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Recently, I've been wondering to what extent certain types of pathologies can arise in finite CW complexes -- notice that I do not want to assume that I'm in the PL category or that the CW complexes are regular, but do want to insist on there only being finitely many cells.

Specific question: is there a finite CW complex homeomorphic to a sphere such that one of its maximal cells has as its closure a ball whose boundary is embedded in the CW-sphere as an Alexander horned sphere?

Follow-up question: if the answer is no, is there a finite CW complex homeomorphic to a sphere such that the closure of one of the maximal cells is a ball, but the closure of its complement is not a ball?

I don't have a specific reason I need to know this, but recently have been working a lot with finite CW complexes that are not in the PL category, or at least not obviously so, and would like to understand better the interaction between the finiteness of the number of open cells and the potential weirdness of the attaching maps. Information about other pathologies that can or cannot occur in this setting would also be much appreciated.

Thanks in advance for your help with this!

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I'm not entirely sure I understand your first question. What exactly does it mean for the closure of a cell to be a ball "with the Alexander horned sphere as its boundary"? The Alexander horned sphere is topologically a sphere -- it's only the embedding of the sphere in $\mathbb{R}^3$ that makes it special. Is the given CW complex embedded in $\mathbb{R}^3$? –  Jim Belk Jun 4 '12 at 3:26
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One point which may be relevant is that the fundamental group of the complement of the Alexander horned ball is not finitely generated. Therefore, the complement of an Alexander horned ball is not homotopy equivalent to a finite CW complex. –  Jim Belk Jun 4 '12 at 3:31
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It was pointed out to me by a topologist I consulted before posting the question that while the complement of the Alexander horned ball has infinitely generated fundamental group, the closure of the complement does not (since the Alexander horned sphere itself is a sphere, hence has trivial fundamental group). –  Patricia Hersh Jun 4 '12 at 3:46
    
Jim, thanks for your comments. I've rephrased the first question to speak in terms of the embedding. –  Patricia Hersh Jun 4 '12 at 5:00
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The closure of the complement of the Alexander horned ball is even contractible, because it is a deformation retract of $S^3$ minus a point in the interior of the horned ball. Also, Bing proved that if you double the closure of the complement of the horned ball, doubling it along the horned sphere, the resulting topological space is homeomorphic to $S^3$. All of this makes Sergey's answer to 1) even more interesting to me. –  Lee Mosher Jun 4 '12 at 15:13
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1 Answer 1

up vote 17 down vote accepted

Specific question: is there a finite CW complex homeomorphic to a sphere such that one of its maximal cells has as its closure a ball whose boundary is embedded in the CW-sphere as an Alexander horned sphere?

1) No. If $K$ is the $2$-skeleton of the CW complex, then $K$ has a mapping cylinder neighborhood in $S^3$, and hence by Nicholson's theorem it is tame, that is, equivalent to a subpolyhedron of $S^3$ by a homeomorphism $h$ of $S^3$. Since $K$ is $2$-dimensional, it is not hard to show that $h$ also takes any $2$-sphere in $K$ onto a subpolyhedron of $S^3$. So $K$ cannot contain the horned sphere.

2) Yes, if you allow an unspecified wild codimension one sphere in place of the Alexander horned sphere. By Example 7.11.2 on p.419 in the Daverman-Venema book (this seems to be among a few original results in the book, perhaps the most important one), $S^n$ for $n\ge 6$ contains a wildly embedded sphere $\Sigma$ with a mapping cylinder neighborhood $N$. By construction, the complement to the interior of $N$ consists of two closed $n$-balls. It follows that the closures of the complementary domains of $\Sigma$ are the mapping cones of some self-maps of $S^{n-1}$. So we get a CW-complex with one $0$-cell, one $(n-1)$-cell and two $n$-cells, which is homeomorphic to $S^n$, and has a wild $(n-1)$-skeleton.

Follow-up question: if the answer is no, is there a finite CW complex homeomorphic to a sphere such that the closure of one of the maximal cells is a ball, but the closure of its complement is not a ball?

1) Yes. You can glue one of the complementary domains of $\Sigma$ and an $n$-ball along their boundary sphere. The result is again $S^n$, according to Proposition 7.10.1 in Daverman-Venema.

2) I should also mention a simpler but somewhat similar example. Using the Edwards-Cannon theorem, it is not hard to construct a finite regular CW-complex $K$ that is homeomorphic to $S^5$, even though the boundary of some $2$-cell (in fact, of each $2$-cell) of $K$ is wild, viewed as a copy of $S^1$ in $S^5$. In more detail, if $H$ is a traingulation of a non-simply-connected homology $3$-sphere, then the double suspension $S^0*S^0*H$ is a simplicial complex homeomorphic to $S^5$; the desired CW-complex is the 'prejoin' $(S^0*S^0)+H$, which is PL homeomorphic to $S^0*S^0*H$ and has all its $2$-cells attached to the suspension circle $S^0*S^0$. On identifying regular CW-complexes with their posets of nonempty faces, the prejoin $P+Q$ of two posets is defined by placing all the elements of $P$ below all the elements of $Q$ in the Hasse diagram, and keeping the original order within $P$ and within $Q$. The order complex of $P+Q$ is easily seen to be isomorphic to the join of the order complexes of $P$ and of $Q$. As an example, $S^0*S^0*pt$ is a simplicial complex with $4$ of $2$-simplices, whereas $(S^0*S^0)+pt$ is a cell complex with only one (quadrilateral) $2$-cell.

Beware that $K$ itself is a PL CW-complex, with PL attaching maps; the only trouble is with the homeomorphism between $K$ and $S^5$.

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This is great. I did not know about Nicholson's Theorem. –  Lee Mosher Jun 4 '12 at 14:43
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It seems worth adding to the Cannon-Edwards example that the complement of the circle in $S^1 * H$ is $R^2 \times H$, thus not simply connected, and this is how to see that the circle is wild. Alternatively, the link of the circle is $H$. –  Ben Wieland Jun 4 '12 at 15:59
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Sergey, thank you for taking the time to write this very helpful answer! I also want to thank Tyler Lawson who helped me offline before I posted the question -- pointing out that the complement of the solid Alexander horned sphere has infinitely generated fundamental group and isn't homotopy equivalent to a finite CW complex whereas its closure only has finitely generated fundamental group. –  Patricia Hersh Jun 5 '12 at 1:08
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