Are the higher homotopy groups of the Hawaiian earring trivial?

Question

The fundamental group of the Hawaiian earring is very complicated, but since it's "1-dimensional" one might guess that the higher homotopy groups vanish. Do they? Since the Hawaiian earring does not have a universal cover, the standard approach to showing that higher homotopy groups of graphs vanish does not apply.

Answer 1

Despite the non-existence of a universal covering space, there is still an object that acts like a universal covering space for the earring space. It is actually quite useful for understanding the structure of the fundamental group as a subgroup of the inverse limit $\varprojlim F_n$ of free groups. The "generalized universal covering" still provides all the same unique path lifting properties for maps out of locally path connected spaces that one might want. The study of these particular coverings for wild spaces is mainly due to Hanspeter Fischer and Andreas Zastrow. I think you should be able to use this to mimic the argument for graphs.

Let $\mathbb{E}$ be the earring space, which is a shrinking wedge of a sequence of circles. Let $\widetilde{\mathbb{E}}$ be the set of homotopy classes of paths starting at the basepoint. Take $p:\widetilde{\mathbb{E}}\to \mathbb{E}$, $p([\alpha])=\alpha(1)$ to be the endpoint projection and give $\widetilde{\mathbb{E}}$ the usual "whisker" topology that you do to construct universal covers: the basic neighborhoods are $B([\alpha],U)=\{[\alpha\cdot\epsilon]|\epsilon([0,1])\subset U\}$ where $U$ is open in $\mathbb{E}$. It turns out that $p:\widetilde{\mathbb{E}}\to \mathbb{E}$ is an open map which has unique lifting of all paths and homotopies of paths. Moreover, if $Y$ is path connected, locally path connected, and simply connected, then any based map $f:Y\to \mathbb{E}$ has a unique lift $\tilde{f}:Y\to \widetilde{\mathbb{E}}$ such that $p\tilde{f}=f$. Consequently, $p$ induces isomorphisms on higher homotopy groups.

It also turns out that $\widetilde{\mathbb{E}}$ has the structure of an $\mathbb{R}$-tree (uniquely arcwise connected metric space where each arc is isometric to a sub-arc of the reals), and these are known to be contractible (this is due to J. Morgan I believe).

Now you should have all the usual ingredients to mimic the usual argument for graphs. In fact, the earring is not special here. This should work out for all 1-dimensional Peano continua.

Answer 2

I did not read this paper, but it claims that the answer is yes: One-dimensional sets and planar sets are aspherical by J.W. Cannon, G.R. Conner and Andreas Zastrow.

Answer 3

Here is a short natural argument that planar continua are aspherical, different from the technique of Cannon/Conner/Zastrow, and straight forwardly applied to the Hawaiian earring.

The Hawaiian earring (and more generally any planar compactum $X$) is the nested intersection of planar polyhedra $X_n$.

The planar Euclidean metric naturally induces a length structure on $X_n$ so that $X_n$ is locally CAT(0).

In particular there is a natural proof $X_n$ is aspherical, since inessential loops in $X_n$ can be canonically extended to maps of the disk, and since the m-ball $B_{m}$ is naturally fibred by disks, so that the boundaries of the disks fibre the m-sphere $S_{m}$.

To see $X$ is aspherical take a map $f: S_{m} \rightarrow X$, and obtain the natural extensions $F_{n}: B_{m} \rightarrow X_{n}$.

Ascoli's theorem ensures the existence of subsequential limit and hence $X$ is aspherical.

In Cannon/Conner/Zastrow, inessential loops in $X$ are contracted internally, within their own image. In contrast, in the argument at hand, roughly speaking, inessential loops in $X$ bound disks fibred by limits of external geodesic chords.

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To apply Ascoli's theorem we must gain global control of the equicontinuity data.

It is useful to show there exists a universal constant K>0 so that if P is a planar polyhedron, if $f: [0,1] \rightarrow P$ is a path, if C is the local geodesic path homotopic to f, (and if g is the parameterization of C induced by projection of f), then if (d,e) is uniform continuity data for f with respect to absolute value, then (d,Ke) is uniform continuity data for g (also with respect to absolute value). K=1.5 is apparently the worst case.

However, in the special case of the Hawaiian earring we may use K=1, since $X_n$ is the union of a finite bouquet of loops and a convex disk.