Lemma 20.15.1. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\varphi : f^*\mathcal{G} \to \mathcal{F}$ be an $f$-map from an $\mathcal{O}_ Y$-module $\mathcal{G}$ to an $\mathcal{O}_ X$-module $\mathcal{F}$. Let $\mathcal{U} : X = \bigcup _{i \in I} U_ i$ and $\mathcal{V} : Y = \bigcup _{j \in J} V_ j$ be open coverings. Assume that $\mathcal{U}$ is a refinement of $f^{-1}\mathcal{V} : X = \bigcup _{j \in J} f^{-1}(V_ j)$. In this case there exists a commutative diagram
\[ \xymatrix{ \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \ar[r] & R\Gamma (X, \mathcal{F}) \\ \check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{G}) \ar[r] \ar[u]^\gamma & R\Gamma (Y, \mathcal{G}) \ar[u] } \]
in $D^{+}(\mathcal{O}_ X(X))$ with horizontal arrows given by Lemma 20.11.2 and right vertical arrow by (20.14.1.1). In particular we get commutative diagrams of cohomology groups
\[ \xymatrix{ \check{H}^ p(\mathcal{U}, \mathcal{F}) \ar[r] & H^ p(X, \mathcal{F}) \\ \check{H}^ p(\mathcal{V}, \mathcal{G}) \ar[r] \ar[u]^\gamma & H^ p(Y, \mathcal{G}) \ar[u] } \]
where the right vertical arrow is (20.14.1.2)
Proof.
We first define the left vertical arrow. Namely, choose a map $c : I \to J$ such that $U_ i \subset f^{-1}(V_{c(i)})$ for all $i \in I$. In degree $p$ we define the map by the rule
\[ \gamma (s)_{i_0 \ldots i_ p} = \varphi (s)_{c(i_0) \ldots c(i_ p)} \]
This makes sense because $\varphi $ does indeed induce maps $\mathcal{G}(V_{c(i_0) \ldots c(i_ p)}) \to \mathcal{F}(U_{i_0 \ldots i_ p})$ by assumption. It is also clear that this defines a morphism of complexes. Choose injective resolutions $\mathcal{F} \to \mathcal{I}^\bullet $ on $X$ and $\mathcal{G} \to J^\bullet $ on $Y$. According to the proof of Lemma 20.11.2 we introduce the double complexes $A^{\bullet , \bullet }$ and $B^{\bullet , \bullet }$ with terms
\[ B^{p, q} = \check{\mathcal{C}}^ p(\mathcal{V}, \mathcal{J}^ q) \quad \text{and} \quad A^{p, q} = \check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{I}^ q). \]
As in Remark 20.14.2 above we also choose an injective resolution $f_*\mathcal{I} \to (\mathcal{J}')^\bullet $ on $Y$ and a morphism of complexes $\beta : \mathcal{J} \to (\mathcal{J}')^\bullet $ making (20.14.2.1) commutes. We introduce some more double complexes, namely $(B')^{\bullet , \bullet }$ and $(B'')^{\bullet , \bullet }$ with
\[ (B')^{p, q} = \check{\mathcal{C}}^ p(\mathcal{V}, (\mathcal{J}')^ q) \quad \text{and} \quad (B'')^{p, q} = \check{\mathcal{C}}^ p(\mathcal{V}, f_*\mathcal{I}^ q). \]
Note that there is an $f$-map of complexes from $f_*\mathcal{I}^\bullet $ to $\mathcal{I}^\bullet $. Hence it is clear that the same rule as above defines a morphism of double complexes
\[ \gamma : (B'')^{\bullet , \bullet } \longrightarrow A^{\bullet , \bullet }. \]
Consider the diagram of complexes
\[ \xymatrix{ \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \ar[r] & \text{Tot}(A^{\bullet , \bullet }) & & & \Gamma (X, \mathcal{I}^\bullet ) \ar[lll]^{qis} \ar@{=}[ddl]\\ \check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{G}) \ar[r] \ar[u]^\gamma & \text{Tot}(B^{\bullet , \bullet }) \ar[r]^\beta & \text{Tot}((B')^{\bullet , \bullet }) & \text{Tot}((B'')^{\bullet , \bullet }) \ar[l] \ar[llu]_{s\gamma } \\ & \Gamma (Y, \mathcal{J}^\bullet ) \ar[u]^{qis} \ar[r]^\beta & \Gamma (Y, (\mathcal{J}')^\bullet ) \ar[u] & \Gamma (Y, f_*\mathcal{I}^\bullet ) \ar[u] \ar[l]_{qis} } \]
The two horizontal arrows with targets $\text{Tot}(A^{\bullet , \bullet })$ and $\text{Tot}(B^{\bullet , \bullet })$ are the ones explained in Lemma 20.11.2. The left upper shape (a pentagon) is commutative simply because (20.14.2.1) is commutative. The two lower squares are trivially commutative. It is also immediate from the definitions that the right upper shape (a square) is commutative. The result of the lemma now follows from the definitions and the fact that going around the diagram on the outer sides from $\check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{G})$ to $\Gamma (X, \mathcal{I}^\bullet )$ either on top or on bottom is the same (where you have to invert any quasi-isomorphisms along the way).
$\square$
Comments (2)
Comment #7505 by Xiaolong Liu on
Comment #7644 by Stacks Project on
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