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I'm working through Hatcher book and done $\pi_1(\mathbb{R}^2 - \mathbb{Q}^2)$ is uncountable. It's easy to see that it's true as you can imagine only non trivial maps contract in the space.

But, was wondering has anyone worked out $\pi_2(\mathbb{R}^2 - \mathbb{Q}^2)$. $\pi_2(\mathbb{R}^3 - \mathbb{Q}^3)$ I imagine this isn't that hard, shouldn't it be uncountable aswell.

Just doing a project on higher homotopy theory, but fundamental group stuff seems hard already. Like $\pi_2$ is that just a sphere doing a weird thing and looping back on itself. I know the definition but can't really see it. Plus the calculation aren't that easy.

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$\pi_2(\mathbb R^3\setminus\mathbb Q^3)$ is certainly uncountable. $\pi_2(\mathbb R^2\setminus\mathbb Q^2)$ is most likely trivial, but I don't see a quick argument. –  Grumpy Parsnip Feb 16 '12 at 18:42
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I find the standard definition of $\pi_2$ fairly unenlightening. The definition that to me works much better is that it is the group of homotopies between null-homotopic paths from a basepoint to itself, up to homotopy. In other words, it's the fundamental group of the loop space (en.wikipedia.org/wiki/Loop_space). –  Qiaochu Yuan Feb 16 '12 at 18:52
    
These things can be counter-intuitive. The hawaiian earring space is a retract of $\mathbb R^2\setminus\mathbb Q^2$, and if you look at the 2D version of the hawaiian earring space, $X$, Barratt and Milnor showed that $H_3(X)\neq 0$. The plane is usually exempt from this sort of exotic behavior, so I'll bet $\pi_2=0$ in that case, but I also think the proof may be quite difficult. –  Grumpy Parsnip Feb 16 '12 at 19:26
    
@QiaochuYuan I will have to look at that in more details. I can see $\pi_1$ and $\pi_2$ seems like a natural generalization. –  simplicity Feb 16 '12 at 19:38
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@Qiaochu, I find it almost incredible that that definition (which is of course very very nice technically) can be considered more enlightening that the definition of $\pi_2$ as homotopy classes of maps from a sphere! –  Mariano Suárez-Alvarez Feb 17 '12 at 2:51

2 Answers 2

I asked my colleague Jurek Dydak about this, and he pointed out the following paper

Topology and its Applications Volume 120, Issues 1–2, 15 May 2002, Pages 23–45 One-dimensional sets and planar sets are aspherical J. W. Cannon , G. R. Conner, and Andreas Zastrow

which shows that every subset of the plane has trivial $\pi_k, k\geq 2$. So in particular $\pi_2(\mathbb R^2\setminus \mathbb Q^2)=0$.

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Amazing result! –  Mariano Suárez-Alvarez Feb 17 '12 at 2:48

My algebraic topology is rusty, so take this with a grain of salt.

Any map from $S^2$ to $\mathbb{R}^2-\mathbb{Q}^2$ could be contracted through its own image, since the image would necessarily not contain any open subsets of $\mathbb{R}^2$.

In $\mathbb{R}^3$, its easy to imagine cube-like surfaces that avoid rational points. These interior of such surfaces contain rational points and so these surfaces cannot be contracted. So this provides infinitely many elements of $\pi_2$. To see that this provides uncountably many, take cube sides of irrational length.

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My problem is that I don't see how to make the first part precise. –  Grumpy Parsnip Feb 16 '12 at 19:23
    
Hmm...does it help to only consider smooth maps from $S^2$ to $\mathbb{R}^2-\mathbb{Q}^2$? Some thoughts: When I try to imagine a map whose image is not simply connected, I see overlapping tendrils. At the tips of those tendrils, some kind of derivative is zero. The preimage is a point on the sphere with an isotherm surrounding it. We could contract to the image of that isotherm. –  alex.jordan Feb 16 '12 at 21:59

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