Viewing the torus as $T^{2} = \mathbb{R}^{2} / \mathbb{Z}^{2} $, let $L$ be a straight line in $\mathbb{R}^{2}$ through the origin. How many path components does $T^{2} \setminus L$ have in terms of the slope of L?
Hint: consider the case where the slope is rational first.
I'm not sure how to solve this. I have read somewhere that the image of $L$ in the torus is dense if and only if the slope is irrational, which seems to make sense, but not entirely sure if this helps solve the question.
If the slope of $L$ is rational then the image of $L$ in $T^2$ is a circle, and the complement of the image is an open annulus hence is connected.
If the slope of $L$ is irrational then the covering map is injective on $L$, and indeed on every line parallel to $L$. Let's consider all covering translates of $L$: $$L_{(m,n)} = \{(x+m,y+n) \mid (x,y) \in L, (m,n) \in \mathbb Z^2\} $$ There's only countably many of these covering translates, and each of them has the same injective image in $T^2$.
What about the uncountably many other lines $L'$ in $\mathbb R^2$ that are parallel to $L$? Well, each of them has countably many covering translates $L'_{(m,n)} = \{(x+m,y+n) \mid (x,y) \in L', (m,n) \in \mathbb Z^2\}$, and each of those translates has the same image in $T^2$. Thus, the uncountable set of lines in $\mathbb R^2$ parallel to $L$ is partitioned into subsets of countably infinite size, each partition element representing lines with the same image in $T^2$, different partition elements representing lines with disjoint image in $T^2$. There are uncountably many partition elements, and hence uncountably many distinct component of $T^2$ minus the image of $L$.
So the upshot is: $T^2$ minus the image of $L$ has either one component (when the slope of $L$ is rational), or an uncountably infinite number of components (when the slope of $L$ is irrational).