Show that $\mathbb{R} \times \mathbb{Z} \cup \mathbb{Z} \times \mathbb{R}$ is not simply connected

Question

I am trying to prove that $E=\mathbb{R} \times \mathbb{Z} \cup \mathbb{Z} \times \mathbb{R}$ isn't simply connected. I tried something, but I got something wrong. This is what I had thought: if I act on $E$ with integer translations on both axis, that is with the group $\mathbb{Z} \times \mathbb{Z}$, I have a properly discontinuous action. The quotient space (let's say the projection is $p$) should be the sides of a square. Then I should have (since $E$ is connected) that $\frac{\pi_1 ( \text{square})}{p_{*}(\pi_1 (E))}$ is isomorphic to $\mathbb{Z} \times \mathbb{Z}$, where $p_{*}$ is the inducted homomorphism of $p$. Now, it is obvious that the fundamental group of the square is $\mathbb{Z}$, but then something doesn't work, because I have that the quotient of $\mathbb{Z}$ is isomorphic to $\mathbb{Z} \times \mathbb{Z}$

Answer 1

Your space is not simply-connected because it contains the boundary of the square$$\{(x,y)\in[0,1]^2\,|\,x\in\{0,1\}\vee y\in\{0,1\}\},$$but no point of its interior.