# Can a subset of the plane have nontrivial $H_2$ or $\pi_2$?

This is a question that occurred to me years ago when I was first learning algebraic topology. I've since learned that it's a somewhat aesthetically displeasing question, but I'm still curious about the answer.

Is it possible for a subset of $\mathbb R^2$ to have a nontrivial singular homology group $H_2$? What about a nontrivial homotopy group $\pi_2$?

• I assume you're looking for pathological examples. By Alexander duality, $H_2$ must be trivial if the subset is compact and locally contractible. – Fernando Muro Dec 9 '14 at 21:33
• Read the introduction in imfm.si/preprinti/PDF/00863.pdf. – Igor Belegradek Dec 9 '14 at 21:57
• The link to the preprint cited by Igor Belegradek is now here preprinti.imfm.si/PDF/00863.pdf – j.c. Jun 14 '18 at 16:29

Apparently the asphericity is due to Zastrow (see Cannon-Conner-Zastrow).

Also apparently the result that the higher homology groups vanish is due to Zastrow, but his habilitation thesis never seems to have appeared.

The higher-dimensional analog has the surprising answer "yes". Namely, for $n\geq 2$, the $n$-dimensional Hawaiian earring $H_n = \bigcup_{k=1}^\infty S(k)$, where $S(k)\subseteq \mathbb{R}^{n+1}$ is the $n$-sphere with center ${1\over 2k}\mathbf{e}_1$ and radius ${1\over 2k}$ has nonzero homology in arbitrarily high dimensions. This is a result of Barratt and Milnor (An Example of Anomalous Singular Homology).