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A topological manifold (without boundary) is a locally euclidean Hausdorff space, which is second countable (or maybe only paracompact, that is to say make the definition as strict as you like).

A good open cover of a space $X$ is an open cover $(U_i)_i$, where every finite intersection $U_{i_1} \cap ... \cap U_{i_k}$ is either empty or contractible.

I define a sufficient open cover to be a cover, where every finite intersection is either empty or a disjoint union of contractible spaces.

Does every topological $n$-manifold admit a good open cover or a sufficient open cover?

All the information I could find about good covers of manifolds involved smooth or differentiable manifolds. On ncatlab it is only stated that the methods involved for proving the existence of good covers in the smooth case don't work in the topological case. No reference to a counterexample or other disprove is given.

Note that I am not really (just sort of) interested in the cover being finite. As explained in this answer every $n$-manifold admits a cover by $n+1$ charts. However there is no reason, why this cover should possibly be a good cover. The closest reference for finite good covers is (again according to ncatlab) this paper by Osborn Stern. However the connectivity conditions confuse me and I don't quite understand how they replace the contractibility condition.

My goal is to show in an axiomatic way (see my question here) that for every closed $n$-manifold the homology groups of order $>n$ vanish. I think the existence of a (possibly infinite) sufficient cover would open a way for showing this by induction on the (by compactness) finite number of sufficient opens involved. Of cause, the general existence of a finite sufficient cover would (if my inductive argument is correct) imply that for arbitrary topological $n$-manifolds the higher homology groups vanish...

I would not be surprised, if there is a good reason, why there is no good or sufficient cover in general. However, as I am fairly new to the subject of topological manifolds, I remain hopeful that there is one.

Thank you for your time and efforts.

Jonas Linssen
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    This appears to be an open problem, https://mathoverflow.net/questions/165850/good-covers-of-manifolds – Moishe Kohan Feb 20 '20 at 01:38
  • As for "sufficient cover", every set is the disjoint union of singletons, each of which is contractible. You should rethink your definition. – Moishe Kohan Feb 20 '20 at 02:11
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    @MoisheKohan Topological space are in general not given by the disjoint union of singleton spaces corresponding to their points, as this would make the space discrete... – Jonas Linssen Feb 20 '20 at 07:22
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    @MoisheKohan The "disjoint union" of topological spaces is usually understood to mean the coproduct of those spaces. For example, if someone says that $A = B \amalg C$ for topological spaces $A$, $B$, and $C$, they usually mean that $A$ is the set-theoretic disjoint union of $B$ and $C$ and for all $U \subseteq A$, $U$ is open iff $U \cap B$ and $U \cap C$ are open in $B$ and $C$, respectively. – diracdeltafunk Feb 20 '20 at 07:41

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