Mayer-Vietoris homology sequence

This article defines a long exact sequence of homology groups, for topological spaces or pairs of topological spaces

This fact is related to: excisive triads

Definition

Suppose $X$ is a topological space, and $U$ and $V$ are subsets of $X$ such that the union of the interiors of $U$ and $V$ cover $X$. Then we get a long exact sequence of homology:

$\ldots \to H_{n}(U\cap V)\to H_{n}(U)\oplus H_{n}(V)\to H_{n}(X)\to H_{n-1}(U\cap V)\to \ldots$

where the maps are as follows. Let $i,j$ be the inclusions from $U\cap V$ to $U$ and $k,l$ be the inclusions from $U,V$ into $X$.

Then the map from the homology of $U\cap V$ is:

$f\mapsto (H_{n}(i)(f),H_{n}(j)(f))$

And the map from $H_{n}(U)\oplus H_{n}(V)$ is:

$(g,h)\mapsto H_{n}(k)g-H_{n}(l)h$

Interpretation in different homology theories

For singular homology

For singular homology, the Mayer-Vietoris homology sequence can be viewed as the long exact sequence of homology of a short exact sequence of chain complexes, namely:

$sd^{m}C_{n}(U\cap V)\to sd^{m}C_{n}(U)\oplus sd^{m}C_{n}(V)\to sd^{m}(C_{n}(X)$

where $sd$ denotes the barycentric subdivision operator. Since $sd$ is homotopic to the identity map, the homologies of this are the homologies of the original chain complexes.

The rough idea is that by subdividing sufficiently, we can make sure that each simplex goes either entirely within $U$ or entirely within $V$.