Lemma 20.8.1. Let $X$ be a ringed space. Let $U' \subset U \subset X$ be open subspaces. For any injective $\mathcal{O}_ X$-module $\mathcal{I}$ the restriction mapping $\mathcal{I}(U) \to \mathcal{I}(U')$ is surjective.
20.8 Mayer-Vietoris
Below will construct the Čech-to-cohomology spectral sequence, see Lemma 20.11.5. A special case of that spectral sequence is the Mayer-Vietoris long exact sequence. Since it is such a basic, useful and easy to understand variant of the spectral sequence we treat it here separately.
Proof. Let $j : U \to X$ and $j' : U' \to X$ be the open immersions. Recall that $j_!\mathcal{O}_ U$ is the extension by zero of $\mathcal{O}_ U = \mathcal{O}_ X|_ U$, see Sheaves, Section 6.31. Since $j_!$ is a left adjoint to restriction we see that for any sheaf $\mathcal{F}$ of $\mathcal{O}_ X$-modules
see Sheaves, Lemma 6.31.8. Similarly, the sheaf $j'_!\mathcal{O}_{U'}$ represents the functor $\mathcal{F} \mapsto \mathcal{F}(U')$. Moreover there is an obvious canonical map of $\mathcal{O}_ X$-modules
which corresponds to the restriction mapping $\mathcal{F}(U) \to \mathcal{F}(U')$ via Yoneda's lemma (Categories, Lemma 4.3.5). By the description of the stalks of the sheaves $j'_!\mathcal{O}_{U'}$, $j_!\mathcal{O}_ U$ we see that the displayed map above is injective (see lemma cited above). Hence if $\mathcal{I}$ is an injective $\mathcal{O}_ X$-module, then the map
is surjective, see Homology, Lemma 12.27.2. Putting everything together we obtain the lemma. $\square$
Lemma 20.8.2 (Mayer-Vietoris). Let $X$ be a ringed space. Suppose that $X = U \cup V$ is a union of two open subsets. For every $\mathcal{O}_ X$-module $\mathcal{F}$ there exists a long exact cohomology sequence This long exact sequence is functorial in $\mathcal{F}$.
Proof. The sheaf condition says that the kernel of $(1, -1) : \mathcal{F}(U) \oplus \mathcal{F}(V) \to \mathcal{F}(U \cap V)$ is equal to the image of $\mathcal{F}(X)$ by the first map for any abelian sheaf $\mathcal{F}$. Lemma 20.8.1 above implies that the map $(1, -1) : \mathcal{I}(U) \oplus \mathcal{I}(V) \to \mathcal{I}(U \cap V)$ is surjective whenever $\mathcal{I}$ is an injective $\mathcal{O}_ X$-module. Hence if $\mathcal{F} \to \mathcal{I}^\bullet $ is an injective resolution of $\mathcal{F}$, then we get a short exact sequence of complexes
Taking cohomology gives the result (use Homology, Lemma 12.13.12). We omit the proof of the functoriality of the sequence. $\square$
Lemma 20.8.3 (Relative Mayer-Vietoris). Let $f : X \to Y$ be a morphism of ringed spaces. Suppose that $X = U \cup V$ is a union of two open subsets. Denote $a = f|_ U : U \to Y$, $b = f|_ V : V \to Y$, and $c = f|_{U \cap V} : U \cap V \to Y$. For every $\mathcal{O}_ X$-module $\mathcal{F}$ there exists a long exact sequence This long exact sequence is functorial in $\mathcal{F}$.
Proof. Let $\mathcal{F} \to \mathcal{I}^\bullet $ be an injective resolution of $\mathcal{F}$. We claim that we get a short exact sequence of complexes
Namely, for any open $W \subset Y$, and for any $n \geq 0$ the corresponding sequence of groups of sections over $W$
was shown to be short exact in the proof of Lemma 20.8.2. The lemma follows by taking cohomology sheaves and using the fact that $\mathcal{I}^\bullet |_ U$ is an injective resolution of $\mathcal{F}|_ U$ and similarly for $\mathcal{I}^\bullet |_ V$, $\mathcal{I}^\bullet |_{U \cap V}$ see Lemma 20.7.1. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #933 by correction_bot on
Comment #934 by correction_bot on