When does $f_{a,b}$ become immersion or embedding?

Question

For $a, b \in \mathbb{Z}$, we define $f_{a, b}: S^1 = \mathbb{R}/\mathbb{Z} \rightarrow T^2 = \mathbb{R}^2/\mathbb{Z}^2; x \bmod \mathbb{Z} \mapsto (ax, bx) \bmod \mathbb{Z}^2$. When does $f_{a,b}$ become immersion or embedding?

[My consideration so far]

$(df_{a, b})_x$ seems to be injective if and only if $(a, b) \neq (0, 0)$. So, $f_{a,b}$ is immersion when $(a, b) \neq (0, 0)$. When $f_{a,b}$ is embedding, $f_{a,b}$ is immersion and injective. So, it seems we only need to consider the cases when $(a, b) = (0, \pm 1), (\pm 1, 0), (m, n)$. Here, $m$ and $n$ are coprime. Does $f_{a,b}$ really become embedding in all the cases above?

Answer 1

$f$ is injective if and only if $gcd(a,b)=1$.

Suppose that $f$ is injective and $gcd(a,b)\neq 1$ Let $d$ be a divisor of $a$ and $b$, $f([{1\over d}])=[({a\over d},{b\over d})]=f([(0,0)])$. Contradiction.

Suppose that $gcd(a,b)=1$ and $f([x])=f([y])$ this implies that $a(x-y), b(x-y)\in\mathbb{Z}$ we deduce that $x-y={p\over q}$ and $q$ is a divisor of $a$ and $b$, since $gcd(a,b)=1,q=1$ and $[x]=[y]$.