It is well-known and easy to prove (see for instance this post) that every smooth manifold admits a good cover, i.e., a locally finite cover by open balls such that all nonempty intersections of the elements of this cover are still homeomorphic to open balls. The same is true, more generally, for PL manifolds.

Question: Do all topological $n$-manifolds admit good covers?

I am especially curious about the case $n=4$.

Here is my guess: a good cover is too close to a handle decomposition for gauge-theoretic obstructions to rule them out in dimension 4. In higher dimensions, I do not even have a guess.

Edit: The notion of a good cover has a homotopy analogue, a homotopy good cover where the requirement is that all elements of the covering and all their intersections are contractible. Here is a couple of observations about existence of homotopy good coverings:

  1. Every simplicial complex, of course, admits a homotopy good cover; in particular, all topological manifolds which admit triangulations (not necessarily PL) admit homotopy good coverings.

  2. Many (if not all) simply connected closed 4-manifolds admit homotopy good covers; this is a corollary of Freedman's work. Existence of homotopy good covers for all simply connected manifolds essentially hinges on the case of $*CP^2$.

  • 1
    $\begingroup$ May I ask why this question is tagged 'ag.algebraic-geometry'. Also, a further tag 'topological-manifolds' is appropriate here. $\endgroup$ May 11, 2014 at 17:48
  • $\begingroup$ @RicardoAndrade: That's because I clicked a wrong button. Corrected now. $\endgroup$
    – Misha
    May 11, 2014 at 17:52
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    $\begingroup$ @UrsSchreiber: The nLab is wrong on this one. Their proof assumes smoothess (the exitence of a Riemannian metric). $\endgroup$ May 11, 2014 at 22:07
  • $\begingroup$ Sorry, I shouldn't post late at night. $\endgroup$ May 11, 2014 at 22:25
  • $\begingroup$ Yes, I checked nLab before posting (as well as a number of other sources). $\endgroup$
    – Misha
    May 11, 2014 at 23:46

2 Answers 2


I am not aware of a general result regarding the existence of good covers (and would guess that the general answer is negative). However, if you are willing to make certain sacrifices in terms of additional hypothesis and weaker conclusions, then you may seek solace in the article

RP Osborne and JL Stern. Covering Manifolds with Cells, Pacific Journal of Mathematics, Vol 30, No. 1, 1969.

And here, for ready reference, is the main theorem.

Let $M$ be a $k$-connected topological $n$-manifold without boundary, and define $q = \min(k,n-3)$. $M$ admits a cover by $p$ open cells if $p(q+1) > n$ and moreover, one can arrange that all nonempty intersections of the covering cells are $(q-1)$-connected.

What's nice here is that one actually gets a bound on the size of the open cover, but of course the intersections need not be contractible and you require many homotopy groups of the original manifold to vanish.


A remark about the four-dimensional case: Every open topological $4$-manifold is smoothable (see Theorem 8.2 in Freedman/Quinn: Topology of $4$-manifolds) and thus admits a good cover. Hence every closed connected topological $4$-manifold $M$ admits a finite cover by open balls $B_1,\dots,B_k$ such that every nonempty intersection of $B_2,\dots,B_k$ is homeomorphic to an open ball: equip the complement of a point in $M$ with a good cover, cover the point with an additional ball $B_1$, then choose a finite subcover.

I have no idea whether one can arrange the intersections with $B_1$ to be nice somehow (e.g. connected). The boundary of $B_1$ is usually very rough with respect to any smooth structure on the complement in $M$ of a point in $B_1$. (The Osborne/Stern theorem cited above says that every connected simply connected closed topological $4$-manifold, e.g. $*CP^2$, admits a cover by open balls $B_1,B_2,B_3$ all of whose intersections are connected.)


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