Suppose \(X\) is a space which can be written as a union of non-empty open sets \(A_1,\dots,A_n\) so that for each \(1\le k\le n\), the intersection of any \(k\) of the sets is either empty or contractible. Show that \(\widetilde H_i(X)=0\) for \(i\ge n-1\), and give an example showing that this inequality is sharp.
Solution
1. For the finite open cover \(\{A_1,\dots,A_n\}\) there is a Mayer–Vietoris spectral sequence\[ E^1_{p,q}=\bigoplus_{i_0<\cdots<i_p}H_q(A_{i_0}\cap\cdots\cap A_{i_p})\Longrightarrow H_{p+q}(X). \]
This is the standard iterated Mayer–Vietoris (Čech-to-homology) spectral sequence of a finite open cover, obtained from the double complex whose horizontal direction is the alternating Čech differential and whose vertical direction is the singular boundary.
2. By hypothesis every nonempty intersection is contractible, so after passing to reduced homology the spectral sequence is concentrated in row \(q=0\) and identifies \(\widetilde H_*(X)\) with the reduced simplicial homology of the nerve \(N\) of the cover.
A nonempty contractible intersection has \(H_q=0\) for \(q>0\) and \(H_0\cong\mathbb Z\). The horizontal \(q=0\) complex is precisely the simplicial chain complex of the nerve. Empty intersections contribute no summand.
3. A simplicial complex on \(n\) vertices has no reduced homology in degrees \(i\ge n-1\).
If the full \((n-1)\)-simplex is absent, the nerve has dimension at most \(n-2\). If it is present, then every subset of the \(n\) vertices is present and the nerve is the full simplex, hence contractible. In either case \(\widetilde H_i(N)=0\) for \(i\ge n-1\).
4. Consequently\[ \boxed{\widetilde H_i(X)=0\quad\text{for all }i\ge n-1.} \]
Combine 2 and 3.
5. The bound is sharp: there are good covers of \(S^{n-2}\) by \(n\) open sets whose nerve is the boundary of an \((n-1)\)-simplex.
Take the standard cover obtained from sufficiently small open stars of the vertices in the barycentric subdivision of the boundary of an \((n-1)\)-simplex. Every nonempty finite intersection is contractible, the total intersection of all \(n\) sets is empty, and the nerve is the boundary sphere \(S^{n-2}\). Thus \(\widetilde H_{n-2}(X)\cong\mathbb Z\ne0\), so the vanishing threshold cannot be lowered from \(n-1\) to \(n-2\).