58
$\begingroup$

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 $\pi_2$?

Improve this question
$\endgroup$
3
  • 7
    $\begingroup$ I assume you're looking for pathological examples. By Alexander duality, $H_2$ must be trivial if the subset is compact and locally contractible. $\endgroup$
    – Fernando Muro
    Commented Dec 9, 2014 at 21:33
  • 9
    $\begingroup$ Read the introduction in imfm.si/preprinti/PDF/00863.pdf. $\endgroup$
    – Igor Belegradek
    Commented Dec 9, 2014 at 21:57
  • 3
    $\begingroup$ The link to the preprint cited by Igor Belegradek is now here preprinti.imfm.si/PDF/00863.pdf $\endgroup$
    – j.c.
    Commented Jun 14, 2018 at 16:29

2 Answers 2

Reset to default
42
$\begingroup$

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.

Improve this answer
$\endgroup$
1
69
$\begingroup$

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).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .