Mayer–Vietoris sequence

The Mayer–Vietoris sequence is a long exact homology sequence for a space decomposed into two suitable subspaces.

Identifying two distinct points of $S^2$ produces a space with homology groups $H_0=H_1=H_2=\mathbb{Z}$ and all higher groups zero. Cover the identified point by the image of two small disks, a contractible wedge of disks, and take as the other open set the complement of the point, a cylinder. Their intersection is two annuli. In the Mayer–Vietoris sequence, $H_1$ of this intersection maps onto $H_1$ of the cylinder with a rank-one kernel, producing $H_2=\mathbb{Z}$. The kernel at $H_0$ has rank one and produces $H_1=\mathbb{Z}$.

Attach two disks to a simply connected space $A$ along disjoint null-homotopic boundary circles, and additionally identify one interior point of each disk. Assume the attachments are CW complexes or admit compatible triangulations. The extra interior identification contributes a copy of $\mathbb{Z}$ to the first homology group, in addition to the two new second-homology classes. Indeed, after the first attachment $Y=A\cup D_1$, the intersection with $D_2$ has two components, $S^1\bigsqcup \{p\}$. In the Mayer–Vietoris sequence, the map $H_0(S^1\bigsqcup \{p\})\to H_0(Y)\oplus H_0(D_2)$ sends $(m,n)$ to $(m+n,-m-n)$, whose kernel is $\mathbb{Z}$. If $H_1(A)=0$, exactness gives $H_1(Y\cup D_2)=\mathbb{Z}$. The degree-two sequence is $0\to H_2(Y)\to H_2(Y\cup D_2)\to \mathbb{Z}\to 0$, which splits because $\mathbb{Z}$ is free. For $A=S^3$, the integral homology groups are $\mathbb{Z},\mathbb{Z},{\mathbb{Z}}^2,\mathbb{Z}$ in degrees zero through three.

If a space is a union of $n\ge 2$ subcomplexes and every nonempty intersection is acyclic, its homology vanishes in degrees at least $n-1$. Induction with the Mayer–Vietoris sequence proves the bound; the boundary of an $(n-1)$-simplex covered by its facets shows sharpness.