The Stacks project

21.26 Mayer-Vietoris

For the usual statement and proof of Mayer-Vietoris, please see Cohomology, Section 20.8.

Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Consider a commutative diagram

\[ \xymatrix{ E \ar[d] \ar[r] & Y \ar[d] \\ Z \ar[r] & X } \]

in the category $\mathcal{C}$. In this situation, given an object $K$ of $D(\mathcal{O})$ we get what looks like the beginning of a distinguished triangle

\[ R\Gamma (X, K) \to R\Gamma (Z, K) \oplus R\Gamma (Y, K) \to R\Gamma (E, K) \]

In the following lemma we make this more precise.

Lemma 21.26.1. In the situation above, choose a K-injective complex $\mathcal{I}^\bullet $ of $\mathcal{O}$-modules representing $K$. Using $-1$ times the canonical map for one of the four arrows we get maps of complexes

\[ \mathcal{I}^\bullet (X) \xrightarrow {\alpha } \mathcal{I}^\bullet (Z) \oplus \mathcal{I}^\bullet (Y) \xrightarrow {\beta } \mathcal{I}^\bullet (E) \]

with $\beta \circ \alpha = 0$. Thus a canonical map

\[ c^ K_{X, Z, Y, E} : \mathcal{I}^\bullet (X) \longrightarrow C(\beta )^\bullet [-1] \]

This map is canonical in the sense that a different choice of K-injective complex representing $K$ determines an isomorphic arrow in the derived category of abelian groups. If $c^ K_{X, Z, Y, E}$ is an isomorphism, then using its inverse we obtain a canonical distinguished triangle

\[ R\Gamma (X, K) \to R\Gamma (Z, K) \oplus R\Gamma (Y, K) \to R\Gamma (E, K) \to R\Gamma (X, K)[1] \]

All of these constructions are functorial in $K$.

Proof. This lemma proves itself. For example, if $\mathcal{J}^\bullet $ is a second K-injective complex representing $K$, then we can choose a quasi-isomorphism $\mathcal{I}^\bullet \to \mathcal{J}^\bullet $ which determines quasi-isomorphisms between all the complexes in sight. Details omitted. For the construction of cones and the relationship with distinguished triangles see Derived Categories, Sections 13.9 and 13.10. $\square$

Lemma 21.26.2. In the situation above, let $K_1 \to K_2 \to K_3 \to K_1[1]$ be a distinguished triangle in $D(\mathcal{O})$. If $c^{K_ i}_{X, Z, Y, E}$ is a quasi-isomorphism for two $i$ out of $\{ 1, 2, 3\} $, then it is a quasi-isomorphism for the third $i$.

Proof. By rotating the triangle we may assume $c^{K_1}_{X, Z, Y, E}$ and $c^{K_2}_{X, Z, Y, E}$ are quasi-isomorphisms. Choose a map $f : \mathcal{I}^\bullet _1 \to \mathcal{I}^\bullet _2$ of K-injective complexes of $\mathcal{O}$-modules representing $K_1 \to K_2$. Then $K_3$ is represented by the K-injective complex $C(f)^\bullet $, see Derived Categories, Lemma 13.31.3. Then the morphism $c^{K_3}_{X, Z, Y, E}$ is an isomorphism as it is the third leg in a map of distinguished triangles in $K(\textit{Ab})$ whose other two legs are quasi-isomorphisms. Some details omitted; use Derived Categories, Lemma 13.4.3. $\square$

Let us give a criterion for when this does produce a distinguished triangle.

Lemma 21.26.3. In the situation above assume

  1. $h_ X^\# = h_ Y^\# \amalg _{h_ E^\# } h_ Z^\# $, and

  2. $h_ E^\# \to h_ Y^\# $ is injective.

Then the construction of Lemma 21.26.1 produces a distinguished triangle

\[ R\Gamma (X, K) \to R\Gamma (Z, K) \oplus R\Gamma (Y, K) \to R\Gamma (E, K) \to R\Gamma (X, K)[1] \]

functorial for $K$ in $D(\mathcal{C})$.

Proof. We can represent $K$ by a K-injective complex whose terms are injective abelian sheaves, see Section 21.19. Thus it suffices to show: if $\mathcal{I}$ is an injective abelian sheaf, then

\[ 0 \to \mathcal{I}(X) \to \mathcal{I}(Z) \oplus \mathcal{I}(Y) \to \mathcal{I}(E) \to 0 \]

is a short exact sequence. The first arrow is injective because by condition (1) the map $h_ Y \amalg h_ Z \to h_ X$ becomes surjective after sheafification, which means that $\{ Y \to X, Z \to X\} $ can be refined by a covering of $X$. The last arrow is surjective because $\mathcal{I}(Y) \to \mathcal{I}(E)$ is surjective. Namely, we have $\mathcal{I}(E) = \mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}_ E^\# , \mathcal{I})$, $\mathcal{I}(Y) = \mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}_ Y^\# , \mathcal{I})$, the map $\mathbf{Z}_ E^\# \to \mathbf{Z}_ Y^\# $ is injective by (2), and $\mathcal{I}$ is an injective abelian sheaf. Please compare with Modules on Sites, Section 18.5. Finally, suppose we have $s \in \mathcal{I}(Y)$ and $t \in \mathcal{F}(Z)$ mapping to the same element of $\mathcal{I}(E)$. Then $s$ and $t$ define a map

\[ s \amalg t : h_ Y^\# \amalg h_ Z^\# \longrightarrow \mathcal{I} \]

which by assumption factors through $h_ Y^\# \amalg _{h_ E^\# } h_ Z^\# $. Thus by assumption (1) we obtain a unique map $h_ X^\# \to \mathcal{I}$ which corresponds to an element of $\mathcal{I}(X)$ restricting to $s$ on $Y$ and $t$ on $Z$. $\square$

Lemma 21.26.4. Let $\mathcal{C}$ be a site. Consider a commutative diagram

\[ \xymatrix{ \mathcal{D} \ar[r] \ar[d] & \mathcal{F} \ar[d] \\ \mathcal{E} \ar[r] & \mathcal{G} } \]

of presheaves of sets on $\mathcal{C}$ and assume that

  1. $\mathcal{G}^\# = \mathcal{E}^\# \amalg _{\mathcal{D}^\# } \mathcal{F}^\# $, and

  2. $\mathcal{D}^\# \to \mathcal{F}^\# $ is injective.

Then there is a canonical distinguished triangle

\[ R\Gamma (\mathcal{G}, K) \to R\Gamma (\mathcal{E}, K) \oplus R\Gamma (\mathcal{F}, K) \to R\Gamma (\mathcal{D}, K) \to R\Gamma (\mathcal{G}, K)[1] \]

functorial in $K \in D(\mathcal{C})$ where $R\Gamma (\mathcal{G}, -)$ is the cohomology discussed in Section 21.13.

Proof. Since sheafification is exact and since $R\Gamma (\mathcal{G}, -) = R\Gamma (\mathcal{G}^\# , -)$ we may assume $\mathcal{D}, \mathcal{E}, \mathcal{F}, \mathcal{G}$ are sheaves of sets. Moreover, the cohomology $R\Gamma (\mathcal{G}, -)$ only depends on the topos, not on the underlying site. Hence by Sites, Lemma 7.29.5 we may replace $\mathcal{C}$ by a “larger” site with a subcanonical topology such that $\mathcal{G} = h_ X$, $\mathcal{F} = h_ Y$, $\mathcal{E} = h_ Z$, and $\mathcal{D} = h_ E$ for some objects $X, Y, Z, E$ of $\mathcal{C}$. In this case the result follows from Lemma 21.26.3. $\square$

Comments (0)

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EVX. Beware of the difference between the letter 'O' and the digit '0'.