## 20.33 Unbounded Mayer-Vietoris

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

Lemma 20.33.1. 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 20.32 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 20.33.2. 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 20.32.1). 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 20.8.2). $\square$

Lemma 20.33.3. 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 20.33.1 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 20.32.8. $\square$

Lemma 20.33.4. 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 20.32.1. 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 20.8.2. The final statement follows from the functoriality of the construction in Injectives, Theorem 19.12.6. $\square$

Lemma 20.33.5. 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 20.32.1. 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 20.8.3 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 20.33.6. 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 20.32.4 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 20.32.4 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. $\square$

Lemma 20.33.7. Let $(X, \mathcal{O}_ X)$ be a ringed space. Set $A = \Gamma (X, \mathcal{O}_ X)$. Suppose that $X = U \cup V$ is a union of two open subsets. For objects $K$ and $M$ of $D(\mathcal{O}_ X)$ we have a map of distinguished triangles

$\xymatrix{ R\Gamma (X, K) \otimes _ A^\mathbf {L} R\Gamma (X, M) \ar[r] \ar[d] & R\Gamma (X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) \ar[d] \\ R\Gamma (X, K) \otimes _ A^\mathbf {L} (R\Gamma (U, M) \oplus R\Gamma (V, M)) \ar[r] \ar[d] & R\Gamma (U, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) \oplus R\Gamma (V, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M)) \ar[d] \\ R\Gamma (X, K) \otimes _ A^\mathbf {L} R\Gamma (U \cap V, M) \ar[r] \ar[d] & R\Gamma (U \cap V, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) \ar[d] \\ R\Gamma (X, K) \otimes _ A^\mathbf {L} R\Gamma (X, M)[1] \ar[r] & R\Gamma (X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M)[1] }$

where

1. the horizontal arrows are given by cup product,

2. on the right hand side we have the distinguished triangle of Lemma 20.33.4 for $K \otimes _{\mathcal{O}_ X}^\mathbf {L} M$, and

3. on the left hand side we have the exact functor $R\Gamma (X, K) \otimes _ A^\mathbf {L} -$ applied to the distinguished triangle of Lemma 20.33.4 for $M$.

Proof. Choose a K-flat complex $T^\bullet$ of flat $A$-modules representing $R\Gamma (X, K)$, see More on Algebra, Lemma 15.59.10. Denote $T^\bullet \otimes _ A \mathcal{O}_ X$ the pullback of $T^\bullet$ by the morphism of ringed spaces $(X, \mathcal{O}_ X) \to (pt, A)$. There is a natural adjunction map $\epsilon : T^\bullet \otimes _ A \mathcal{O}_ X \to K$ in $D(\mathcal{O}_ X)$. Observe that $T^\bullet \otimes _ A \mathcal{O}_ X$ is a K-flat complex of $\mathcal{O}_ X$-modules with flat terms, see Lemma 20.26.8 and Modules, Lemma 17.20.2. By Lemma 20.26.17 we can find a morphism of complexes

$T^\bullet \otimes _ A \mathcal{O}_ X \longrightarrow \mathcal{K}^\bullet$

of $\mathcal{O}_ X$-modules representing $\epsilon$ such that $\mathcal{K}^\bullet$ is a K-flat complex with flat terms. Namely, by the construction of $D(\mathcal{O}_ X)$ we can first represent $\epsilon$ by some map of complexes $e : T^\bullet \otimes _ A \mathcal{O}_ X \to \mathcal{L}^\bullet$ of $\mathcal{O}_ X$-modules representing $\epsilon$ and then we can apply the lemma to $e$. Choose a K-injective complex $\mathcal{I}^\bullet$ whose terms are injective $\mathcal{O}_ X$-modules representing $M$. Finally, choose a quasi-isomorphism

$\text{Tot}(\mathcal{K}^\bullet \otimes _\mathcal {O} \mathcal{I}^\bullet ) \longrightarrow \mathcal{J}^\bullet$

into a K-injective complex whose terms are injective $\mathcal{O}_ X$-modules. Observe that source and target of this arrow represent $K \otimes _{\mathcal{O}_ X}^\mathbf {L} M$ in $D(\mathcal{O}_ X)$. At this point, for any open $W \subset X$ we obtain a map of complexes

$\text{Tot}(T^\bullet \otimes _ A \mathcal{I}^\bullet (W)) \to \text{Tot}(\mathcal{K}^\bullet (W) \otimes _ A \mathcal{I}^\bullet (W)) \to \mathcal{J}^\bullet (W)$

of $A$-modules whose composition represents the map

$R\Gamma (X, K) \otimes _ A^\mathbf {L} R\Gamma (W, M) \longrightarrow R\Gamma (W, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M)$

in $D(A)$. Clearly, these maps are compatible with restriction mappings. OK, so now we can consider the following commutative(!) diagram of complexes of $A$-modules

$\xymatrix{ 0 \ar[d] & 0 \ar[d] \\ \text{Tot}(T^\bullet \otimes _ A \mathcal{I}^\bullet (X)) \ar[d] \ar[r] & \mathcal{J}^\bullet (X) \ar[d] \\ \text{Tot}(T^\bullet \otimes _ A (\mathcal{I}^\bullet (U) \oplus \mathcal{I}^\bullet (V)) \ar[d] \ar[r] & \mathcal{J}^\bullet (U) \oplus \mathcal{J}^\bullet (V) \ar[d] \\ \text{Tot}(T^\bullet \otimes _ A \mathcal{I}^\bullet (U \cap V)) \ar[r] \ar[d] & \mathcal{J}^\bullet (U \cap V) \ar[d] \\ 0 & 0 }$

By the proof of Lemma 20.8.2 the columns are exact sequences of complexes of $A$-modules (this also uses that $\text{Tot}(T^\bullet \otimes _ A -)$ transforms short exact sequences of complexes of $A$-modules into short exact sequences as the terms of $T^\bullet$ are flat $A$-modules). Since the distinguished triangles of Lemma 20.33.4 are the distinguished triangles associated to these short exact sequences of complexes, the desired result follows from the functoriality of “taking the associated distinguished triangle” discussed in Derived Categories, Section 13.12. $\square$

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