Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This should be a very basic algebraic topology question. The other day I was thinking about the fact that $P^2(R)$ has $\pi_1 = Z/2Z$.

On the other hand I thought to myself how something like this can never happen for, say, an open subset of the real plane $R^2$. It's a very intuitive fact but I can't prove it.

So I guess my general question is: can open subsets of $R^2$ have torsion elements in $\pi_1$?

Related to this: is there a classification of the homotopy types of open subsets of the plane?

share|cite|improve this question
I think you want to ask about weak homotopy types, not homotopy types: I shudder to think about what the homotopy type of say, $\mathbb{R}^2 \setminus \mathbb{Q}^2$ is... – Qiaochu Yuan Feb 16 '14 at 5:51
What do you mean by weak homotopy type? Just that there is a weak homotopy equivalence between the two? Also, $Q^2$ is not closed in $R^2$ so $R^2 - Q^2$ shouldn't really enter the picture (but I guess you could take $R \times C$, where $C$ is the Cantor set). – user125763 Feb 16 '14 at 5:59
Sorry, yes, I meant something like $\mathbb{R}^2$ minus a $2$-dimensional Cantor set. And yes, that's what I mean by weak homotopy type. – Qiaochu Yuan Feb 16 '14 at 5:59
QY: Do you have a good example to understand the difference between weak and strong homotopy type for a pathological space? I'm not too familiar with this - whenever I think about homotopy theory I always assume everything to be a CW complex. (also - I cannot comment on MO so I might as well ask here: do you have a concise reference to learn about this stuff about Stone-Cech and $C^*$-algebras?) – user125763 Feb 16 '14 at 6:07
Weak homotopy type is the concept that lets you assume that everything is a CW complex (every space is weakly homotopy equivalent to a CW complex). For example, any space which is totally disconnected has the weak homotopy type of a discrete space (e.g. the Cantor set). I don't have a reference. – Qiaochu Yuan Feb 16 '14 at 6:10
up vote 4 down vote accepted

The new question is very different from the original one. The answer to it is that every noncompact connected surface is homotopy-equivalent to a bouquet of circles, see proofs and references here. In particular, fundamental group is free and, hence, torsion-free.

Thus, the answer to the last question is: Yes, there is a description, namely they all are disjoint unions of bouquets of circles.

A (quite a bit) more difficult theorem is that the fundamental group of every open connected subset of $R^3$ is still torsion-free. In dimensions $\ge 4$ this is, of course, false.

share|cite|improve this answer
thanks - I wanted to strike through the old question, rather than delete it, but I couldn't find how. – user125763 Feb 16 '14 at 6:20
I don't believe that this is true in the generality you're claiming. An open subset of $\mathbb{R}^2$ need not be homotopy equivalent to the disjoint union of its connected components; for example, as mentioned in the comments, it could be $\mathbb{R}^2$ minus $\mathbb{R}$ times a Cantor set. – Qiaochu Yuan Feb 16 '14 at 8:21
@Qiaochu: What I said is correct, proven by Whitehead in much greater generality than I stated here. He also proved that any open n-manifold is homotopy equivalent to its subcomplex of dimension $n-1$. This is from his paper on immersion a of 3-manifolds (around 1960, I will find the exact date) but the result I mentioned is general. – studiosus Feb 16 '14 at 10:31
@Quiaochu: Whitehead's paper is from Proceeding of LMS, 1961, lemma 2.1. – studiosus Feb 16 '14 at 10:42
Incidentally, complement to R times cantor set has homotopy type of a countable discrete set. This is completely trivial. In my bouquet of circles I allow no circles, i.e. just a single point. Maybe this was unclear. Pathological examples you are worrying about do not occur in the context of manifolds. – studiosus Feb 16 '14 at 15:16

I think it is still possible to have $2$-torsion. The special orthogonal group $SO(n)$ is orientable, yet has a fundamental group of $\mathbb{Z}/2\mathbb{Z}$ for all $n\geq 3$. To see that $SO(n)$ is orientable, note that $SO(n)$ has a Lie group structure, so its tangent bundle is trivial. Hence, the top exterior power of the cotangent bundle is trivial, meaning that $SO(n)$ has a non-vanishing form of top-degree.

share|cite|improve this answer
ah crap. But is what I think makes sense in $R^2$ true? Can opens inside it have any $2$-torsion? I was hoping this wasn't something which boiled down to a big theorem. (like a classification of the homotopy types of open subsets of the real plane) – user125763 Feb 16 '14 at 5:30
Note that $\text{SO}(3) \cong \mathbb{RP}^3$ and all odd-dimensional real projective spaces are orientable as well (as well as all having fundamental group $\mathbb{Z}_2$ in dimension at least $2$). – Qiaochu Yuan Feb 16 '14 at 5:34
Thanks guys - I guess that's why I couldn't remember the connection between the two things: there isn't one! – user125763 Feb 16 '14 at 5:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.