
## 20.31 Unbounded Mayer-Vietoris

There is a Mayer-Vietoris sequence for unbounded cohomology as well.

Lemma 20.31.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $X = U \cup V$ be the union of two open subspaces. For any object $E$ of $D(\mathcal{O}_ X)$ we have a distinguished triangle

$j_{U \cap V!}E|_{U \cap V} \to j_{U!}E|_ U \oplus j_{V!}E|_ V \to E \to j_{U \cap V!}E|_{U \cap V}[1]$

in $D(\mathcal{O}_ X)$.

Proof. We have seen in Section 20.30 that the restriction functors and the extension by zero functors are computed by just applying the functors to any complex. Let $\mathcal{E}^\bullet$ be a complex of $\mathcal{O}_ X$-modules representing $E$. The distinguished triangle of the lemma is the distinguished triangle associated (by Derived Categories, Section 13.12 and especially Lemma 13.12.1) to the short exact sequence of complexes of $\mathcal{O}_ X$-modules

$0 \to j_{U \cap V!}\mathcal{E}^\bullet |_{U \cap V} \to j_{U!}\mathcal{E}^\bullet |_ U \oplus j_{V!}\mathcal{E}^\bullet |_ V \to \mathcal{E}^\bullet \to 0$

To see this sequence is exact one checks on stalks using Sheaves, Lemma 6.31.8 (computation omitted). $\square$

Lemma 20.31.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $X = U \cup V$ be the union of two open subspaces. For any object $E$ of $D(\mathcal{O}_ X)$ we have a distinguished triangle

$E \to Rj_{U, *}E|_ U \oplus Rj_{V, *}E|_ V \to Rj_{U \cap V, *}E|_{U \cap V} \to E[1]$

in $D(\mathcal{O}_ X)$.

Proof. Choose a K-injective complex $\mathcal{I}^\bullet$ representing $E$ whose terms $\mathcal{I}^ n$ are injective objects of $\textit{Mod}(\mathcal{O}_ X)$, see Injectives, Theorem 19.12.6. We have seen that $\mathcal{I}^\bullet |U$ is a K-injective complex as well (Lemma 20.30.1). Hence $Rj_{U, *}E|_ U$ is represented by $j_{U, *}\mathcal{I}^\bullet |_ U$. Similarly for $V$ and $U \cap V$. Hence the distinguished triangle of the lemma is the distinguished triangle associated (by Derived Categories, Section 13.12 and especially Lemma 13.12.1) to the short exact sequence of complexes

$0 \to \mathcal{I}^\bullet \to j_{U, *}\mathcal{I}^\bullet |_ U \oplus j_{V, *}\mathcal{I}^\bullet |_ V \to j_{U \cap V, *}\mathcal{I}^\bullet |_{U \cap V} \to 0.$

This sequence is exact because for any $W \subset X$ open and any $n$ the sequence

$0 \to \mathcal{I}^ n(W) \to \mathcal{I}^ n(W \cap U) \oplus \mathcal{I}^ n(W \cap V) \to \mathcal{I}^ n(W \cap U \cap V) \to 0$

is exact (see proof of Lemma 20.9.2). $\square$

Lemma 20.31.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $X = U \cup V$ be the union of two open subspaces of $X$. For objects $E$, $F$ of $D(\mathcal{O}_ X)$ we have a Mayer-Vietoris sequence

$\xymatrix{ & \ldots \ar[r] & \mathop{\mathrm{Ext}}\nolimits ^{-1}(E_{U \cap V}, F_{U \cap V}) \ar[lld] \\ \mathop{\mathrm{Hom}}\nolimits (E, F) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (E_ U, F_ U) \oplus \mathop{\mathrm{Hom}}\nolimits (E_ V, F_ V) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (E_{U \cap V}, F_{U \cap V}) }$

where the subscripts denote restrictions to the relevant opens and the $\mathop{\mathrm{Hom}}\nolimits$'s and $\mathop{\mathrm{Ext}}\nolimits$'s are taken in the relevant derived categories.

Proof. Use the distinguished triangle of Lemma 20.31.1 to obtain a long exact sequence of $\mathop{\mathrm{Hom}}\nolimits$'s (from Derived Categories, Lemma 13.4.2) and use that

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(j_{U!}E|_ U, F) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(E|_ U, F|_ U)$

by Lemma 20.30.8. $\square$

Lemma 20.31.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Suppose that $X = U \cup V$ is a union of two open subsets. For an object $E$ of $D(\mathcal{O}_ X)$ we have a distinguished triangle

$R\Gamma (X, E) \to R\Gamma (U, E) \oplus R\Gamma (V, E) \to R\Gamma (U \cap V, E) \to R\Gamma (X, E)[1]$

and in particular a long exact cohomology sequence

$\ldots \to H^ n(X, E) \to H^ n(U, E) \oplus H^0(V, E) \to H^ n(U \cap V, E) \to H^{n + 1}(X, E) \to \ldots$

The construction of the distinguished triangle and the long exact sequence is functorial in $E$.

Proof. Choose a K-injective complex $\mathcal{I}^\bullet$ representing $E$. We may assume $\mathcal{I}^ n$ is an injective object of $\textit{Mod}(\mathcal{O}_ X)$ for all $n$, see Injectives, Theorem 19.12.6. Then $R\Gamma (X, E)$ is computed by $\Gamma (X, \mathcal{I}^\bullet )$. Similarly for $U$, $V$, and $U \cap V$ by Lemma 20.30.1. Hence the distinguished triangle of the lemma is the distinguished triangle associated (by Derived Categories, Section 13.12 and especially Lemma 13.12.1) to the short exact sequence of complexes

$0 \to \mathcal{I}^\bullet (X) \to \mathcal{I}^\bullet (U) \oplus \mathcal{I}^\bullet (V) \to \mathcal{I}^\bullet (U \cap V) \to 0.$

We have seen this is a short exact sequence in the proof of Lemma 20.9.2. The final statement follows from the functoriality of the construction in Injectives, Theorem 19.12.6. $\square$

Lemma 20.31.5. 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 object $E$ of $D(\mathcal{O}_ X)$ there exists a distinguished triangle

$Rf_*E \to Ra_*(E|_ U) \oplus Rb_*(E|_ V) \to Rc_*(E|_{U \cap V}) \to Rf_*E[1]$

This triangle is functorial in $E$.

Proof. Choose a K-injective complex $\mathcal{I}^\bullet$ representing $E$. We may assume $\mathcal{I}^ n$ is an injective object of $\textit{Mod}(\mathcal{O}_ X)$ for all $n$, see Injectives, Theorem 19.12.6. Then $Rf_*E$ is computed by $f_*\mathcal{I}^\bullet$. Similarly for $U$, $V$, and $U \cap V$ by Lemma 20.30.1. Hence the distinguished triangle of the lemma is the distinguished triangle associated (by Derived Categories, Section 13.12 and especially Lemma 13.12.1) to the short exact sequence of complexes

$0 \to f_*\mathcal{I}^\bullet \to a_*\mathcal{I}^\bullet |_ U \oplus b_*\mathcal{I}^\bullet |_ V \to c_*\mathcal{I}^\bullet |_{U \cap V} \to 0.$

This is a short exact sequence of complexes by Lemma 20.9.3 and the fact that $R^1f_*\mathcal{I} = 0$ for an injective object $\mathcal{I}$ of $\textit{Mod}(\mathcal{O}_ X)$. The final statement follows from the functoriality of the construction in Injectives, Theorem 19.12.6. $\square$

Lemma 20.31.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $j : U \to X$ be an open subspace. Let $T \subset X$ be a closed subset contained in $U$.

1. If $E$ is an object of $D(\mathcal{O}_ X)$ whose cohomology sheaves are supported on $T$, then $E \to Rj_*(E|_ U)$ is an isomorphism.

2. If $F$ is an object of $D(\mathcal{O}_ U)$ whose cohomology sheaves are supported on $T$, then $j_!F \to Rj_*F$ is an isomorphism.

Proof. Let $V = X \setminus T$ and $W = U \cap V$. Note that $X = U \cup V$ is an open covering of $X$. Denote $j_ W : W \to V$ the open immersion. Let $E$ be an object of $D(\mathcal{O}_ X)$ whose cohomology sheaves are supported on $T$. By Lemma 20.30.4 we have $(Rj_*E|_ U)|_ V = Rj_{W, *}(E|_ W) = 0$ because $E|_ W = 0$ by our assumption. On the other hand, $Rj_*(E|_ U)|_ U = E|_ U$. Thus (1) is clear. Let $F$ be an object of $D(\mathcal{O}_ U)$ whose cohomology sheaves are supported on $T$. By Lemma 20.30.4 we have $(Rj_*F)|_ V = Rj_{W, *}(F|_ W) = 0$ because $F|_ W = 0$ by our assumption. We also have $(j_!F)|_ V = j_{W!}(F|_ W) = 0$ (the first equality is immediate from the definition of extension by zero). Since both $(Rj_*F)|_ U = F$ and $(j_!F)|_ U = F$ we see that (2) holds. $\square$

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