Fundamental group of the grid on $\mathbb{R}^\mathbb{N}$

Question

The grid on $\mathbb{R}^2$ is defined by the set of points such that at most one coordinate is not an integer. With this in mind, e endow $\mathbb{R}^\mathbb{N}$ with the product topology, where $\mathbb{R}$ carries the Euclidean topology and consider the "infinite dimensional grid" $$X = \Big\{x\in \mathbb{R}^\mathbb{N}: (\exists k\in\mathbb{N}):(\forall i\in\mathbb{N}\setminus\{k\}): x(i) \in\mathbb{Z}\Big\}$$ with the subspace topology inherited from $\mathbb{R}^\mathbb{N}$.

What is the fundamental group $\pi_1(X)$?

Answer 1

Let $G_n$ be your "grid" in $\mathbb{R}^n$ for $n\geq 2$. Let $p_{n,n-1}:G_n\to G_{n-1}$ be the usual projection map. Then $X$ can be identified canonically with the inverse limit $\varprojlim_{n}(G_n,p_{n,n-1})$. One also has an inverse system of fundamental groups and induced homomorphisms $(\pi_1(G_n),p_{n,n-1\#})$. The limit of this system is the first Cech homotopy group $\check{\pi}_1(X)=\varprojlim_{n}(\pi_1(G_n),p_{n,n-1\#})$. The projection maps $p_n:X\to G_n$ induce a canonical homomorphism $\phi:\pi_1(X)\to \varprojlim_{n}\pi_1(G_n)$ to the shape group given by $\phi([\alpha])=([p_1\circ\alpha],[p_2\circ\alpha],[p_3\circ\alpha],\dots)$.

Since each $G_n$ is a graph, $X$ is a one-dimensional metric space (an inverse limit of 1-dimensional spaces is at most 1-dimensional). It is known that fundamental groups of one-dimensional spaces canonically embed into their shape groups by $\phi$. Hence, $\phi:\pi_1(X)\to \varprojlim_{n}\pi_1(G_n)$ is injective. In particular, $\varprojlim_{n}\pi_1(G_n)$ is an inverse limit of free groups. This tells us that $\pi_1(X)$ is residually free and locally free. Moreover, $\phi$ will allow you to distinguish any elements of $\pi_1$ at some finite level: $[\alpha]\neq [\beta]$ in $\pi_1(X)$ if and only if there exists some $n$ where $[p_n\circ\alpha]\neq [p_n\circ\beta]$ in the free group $\pi_1(G_n)$.

One can thus understand $\pi_1(X)$ as a subgroup of an inverse limit of free groups. Exactly what the image of $\phi$ is would require a little more thought but certainly I expect it to be uncountable, much like the fundamental group of Sierpinski Carpet.