# Mayer-Vietoris sequence

Let $X$ is a topological space^{}, and $A,B\subset X$ are such that $X=\mathrm{int}(A)\cup \mathrm{int}(B)$, and $C=A\cap B$. Then there is an exact sequence^{} of homology groups:

$$\begin{array}{ccccccccccc}\hfill \mathrm{\cdots}\hfill & \hfill \to \hfill & \hfill {H}_{n}(C)\hfill & \hfill \stackrel{{i}_{*}\oplus -{j}_{*}}{\to}\hfill & \hfill {H}_{n}(A)\oplus {H}_{n}(B)\hfill & \hfill \stackrel{{j}_{*}+{i}_{*}}{\to}\hfill & \hfill {H}_{n}(X)\hfill & \hfill \stackrel{{\partial}_{*}}{\to}\hfill & \hfill {H}_{n-1}(C)\hfill & \hfill \to \hfill & \hfill \mathrm{\cdots}\hfill \end{array}$$ |

Here, ${i}_{*}$ is induced by the inclusions $i:B\hookrightarrow X$ and ${j}_{*}$ by $j:A\hookrightarrow X$, and ${\partial}_{*}$ is the following map:
if $x$ is in ${H}_{n}(X)$, then it can be written as the sum of a chain in $A$ and one in $B$, $x=a+b$.
$\partial a=-\partial b$, since $\partial x=0$. Thus, $\partial a$ is a chain in $C$, and so represents
a class in ${H}_{n-1}(C)$. This is ${\partial}_{*}x$. One can easily check (by standard diagram chasing) that this map is well defined on the level of homology^{}.

Title | Mayer-Vietoris sequence |
---|---|

Canonical name | MayerVietorisSequence |

Date of creation | 2013-03-22 13:14:52 |

Last modified on | 2013-03-22 13:14:52 |

Owner | bwebste (988) |

Last modified by | bwebste (988) |

Numerical id | 6 |

Author | bwebste (988) |

Entry type | Definition |

Classification | msc 55N10 |