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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?

Motivation/Background

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.

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1 Answer 1

up vote 12 down vote accepted

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.

EDIT:

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.

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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.) –  Martin Tancer Jun 8 '11 at 10:37

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