The Nerve Theorem (see nLab;) asserts that given a finite collection $\cal K$ of compact sets with the property that all non empty intersections of sets in the family are homotopically trivial, then the union $X$ of the sets in the family is homotopically equivalent to the nerve $N({\cal K})$. Here the nerve of the family is the abstract simplicial complex that records the intersection pattern so if ${\cal K}=\{K_1,K_2\dots,K_n\}$ then $N({\cal K})=\{S \subset \{1,2,\dots,n\}: \cap\{K_i : i \in S\} \ne \emptyset$. The nerve theorem is attributed to Borsuk and Leray (and others). There are various variations. For example if you require that all non empty intersections are homologically trivial then the homology groups of the nerve are the same as those of $X$.

The desired extension

Suppose that all non empty intersections of sets in $\cal K$ are homotopically equivalent to a single space $Z$. Under what conditions can you conclude that $X$ is homologically the same as the product of Z and the nerve.

This is usually not the case, so the question is to find conditions (as simple as possible) that will imply it.

Perhaps the kind of condition you would need is about sets in K intersecting "nicely". For example, we can define two sets A and B as "nicely intersecting" if the usual long exact Meyer-Vietoris sequence splits into short exact sequences. So you can require that every intersection of some sets in $\cal K$ intersect "nicely" every intersection of some other sets in $\cal K$. I don't have reason to believe that this is the definition required for the conclusion but it just illustrates what type of conditions are possible.

A possible application is indicated in this question.

Edit: Oscar Randal-Williams' answer shows that my suggestion for "nice" intersection is not sufficient for the desired conclusion. Perhaps the most interesting question in view of Oscar's result is this.

Suppose that you have a family $\cal > K$ of compact subsets of $latex R^d$ so that all non empty intersections of members of $\cal K$ are homotopically equivalent to to the same space $Z$ and every intersection of some sets in $\cal K$ intersect "nicely" every intersection of some other sets in $\cal K$, where "nice intersection" means that the Meyer-Vietoris long exact sequence splits. Is it true that the nerve of $\cal K$ has vanishing $i$th homology groups for $i>d$?

I think that a positive answer would suffice for the Helly-type application.


The condition you suggest is insufficient. For example, let $F \to E \overset{\pi}\to B$ be a fibre bundle, with compact fibre say, (or more generally a local quasi-fibration), and $\mathcal{K}_B$ be a cover of $B$ satisfying the conditions of the Nerve theorem (i.e. all intersections are empty or contractible). Then $$\mathcal{K} := \{ \pi^{-1}(K) \subset E \vert K \in \mathcal{K}_B \}$$ is a cover of $E$ satisfying your Mayer--Vietoris condition, and it is fairly clear that $$N(\mathcal{K}) = N(\mathcal{K}_B) \simeq B,$$ and $E$ will not typically be equivalent to $N(\mathcal{K}) \times F$ if the original bundle is not trivial.

So, to answer a different question: given a cover $\mathcal{K}$ of a space $X$ where all non-empty intersections are equivalent to $Z$, I think there will typically be a homotopy fibre seqeuence $$Z \to X \to N(\mathcal{K}).$$ Whether or not this is trivial is then another problem. By the example above, every local quasi-fibration arises is this way, so there is unlikely to be an especially simple description of when it is possible.

One condition that will give what you want is the following: if you can find a map $f : X \to Z$ such that for each non-empty intersection $K_\alpha = \cap_{i \in \alpha} K_i$ the map $f\vert_{K_\alpha}: K_\alpha \to Z$ is a homotopy equivalence, then $X \simeq Z \times N(\mathcal{K})$. This is probably much too strong for most applications through.

| cite | improve this answer | |
  • $\begingroup$ Dear Oscar, thanks. I will add a possible application. $\endgroup$ – Gil Kalai Oct 19 '11 at 17:05
  • $\begingroup$ Dear Oscar, Is it possible that under my Meyer Vietoris condition E will be equivalent to such an E for some bundle (not necessary trivial)? (Do you think this might be the case even without any M-V conditions)? $\endgroup$ – Gil Kalai Oct 24 '11 at 16:00
  • $\begingroup$ I should have been more careful. In my second paragraph, I should have said "given a cover $\mathcal{K}$ of a space $X$ such that all intersections are homotopy (or homology) equivalent to a space $Z$, and the inclusions between intersections are all homotopy equivalences, then there is a homotopy (or homology) fibre sequence..." In particular, this essentially forces the M-V condition to hold. $\endgroup$ – Oscar Randal-Williams Oct 25 '11 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.