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 $\pi_2$?

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    $\begingroup$ I assume you're looking for pathological examples. By Alexander duality, $H_2$ must be trivial if the subset is compact and locally contractible. $\endgroup$ Dec 9, 2014 at 21:33
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    $\begingroup$ Read the introduction in imfm.si/preprinti/PDF/00863.pdf. $\endgroup$ Dec 9, 2014 at 21:57
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    $\begingroup$ The link to the preprint cited by Igor Belegradek is now here preprinti.imfm.si/PDF/00863.pdf $\endgroup$
    – j.c.
    Jun 14, 2018 at 16:29

2 Answers 2


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).


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