Prove that the fundamental group of a torus is $\mathbb{Z} \times \mathbb{Z}$ using a fundamental covering map.
How would one go about doing this?? Is there a specific covering map to use? Or some quotient via the universal cover $\mathbb{R} \times \mathbb{R}$?
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Here is a bit more general way to see the connection between covering spaces and fundamental groups. Suppose $X$ is any path-connected space admitting a universal (i.e. simply-connected) cover $(\overset{\sim}{X},p)$. Then $\pi_1(X)$ is isomorphic to the group of $\textit{deck transformations}$ $\overset{\sim}{X} \to \overset{\sim}{X}$, i.e. homeomorphisms $f: \overset{\sim}{X} \to \overset{\sim}{X}$ such that $p \circ f = p$.
In our case, a universal cover for $T= \mathbb{S}^1 \times \mathbb{S}^1$ is the space $\mathbb{R}^2$ together with $p=(p_1,p_2): \mathbb{R}^2 \to T$ where the $p_i$'s are the covering maps $\mathbb{R} \to \mathbb{S}^1$. It remains to compute deck transformations $\mathbb{R}^2 \to \mathbb{R}^2$. Since $p_1$ and $p_2$ are the covering maps $\mathbb{R} \to \mathbb{S}^1$, any such deck transformation must consist of a pair of deck transformations $\mathbb{R} \to \mathbb{R}$. But the collection of such transformations is precisely $\pi_1(\mathbb{S}^1) \cong \mathbb{Z}$, so we get $\pi_1(T) \cong \mathbb{Z} \times \mathbb{Z}$.
It's easier than that. Or, at least, the intuition is. Take the quotient map $q:\Bbb R\times \Bbb R \to T^2$, where $T^2$ is the torus. Then the inverse image $q^{-1}(t)$ of a single point $t\in T^2$ naturally looks like $\Bbb Z\times \Bbb Z$.
There are a few more details to this if you want an actual proof (like how a class $x\in \pi_1(T^2,t)$ is naturally linked to a single point of $q^{-1}(t)$, after you choose an origin, and how this lets you determine the group structure) but this is the basic idea behind it.
Some details on the above parenthesis: The universal cover map is simply connected, which means that any two paths with the same end points are homotopic with a homotopy fixing the end points. If the two end points are both in $q^{-1}(t)$, then the composition of that path with $q$ becomes a loop with end point $t$.
If we have a loop $\gamma$ in $T^2$ with end point $t$, and choose a specific starting point $t_0\in q^{-1}(t)$, then $\gamma$ is the image of exactly one path in $\Bbb R^2$ that starts at $t_0$. (And if we choose a different starting point, the result is that the entire path is translated.)
Think a little on this correspondence between loops in $T^2$ with end point $t$ and paths in $\Bbb R^2$ with end points in $q^{-1}(t)$, and you will hopefully see that two loops represent the same class in $\pi_1(T^2,t)$ iff their corresponding paths in $\Bbb R^2$ with starting point in $t_0$ has the same end point.
Finally, for the group structure, we can see what happens when we compose loops. If we have two loops $\gamma_1,\gamma_2$ with end points $t$, then we can compose them to get $\gamma_3 = \gamma_1 * \gamma_2$, which is also a loop with end point $t$. What happens to their corresponding paths in $\Bbb R$? If we say they all start at $t_0$, and say that $\gamma_i$ ends at $t_i$, then we get $t_3=t_1+t_2-t_0$. This gives it the group structure of $\Bbb Z^2$, with $(0,0)$ set at $t_0$.