# Fundamental group of R^2-Q^2

After learning about the fundamental group, and proving that $\mathbb{R}^n$ minus any countable set is path-connected, I started wondering if the fundamental group of $\mathbb{R}^2-\mathbb{Q}^2$ is known. Does anyone know whether or not it is? Or how one might go about determining it?

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This math.stackexchange question might help. math.stackexchange.com/q/16948/1321 – George Lowther Jun 11 '11 at 23:30
An exercise in Hatcher's textbook asks to show that it is uncountable. – Chris Gerig Jun 11 '11 at 23:36
What does it mean for a group to be "known"? Fundamental groups of subsets of the plane have a certain characterization (Eda 1998). But even finitely-presented groups have no classification, and this group isn't even finitely generated... – Ryan Budney Jun 11 '11 at 23:49
Perhaps the simple question of 'what is this fundamental group?' is not a good fit for MO. But I think there is something to be learned from this example, namely, you don't really want the fundamental group in this case, but something else. This 'something else' is detailed in my answer. – David Roberts Jun 12 '11 at 0:03

The space $X = \mathbb{R}^2 - \mathbb{Q}^2$ is not semilocally simply connected, and so in a sense the fundamental group is a poor measure of the homotopy 1-type of $X$. It is an exercise in Hatcher's book Algebraic Topology that this group is uncountable.
Here is an attempt to motivate a different view of the fundamental group: Consider the space $X_{n,m}$ which is the complement in $\mathbb{R}^2$ of all points with coordinates $(p/n,q/m)$ where $n,m$ are positive integers and $q,p$ are integers. This forms a cofiltered diagram $\mathbb{N}^2 \to Top$ whose limit is $X$, and where each space in the diagram is semilocally simply-connected. There is a map $X_{n,m} \to X_{n',m'}$ when $n'|n$ and $m'|m$. The fundamental group of $X_{n,m}$ is free on a countable number of generators (a generator is given by a small circle around each deleted point, and connect this circle to the basepoint, $(\pi,\pi)$, say, by a path), and the maps between them kill off certain generators. Already, you can see how complicated $\pi_1(X)$ is going to be, since it has a countable number of quotients, each of which is free on a countable number of generators (actually, each of them is isomorphic to the free group on generators $\mathbb{Z}\times\mathbb{Z}$).
This system of groups is known as a (strict) progroup, and can be thought of as a formal limit This is usually the 'right' algebraic object to consider when one has a badly behaved space such as this one. It is this object which 'controls' what the covering spaces of $X$ look like, as $X$ doesn't have a universal covering space (which if it did exist, would have fibre $\pi_1(X)$, and other connected covering spaces would be quotients of this one). This gets into the realm of shape theory (wikipedia, nLab), developed to consider spaces with bad local homotopy properties.
@Dan - it's not canonical at all (really just the first construction that came to mind) but my guess from what I know of shape theory is that this choice of 'resolution' of $X$ doesn't matter, up to some sort of weak equivalence in the so-called shape category. @Avi - actually there is a mistake in that paper in that the fundamental group is not always a topological group, only a quasi-topological group (multiplication is separately continuous in each variable). There would be such a structure floating around here, but as always, it depends on what you want to do with the fundamental group.. – David Roberts Jun 12 '11 at 23:43
... If you are interested in (ordinary) covering spaces, then the above is the right way to do it. There are also approaches using topoi and localic fundamental groups (work by Marta Bunge is appropriate here), fibre functors and Galois categories (due to Grothendieck, discussed in a nice paper by Eduardo Debuc) - which is close to what I mention above, but removes the choice of resolving $X$ by the $X_{n,m}$ - it considers all covering spaces. As far as shape theory goes, it's a slippery beast. The 'right' way to do it is how Tim and others have since developed it, and that is abstractly.... – David Roberts Jun 12 '11 at 23:50