What is the fundamental group of $\mathbb{R}^2\setminus\mathbb{Z}^2$

Question

Basically the title. As the plane minus $n$ points has as fundamental group the free product of $n$ copies of $\mathbb{Z}$, is there something like a countable free product? In that case, is it related to the mentioned fundamental group?

Answer 1

Let's base ourselves at, say, somewhere in $(-1,1)^2\setminus\{0\}$.

By an easy compactness argument the fundamental group is isomorphic to $\varinjlim_{N\in\Bbb N}\pi_1((-N,N)^2\setminus(\{-(N-1),\cdots,N-1\})^2)$ which is a countable free product of $\Bbb Z$. And yes, this makes sense. Arbitrary free products of arbitrary sets of groups exist and are groups with the usual universal property (and can be constructed in the same way). A countable free product of $\Bbb Z$ - the free group on one generator would just be the free group on $\Bbb N$ generators. My colimit is just the colimit of $F(1),F(2),F(3),\cdots$ where $F$ is the free group functor - a left adjoint - and left adjoints preserve colimits so abstractly I know this is equivalently just $F(\varinjlim(1\subset2\subset3\subset\cdots))=F(\Bbb N)$.

There is surely a natural map (take $\pi_1(\text{inclusions})$) from this group to $\pi_1(\Bbb R^2\setminus\Bbb Z^2)$ and we just need to check it injects and surjects. But any loop representing an element of the right hand side has compact image and is in the image of the groups defining the left hand side for sufficiently large $N$, so this map surjects. Injectivity is similarly arguable as elements of the colimit are representable from some loop coming from a finite stage $N_1$ any homotopy in $\Bbb R^2\setminus\Bbb Z^2$ killing the image of a loop from the colimit will have compact image and may be applied beyond some stage $N_2$ and for $N>\max(N_1,N_2)$ the loop will vanish in the $N$th group so the represented element of the colimit is just the identity.

Answer 2

Here's a slightly different argument to that of @FShrike.

To make coordinates less awkward, let's consider $\mathbb{R}^2 \setminus ((1/ 2, 1 / 2) + \mathbb{Z}^2)$ instead of $\mathbb{R}^2 \setminus \mathbb{Z}^2$. This space deformation retracts onto a 1-dimensional CW-complex $X$ embedded in $\mathbb{R}^2$ with 0-cells all points $(n, m) \in \mathbb{Z}^2$ and 1-cells the line segments connecting $(n, m)$ with $(n + 1, m)$ or $(n, m + 1)$ (in other words, $X$ is a grid with grid points the points with integer coordinates in $\mathbb{R}^2$). But the fundamental group of $X$ is free on the 1-cells of $X \setminus T$ where $T \subset X$ is a maximal spanning tree (cf. Hatcher Proposition 1A.2), and $X \setminus T$ will have a countably infinite set of 1-cells (I suggest you check this yourself if this is not immediately clear to you), so the claim follows.