# What does it mean that if the cover of $k$-fold intersections is not contractible it takes the form of a spectral sequence of cohomology?

I asked the following question in a previous post: Suppose a CW complex $M$ is given by the union of $n$-spheres, namely $M=\bigcup_{\alpha\in A}S^n$, without knowledge of intersections. The only requirement is that the union covers $M$. Let $\Sigma=\{S^n,\dots,S^n\}$ be a finite collection of sets, with cardinality $|A|$. Then how do we compute homology groups $H_k(M;\mathbb{Z})$ from the nerve $\text{Nrv}(\Sigma)$ for $\Sigma=\{S^n,\dots,S^n\}$ covering the CW complex $M$?

The answer given was that this is not enough to know the nerve. For an explicit counterexample consider $M_1$ and $M_2$ given as the union of two circles, where the circles in $M_1$ intersect twice and the circle in $M_2$ intersect four times.

The nerves are isomorphic but the homology groups are not isomorphic. When the covering is given by open balls, in order to be able to compute homology using the nerve you need to assume strong conditions on the intersections (they should be all contractible or empty) on your cover.

It was suggested that the condition we want is that the $k$-fold intersections, for $k$ sufficiently large, are all empty or contractible. Then the Čech complex of this cover (with constant sheaf $\mathbb{Z}$) recovers the homology of the manifold. The general case where $k$-fold intersections are not contractible instead takes the form of a spectral sequence involving cohomology of the various intersections.

Could this be made mathematically precise with appropriate formal notations?

Any help would be appreciated. Thanks in advance!

• This is explained in Bott and Tu, I think. You get a double complex, with one differential coming from the ifferential on forms, and the other from the simplicial differential on intersections. When your intersections are nice, then say the rows are acyclic, so the acyclic assembly lemma (this is the technical tool here), allows you to jump from Cech to de Rham. – Pedro Tamaroff Aug 26 '18 at 13:21
• Bott and Tu was mentionned in the previous thread. Did you look at it ? – Nicolas Hemelsoet Aug 27 '18 at 7:16