Product of Tori under antipodal equivalence

Question

Let $X=(S^1)^n$ be the product of $n-$circles (where $n>1$) and let $G=\mathbb{Z}/2\mathbb{Z}$ act on $X$ by $(-1).(x_1, \dots, x_n)=(-x_1, \dots, -x_n)$. What is the fundamental group of ${X}/{G}$?

Answer 1

Since $G$ is finite and $\mathbb{R}^n$ is the universal cover of $X$, the universal cover of $X/G$ is the composition $$\mathbb{R}^n\xrightarrow{q} X\xrightarrow{p} X/G,$$ where $q$ is the product of the exponential maps. If we let $\gamma$ be the linear path from $(0,\ldots,0)$ to $(\frac 12,\ldots,\frac 12)$, then $\alpha = (p\circ q)_\ast(\gamma)$ is an element of $\pi_1(X/G)$ which is not in $p_\ast(\pi_1(X))=\mathbb{Z}^n$. Since $p_\ast (\pi_1(X))$ has index $2$, any element of $\pi_1(X/G)$ not in $p_\ast(\pi_1(X))$ has the form $\alpha\cdot \beta$ with $\beta\in p_\ast(\pi_1(X))$. The action of $\pi_1(X/G)$ on $\mathbb{R}^n$ is given by $\alpha\cdot \beta \cdot x = x + (k_1+\frac 12,\ldots,k_n + \frac 12)$, where $\beta$ corresponds to $(k_1,\ldots, k_n)$ under the identification of $\pi_1(X)=\mathbb{Z}^n$. We see that $\pi_1(X/G)$ is abelian and has no torsion, hence is isomorphic to $\mathbb{Z}^n$. In fact, we can identify it with the subgroup of $\mathbb{R}^n$ obtained by adding the generator $(\frac 12,\ldots, \frac 12)$ to $\mathbb{Z}^n$.