Using universal coverings to calculate fundamental group I want to calculate the fundamental group of $T^*=(T\setminus\Delta)/\sim$ where


*

*$T$ is the torus.

*$\Delta$ is the diagonal

*$\sim$ is an equivalence relation defined by $(x,y)\sim (y,x)$
I want to do this without relying on the Seifert-Van Kampen Theorem, since the hint for the problem was to use lifts/universal coverings.
My question is the following: How would one approach a problem like this, using lifts/universal coverings?
I understand that we could find the univeral covering space $\widetilde{(T-\Delta)}$ and then use this to cover $T^*$ by composing $\widetilde{(T-\Delta)}\longrightarrow(T-\Delta)\longrightarrow T^*$, but I fail to see what this would say about the fundamental groups. Or perhaps this is the wrong approach.
I am beginner when it comes to topology and I find these types of problems  particularly difficult, so any advice would be very much appreciated!
 A: Let's start with the universal covering space and deck transformation group of the torus $T$, which I'm sure you know very well:
$$f : \mathbb R^2 \to T = (S^1)^2
$$
$$D_f = \mathbb Z^2 \quad\text{acting by translations on} \quad \mathbb R^2
$$
Of course, $D_f$ is isomrophic to $\pi_1(T)$.
The strategy is to use this well known information to carry out two steps:
Step 1. First, determine the universal covering space and deck transformation group for $T \setminus \Delta$.
Step 2. Second, determine the universal covering space and deck transformation group for $(T \setminus \Delta) / \sim$. 
Of course, that second deck transformation group will be isomorphic to the fundamental group of $(T \setminus \Delta) / \sim$.
For Step 1, downstairs in $T$, we first remove the diagonal
$$\Delta = \{(x,x) \in (S^1)^2 \mid x \in S^1\}
$$ 
Upstairs in $\mathbb R^2$, we remove the "total lift" of the diagonal, which is
$\widetilde\Delta = f^{-1}(\Delta)$, and one can work that set out to be
$$\widetilde\Delta = \{(x,y) \in \mathbb R^2 \mid y-x \in \mathbb Z\}
$$
in other words all lines of slope $1$ passing through integer points.
Now consider the following question: how, from the information just described, can we determine the universal covering space and deck transformation group of $T \setminus \Delta$? 
Answer: First consider two consecutive components of $\widetilde\Delta$, let's say the lines $y=x$ and $y=x+1$. Then consider the strip of $\mathbb R^2 - \widetilde\Delta$ that lies between those two lines, i.e. the solution set of the inequality $x < y < x+1$. Let's denote that set 
$\widetilde{T - \Delta}$. This set is simply connected, and the restricted map
$$f : \widetilde{T - \Delta} \to T \setminus \Delta
$$
is a covering map, so it is a universal covering map. 
Next, consider the subgroup of $D_f = \mathbb Z^2$ that takes the set $\widetilde{T - \Delta}$ to itself. That subgroup is the diagonal subgroup
$$D_{T-\Delta} = \{(k,k) \in \mathbb Z^2\}
$$
and it is isomorphic to $\mathbb Z$. One can check that this is the deck transformation group of the map $\widetilde{T-\Delta} \mapsto T-\Delta$.
For step 2, which I'll cover more briefly, one proceeds similarly but this time with a quotient construction. $T^*$ is the quotient of $T-\Delta$ by the identification $(x,y) \sim (y,x)$. Its universal cover $\widetilde T^*$ is the quotient of $\widetilde{T-\Delta}$ by a quotient construction which one can work out is given by the formula $(x,y) \sim (y - \frac{1}{2},x+\frac{1}{2})$. Using this quotient one sees that $\widetilde T^*$ is homeomorphic to the half-strip $x < y \le x+\frac{1}{2}$ hence is simply connected. Using quotient map theory, one shows that there is an induced universal covering map $\widetilde T^* \to T^*$, and an induced action of $D_{T-\Delta}$ of $\widetilde T^*$ which is the deck transformation action of this map. Thus, $\pi_1(T^*)$ is isomorphic to $D_{T-\Delta} \approx \mathbb Z$.
