How to calculate the Fundamental Group of $\mathbb{R}^2 \setminus \mathbb{Z}^2$.
I know that the Fundamental group of the Plane with $k$ punctures is the free product of $k$ copies of $\mathbb{Z}$. My guess is that $\pi (\mathbb{R}^2 \setminus \mathbb{Z}^2)$ will be the free product of infinitely many copies of $\mathbb{Z}$. But I could not prove it.
Your guess is correct. The standard 'pushing away from the holes' idea gives you that this space is homotopically equivalent to an "infinite grid" (the union of all lines in $\mathbb{R}^2$ of the form $y = n$ or $x = m$ for integers $n, m$). The $\pi_1$ of this is certainly gonna be free (it's a connected graph, so we take a maximal subtree and collapse). This graph only has countably many edges, so the $\pi_1$ is free on at most countably generators, and the maximal tree surely must miss infinitely many edges (every square on the grid gives a cycle, so at least one edge in each square needs to get removed), so this $\pi_1$ is free on (countably) infinitely many generators.
