Let $d$ be an integer. Let $A,B \subseteq \mathbb R^d$ be two sets homeomorphic to an open $d$-ball such that their intersection is again homeomorphic to an open $d$-ball. Does it follow that their union is homeomorphic to an open $d$-ball?


The motivation comes from nerve type theorems. For instance it follows in the above-mentioned case that $A \cup B$ has to be nullhomotopic.

In more general setting, let $A_1, \dots, A_n \subseteq \mathbb R^d$ be a collection of sets homeomorphic to an open $d$-ball such that the intersection of every subcollection is either empty or homeomorphic to an open $d$-ball. Moreover, let us assume that it is determined which subcollections are supposed to have a nonempty intersection. This data can be "stored" as a simplicial $K$ with vertices $A_1, \dots, A_n$ and whose faces are exactly that subcollections which have a nonempty intersection.

By standard nerve theorems, $K$ has to be homotopy equivalent to $A_1 \cup \cdots \cup A_k$; however I wonder whether anyone is aware of any "homeomorphism-type" nerve theorem:

Let $A_1, \dots, A_n \subseteq \mathbb R^d$ and $K$ be described as above. Does it follow that the homeomorphism type of $A_1 \cup \cdots \cup A_n$ is determined by $K$? Does it follow if $K$ is at most $d$-dimensional?

Even if the answer to the question above is negative, an interesting specific case occurs when $A_1, \dots, A_n$ are assumed to be convex. (Note that the answer to first question is positive in this case, since $A \cup B$ is star-convex.)

I came to these questions when I was considering a certain algorithmic result on the collections of sets as described above. With a coauthor, we finally circumvent these questions (it showed up to be more convenient). However, I would be still very curious about the answers.


No to the first question. You can make examples where the "fundamental group at infinity" of $A\cup B$ is nontrivial.

Start with a finite complex $X$ that has trivial homology but nontrivial fundamental group. Embed the suspension $\Sigma X$ in $S^d=\mathbb R^d\cup\infty$. The suspension is contractible and is the union of two cones, also contractible. Let $A$ and $B$ be the complements of the two cones. If $d$ is big enough then $A$, $B$, and $A\cap B$ (the complement of $\Sigma X$) can be shown to be diffeomorphic to $\mathbb R^d$ by the $h$-cobordism theorem. But $A\cup B$, the complement of $X$, is such that the complement of a compact set in it is never simply connected.


To be a bit more precise, the open set $A$ (or $B$ or $A\cap B$) will be simply connected if the codimension of its complement in $S^d$ is at least $3$. (This is clear at least if the complement is embedded nicely enough.) And $A$ will have trivial homology, by Alexander duality, so it will be contractible. Now, Siebenmann's thesis gives sufficient conditions on a noncompact smooth manifold (of not too small dimension) for there to be a compact manifold with boundary having that as its interior. If there are arbitrarily small simply connected neighborhoods of infinity then these conditions are satisfied. Using this you see that $A$ is the interior of a contractible manifold with simply connected boundary. The $h$-cobordism theorem implies that that compact thing is a closed disk, so that $A$ is an open disk.

Or you can take a slightly different approach and avoid Siebenmann's thesis. Embed that suspension of $X$ piecewise linearly, make a regular neighborhood that is the union of regular neighborhoods of the two cones intersecting in a regular neighborhood of $X$. Use the complements of these compact neighborhoods.

Or, better yet, let $Q$ be a compact codimension zero acyclic non simply connected manifold (with boundary) in the interior of $D^d$. The complement in $S^{d+1}=(\mathbb R^d\times \mathbb R)\cup\infty$ of $D^d\times [0,1]\cup Q\times [1,2]$ is an open ball, as is the complement of $Q\times [1,2]\cup D^d\times [2,3]$, as is the complement of their union. But the complement of the intersection is not.

  • $\begingroup$ Thank you! (At the moment, it is not clear to me how $h$-cobordism theorem implies the claims about $A$, $B$ and $A \cap B$, since I am not very familiar with $h$-cobordism theorem. However, I beleive, I will be able to figure it out and your construction seems very beleivable.) $\endgroup$ – Martin Tancer Jun 8 '11 at 10:37

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.