K$ induces this isomorphism. $\square$


Comments (2)

Comment #3605 by Carl on

In the first line of the proof, q should be n.


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