Analyze what happens when the number of handles in the basic building block for the Alexander horned sphere is doubled, as in the figure at the right.
Solution
In the doubled building block shown in Hatcher’s figure, each old handle is replaced by four smaller handles, arranged as two linked pairs. The same limiting construction still gives a horned ball, but the free-group bonding maps change from one commutator to a product of two commutators.
1. After \(n\) stages there are \(4^n\) smallest handles, and the complement \(Y_n\) of the \(n\)th-stage compact set has\[ \pi_1(Y_n)\cong F_{4^n}. \]
For \(n=0\) the initial compact set is a standard solid torus, so its complement has fundamental group \(F_1\cong\mathbb Z\). At each replacement, one handle is replaced by four disjoint small handles. The new complement piece is a ball with four tubes removed, whose fundamental group is free on four meridians. Repeating independently for each of the \(4^n\) handles gives a free group on the meridians of the \(4^n\) smallest handles.
2. For one doubled block, if \(a_1,a_2,a_3,a_4\) are meridians of the four new handles and \(a\) is the old meridian, then the inclusion sends\[ a\longmapsto [a_1,a_2][a_3,a_4]. \]
The attaching annulus of the old complement bounds, in the closure of the new complement piece, a genus-two surface with one boundary component. Choose the standard symplectic generators \(a_1,a_2,a_3,a_4\) for this punctured genus-two surface, represented by meridians of the two linked pairs. The oriented boundary of a genus-two surface represents the word \[ [a_1,a_2][a_3,a_4] \] in its free fundamental group. This boundary is precisely the old meridian \(a\).
3. Every bonding map\[ \pi_1(Y_n)\longrightarrow\pi_1(Y_{n+1}) \]is injective.
The group \(F_{4^{n+1}}\) splits as a free product of \(4^n\) free groups of rank four, one for each old handle. On each cyclic free factor generated by an old meridian, the map sends the generator to the nontrivial reduced word \[ [a_1,a_2][a_3,a_4] \] in the corresponding rank-four free factor. A homomorphism \(\mathbb Z\to F_4\) sending \(1\) to a nontrivial element is injective, and the free product of these injections is injective.
4. Hence the exterior fundamental group is the increasing union\[ \pi_1(\mathbb R^3-B) \cong \varinjlim\left( F_1\hookrightarrow F_4\hookrightarrow F_{16}\hookrightarrow\cdots \right), \]so it is nontrivial.
As in Hatcher’s original construction, the horned ball \(B\) is the intersection of the nested compact sets and its complement is the increasing union of the \(Y_n\). Every loop and every nullhomotopy has compact image, hence lies in some finite stage. Therefore the fundamental group of the union is the direct limit of the stage groups. By 3 all bonding maps are injective, so the initial nontrivial meridian survives.
5. Nevertheless the reduced homology of the exterior is still zero.
On abelianization, each bonding map sends an old generator to \[ [a_1,a_2][a_3,a_4]\mapsto0. \] Thus every \(H_1(Y_n)\) class dies at the next stage, so the direct-limit \(H_1\) is zero. Higher reduced homology vanishes just as in the original handle-complement stages, and the complement is connected. This also agrees with Proposition 2B.1 for the complement of the boundary sphere of the limiting ball.
The limiting compact set is still a topological \(3\)-ball: the same shrinking-horn homeomorphisms used in Hatcher’s construction converge uniformly, now with four horns per old handle instead of two. Thus the doubled construction again produces a wild \(2\)-sphere, but with the faster-growing system of free groups described above.