Proving map si covering map of torus

Question

Let $f:\mathbb{R^2} \rightarrow \mathbb{T^2}$ be defined as $$f(x,y)=(x-[x],y-[y]),$$ where $[x]$ denotes the integer part of $x$.
I would like to show that this is a covering map of the torus $\mathbb{T^2}$. I consider the torus as the set $([0,1] \times [0,1])/\sim$, where the equivalence relation $\sim$ is given by identifying opposite edges of the unit square $[0,1] \times [0,1]$. How can I choose an open cover $\{U_\alpha\}$ of the torus such that $f^{-1}(U_\alpha)$ is a disjoint union of open sets $V_{\alpha \beta}$ and $f:V_{\alpha \beta} \rightarrow U_\alpha$ is a homeomorphism for each $\alpha$? I understand that this is a covering map intuitively, but I am not sure how to write it properly, in particular, how to choose this open cover.

Answer 1

Show that it is surjective (it is easy). Then chose a point $p = (\overline{x},\overline{y}) \in \mathbb{T}^2$. Find a neighboorhood $U$ of $p$ for which the preimage $f^{-1}(U)$ is not connected : for exemple, take a little ball around $p$ with radius $r < \frac{1}{2}$ (draw a picture!).

You can show that on each connected component, $f$ is injective, thus is bijective. It is easy to show that it is a homeomorphism from the very definition of the quotient topology on $\mathbb{T}^2$. Thus you have shown $f$ is a covering map.