There is a situation that comes up regularly in algebraic topology when giving proofs of facts about manifolds, like Poincare duality and the like. The typical sequence goes like this:

  • Prove something for $\mathbb{R}^n$.
  • Then it follows for open disks.
  • Use a Mayer-Vietoris argument to prove it for finite unions of convex sets in $\mathbb{R}^n$.
  • Use a colimit argument to prove it for arbitrary open subsets of $\mathbb{R}^n$.
  • It then follows for open subsets of a manifold admitting a homeomorphism to an open subset of $\mathbb{R}^n$.
  • Use a Mayer-Vietoris argument to prove it for finite unions of such subsets of a manifold.
  • Use a colimit argument to prove it for arbitrary open subsets of a manifold.

Obviously there is some redundancy here, and it makes the technical details in these proofs overwhelm the underlying ideas. In the smooth category one can do better, but usually only by appealing to machinery which is useful but takes extra time to prove.

The answers/comments in the following question point towards a very useful way to compare coordinate charts in different affine covers of a given scheme:

What should be learned in a first serious schemes course?

Namely, the intersection of any two affine opens has a cover by open sets which are simultaneously distinguished in both.

I feel like I should know a reference to whether this is true in the topological category - and I suspect that it's not - but I shamefully don't know. One can phrase this in terms of continuous local homeomorphisms from $\mathbb{R}^n$ to itself, but I'll instead just ask:

Given two coordinate charts on a topological manifold M and a point in their intersection, is there a neighborhood of this point which is simultaneously a convex open set in both charts? Is there a simple counterexample?

  • 2
    $\begingroup$ I think this boils down to: is there a homeomorphism $U\to V$, where $U$ and $V$ are open in $\mathbb{R}^n$ that un-convexes all convex subsets? $\endgroup$
    – Jeff Strom
    Jul 12, 2010 at 17:30
  • $\begingroup$ Yes - but it only need do this to convex sets containing the origin. $\endgroup$ Jul 12, 2010 at 17:36
  • 3
    $\begingroup$ Surely you know this, but any $C^2$-smooth diffeomorphism $U\to V$ in $\mathbb{R}^n$ preserve convexity of small balls, and the proof is simple. $\endgroup$
    – BS.
    Jul 14, 2010 at 15:17

1 Answer 1


There are piecewise linear counterexamples in dimension $2$.

Arrange $2m$ evenly spaced rays $R_i$ around the origin, $m\ge 3$. If $C$ is a convex neighborhood of the origin, let $r_i$ be the reciprocal of the length of the portion of $R_i$ in $C$. For some number $K>0$ (depending on $m$), convexity implies $r_{i+1}+r_{i-1}\ge 2Kr_i$. Thus if $e(C)$ and $o(C)$ are the sums of $r_i$ over even and odd $i$ we have $e(C)\ge Ko(C)$ and $o(C)\ge Ke(C)$. Now apply a homeomorphism $h$ that linearly stretches even-numbered rays by $A>0$ and odd-numbered rays by $B>0$. If $h(C)$ is a convex set then you would have numbers $e(h(C))=e(C)/A$ and $o(h(C))=o(C)/B$, whence $e(C)/A\ge Ko(C)/B$, contradicting $o(C)\ge Ke(C)$ if $A/B$ is chosen big enough.

  • $\begingroup$ Perfect. That's a nice way to detect lack of convexity. $\endgroup$ Jul 12, 2010 at 18:54

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