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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?

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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

2 Answers 2

up vote 3 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.

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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.

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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

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