What is the quotient $\mathbb{T}^3/\mathbb{Z}_2$?

Question

What is the quotient space $\mathbb{T}^3/\mathbb{Z}_2$, When $\mathbb{Z}_2$ acting on 3-dimensional torus $\mathbb{T}^3 := \mathbb{R}^3/\mathbb{Z}^3$ by sending $x$ to $-x$? Does anyone know how to compute its K-theory or cohomology groups?

Answer 1

Let's construct cell structures on $X_n=\mathbb{T}^n/\{\pm1\}$ by induction on $n$. To do this, consider $X_n$ as a quotient of $Y_n=[-1/2,1/2]^n\subset\mathbb{R}^n$, and consider the following cell structure on $Y_n$ defined by induction on $n$: for each closed cell $A$ in the standard cell structure on the cube $[-1/2,1/2]^{n-1}$, let $[-1/2,0]\times A$ and $[0,1/2]\times A$ be closed cells of $Y_n$. For each closed cell $A$ in our cell structure on $Y_{n-1}$, let $\{x\}\times A$ be a closed cell of $Y_n$ for $x=-1/2$, $x=0$, and $x=1/2$. It is straightforward to verify that this indeed defines a cell structure on $Y_n$, and that it descends to a cell structure on the quotient $X_n$. Furthermore, we can see from this construction that the 1-skeleton of $X_n$ consists of two copies of the 1-skeleton of $X_{n-1}$ joined by an edge. By induction, this implies the 1-skeleton of $X_n$ is contractible for all $n$.

Now set $n=3$; it can be checked that our cell structure on $X_3$ has a single $3$-cell and four $2$-cells. Furthermore, the boundary of the 3-cell can be computed to be of the form $2a-2b$, where $a$ and $b$ are two of the $2$-cells. Since the 1-skeleton of $X_3$ is contractible, we can reduce it to a point without changing the homotopy type. It follows that the only nontrivial reduced homology group of $X_3$ is $H_2(X_3)=\mathbb{Z}^3\oplus \mathbb{Z}/2$, and so $X_3$ is a Moore space $M(\mathbb{Z}^3\oplus\mathbb{Z}/2,2)$.

(I recommend you draw pictures of these cell structures for $n\leq 3$ to verify my claims; you can also see easily that $X_2\cong S^2$.)