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Tag 0CRS

Chapter 66: Derived Categories of Spaces > Section 66.10: Mayer-Vietoris

Lemma 66.10.5. Let $S$ be a scheme. Let $(U \subset X, V \to X)$ be an elementary distinguished square of algebraic spaces over $S$. For an object $E$ of $D(\mathcal{O}_X)$ we have a distinguished triangle $$ R\Gamma(X, E) \to R\Gamma(U, E) \oplus R\Gamma(V, E) \to R\Gamma(U \times_X V, E) \to R\Gamma(X, E)[1] $$ and in particular a long exact cohomology sequence $$ \ldots \to H^n(X, E) \to H^n(U, E) \oplus H^n(V, E) \to H^n(U \times_X V, E) \to H^{n + 1}(X, E) \to \ldots $$ 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 $E$ whose terms $\mathcal{I}^n$ are injective objects of $\textit{Mod}(\mathcal{O}_X)$, see Injectives, Theorem 19.12.6. In the proof of Lemma 66.10.2 we found a short exact sequence of complexes $$ 0 \to \mathcal{I}^\bullet \to j_{U, *}\mathcal{I}^\bullet|_U \oplus j_{V, *}\mathcal{I}^\bullet|_V \to j_{U \times_X V, *}\mathcal{I}^\bullet|_{U \times_X V} \to 0 $$ Since $H^1(X, \mathcal{I}^n) = 0$, we see that taking global sections gives an exact sequence of complexes $$ 0 \to \Gamma(X, \mathcal{I}^\bullet) \to \Gamma(U, \mathcal{I}^\bullet) \oplus \Gamma(V, \mathcal{I}^\bullet) \to \Gamma(U \times_X V, \mathcal{I}^\bullet) \to 0 $$ Since these complexes represent $R\Gamma(X, E)$, $R\Gamma(U, E)$, $R\Gamma(V, E)$, and $R\Gamma(U \times_X V, E)$ we get a distinguished triangle by Derived Categories, Section 13.12 and especially Lemma 13.12.1. $\square$

    The code snippet corresponding to this tag is a part of the file spaces-perfect.tex and is located in lines 1817–1836 (see updates for more information).

    \begin{lemma}
    \label{lemma-unbounded-mayer-vietoris}
    Let $S$ be a scheme. Let $(U \subset X, V \to X)$ be an elementary
    distinguished square of algebraic spaces over $S$. For an object $E$
    of $D(\mathcal{O}_X)$ we have a distinguished triangle
    $$
    R\Gamma(X, E) \to R\Gamma(U, E) \oplus R\Gamma(V, E) \to
    R\Gamma(U \times_X V, E) \to R\Gamma(X, E)[1]
    $$
    and in particular a long exact cohomology sequence
    $$
    \ldots \to
    H^n(X, E) \to
    H^n(U, E) \oplus H^n(V, E) \to
    H^n(U \times_X V, E) \to
    H^{n + 1}(X, E) \to \ldots
    $$
    The construction of the distinguished triangle and the
    long exact sequence is functorial in $E$.
    \end{lemma}
    
    \begin{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
    \ref{injectives-theorem-K-injective-embedding-grothendieck}.
    In the proof of Lemma \ref{lemma-exact-sequence-j-star}
    we found a short exact sequence
    of complexes
    $$
    0 \to \mathcal{I}^\bullet \to
    j_{U, *}\mathcal{I}^\bullet|_U \oplus j_{V, *}\mathcal{I}^\bullet|_V \to
    j_{U \times_X V, *}\mathcal{I}^\bullet|_{U \times_X V} \to 0
    $$
    Since $H^1(X, \mathcal{I}^n) = 0$, we see that
    taking global sections gives an exact sequence of complexes
    $$
    0 \to \Gamma(X, \mathcal{I}^\bullet) \to
    \Gamma(U, \mathcal{I}^\bullet) \oplus
    \Gamma(V, \mathcal{I}^\bullet) \to
    \Gamma(U \times_X V, \mathcal{I}^\bullet) \to 0
    $$
    Since these complexes represent
    $R\Gamma(X, E)$, $R\Gamma(U, E)$, $R\Gamma(V, E)$, and
    $R\Gamma(U \times_X V, E)$ we 
    get a distinguished triangle by
    Derived Categories, Section
    \ref{derived-section-canonical-delta-functor} and especially
    Lemma \ref{derived-lemma-derived-canonical-delta-functor}.
    \end{proof}

    Comments (2)

    Comment #2792 by Pieter Belmans (site) on August 31, 2017 a 9:28 am UTC

    There shouldn't be a 0 in the long exact cohomology sequence, rather an $n$.

    Comment #2897 by Johan (site) on October 7, 2017 a 3:19 pm UTC

    THanks. Fixed here.

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