Show that f is covering map and find covering tranformation group

Question

Prove that $f:\mathbb{R}^2\to T^2$ defined by $f(x,y)=(e^{2\pi i x},e^{2\pi iy})$ is covering map and also find covering tranformation group$=\{g:\mathbb{R}^2\to\mathbb{R}^2\mid g$ is diffeomorphism and $f\circ g=f\}$ for the first part f is surjective since $e^{2\pi i x}\in S$ and x can be any real numbers. So now i have to show that for any $ q\in T^2$ there is an open nbhood $U$ of $q$ satisfy the condition "For any connected component $W$ of $f^{-1}(U) ; W$ is an open set in $\mathbb{R}^2$ and $f\mid_W:W\to U$ is diffeomorphism" and what about the second part? I still have no idea how to do it.

Answer 1

Hint: $f\circ g=f$ means that $\ g(x,y)-(x,y)\,\in\Bbb Z\times\Bbb Z$.
Since $g$, and therefore also $(x,y)\,\mapsto\,g(x,y)-(x,y)\ $ is continuous, and $\Bbb Z\times\Bbb Z$ is discrete, we have that it must be a constant $(a,b)\in\Bbb Z\times\Bbb Z$.

Answer 2

Hint for part 2: consider maps $$g_1\colon\mathbb{R}^2\rightarrow\mathbb{R}^2:(x,y)\mapsto (x+1,y)$$ $$g_2\colon\mathbb{R}^2\rightarrow\mathbb{R}^2:(x,y)\mapsto (x,y+1).$$ Can you show that $f\circ g_i=f$?