Confused by step in calculating fundamental groups of cell complexes using Seifert-Van Kampen theorem I understand the statement of the Seifert-Van Kampen theorem - given an open cover ${A,B}$ of a space ${X}$ such that ${A\cap B}$ is non-empty and path-connected, then
$$
\pi_1(A\cup B) = \frac{\pi_1(A)*\pi_1(B)}{N}
$$
where ${*}$ is the free product, and $N$ is the normal subgroup of ${\pi_1(A)*\pi_1(B)}$ given by
$$
N = \{f_*(c)g_*^{-1}(c)\ |\ c \in \pi_1(A\cap B)\}
$$
where ${f_*}$ and ${g_*}$ are induced from the inclusions ${f : A\cap B\to A}$ and ${g : A\cap B\to B}$ respectively. I understand this is essentially to stop us from "overcounting" loops essentially.
Applying this to the torus, looking at the fundamental polygon for it, - we can use the open cover ${A=T^2\setminus \{x\}}$ for $x$ in the interior of the square (so the punctured torus) and ${B =}$ interior of the square of the torus. We would then have ${A,B}$ are open, and ${A\cap B}$ is clearly path-connected and ${A\cap B\simeq S^1}$. This gives us that
$$
\pi_1(T^2) = \frac{(\mathbb{Z}*\mathbb{Z})*\{0\}}{N} = \frac{\mathbb{Z}*\mathbb{Z}}{N}
$$
but then I am confused about how we can find ${N}$. I understand that for every $c$, we have ${g_*(c)=1}$, hence we essentially just have $N$ being generated by ${f_*(c)}$. But how would we find ${f_*(c)}$?
 A: I'll think of $\mathbb{T}^{2}$ as $[0,1]^{2}/\sim$, where $(0,t)\sim(1,t)$ and $(t,0)\sim(t,1)$.
Let $a$ be the loop $[t,0]$, $b$ the loop $[0,t]$ and $c$ a counterclockwise simple circle around $(\frac{1}{2},\frac{1}{2})$.
Notice that $A=\mathbb{T}^{2}\setminus\{[\frac{1}{2},\frac{1}{2}]\}$ deformation retracts into $\partial[0,1]^{2}/\sim$ (the 'border' of the square), which is homeomorphic to a wedge of two circles. Hence, $\pi_{1}(A)=\mathbb{Z}*\mathbb{Z}$ and a system of generators is given by $a$ and $b$.
Now, $B=(0,1)^{2}/\sim$ is contractible, so $\pi_{1}(B)=0$.
Lastly, $A\cap B$ deformation retracts into a circle, so $\pi_{1}(A\cap B)=\langle c\rangle$ is infinite cyclic.
The Van Kampen Theorem tells us that $\pi_{1}(\mathbb{T}^{2})$ is therefore the quotient of $\pi_{1}(A)*\pi_{1}(B)=\mathbb{Z}*\mathbb{Z}$, with generators $a,b$ and some relations.
I believe that it is easier to describe the relations as "The value of $c$ (generator of $\pi_{1}(A\cap B)$) as a loop in $A$"="The value of $c$ as a loop in $B$". We need to see what $c$ is path-homotopic to in both cases.
In $B$, since $B$ has trivial fundamental group, clearly $c=1$.
In $A$, by choosing as a base point, for example, $(\frac{3}{4},\frac{1}{2})$, and $c$ to be the circle of radius $\frac{1}{4}$ parametrized counter-clockwise, one can see (if you wish, I can elaborate on this part), that "bringing $c$ to the edge" gives $c=ba^{-1}b^{-1}a$.
Therefore, the resulting equation is $ba^{-1}b^{-1}a=1$, which is the same as $ab=ba$.
In conclusion,
$$
\pi_{1}(\mathbb{T}^{2})=\frac{\langle a,b \rangle}{\langle aba^{-1}b^{-1} \rangle}=\mathbb{Z}\oplus\mathbb{Z}.
$$
Hope this helps!
EDIT: on the "bringing $c$ to the edge part", I'm adding some (rather horrible) pictures:
First, I'll adjust the generators of our fundamental groups to the base point $p$.
$a$, and the numbers indicate the order in which you must travel along $a$" />
$b$, same deal as before" />
Our generator for the intersection is

We wish to see what $c$ is homotopic to in $A$. We just stretch $c$ to the edge (like when we try to stretch pizza dough to a tray!), and we get something like this:

You can certainly do this in the square (without the identifications), and applying the projection map to the homotopy (in the square), you get a homotopy (in the torus).
Now, we break up $c$ into 4 parts (I'm implicitly adding new paths, but since I travel along them in both directions, they cancel out, so I'm not cheating):

Finally, use the identifications in the torus (and the directions in which $a$ and $b$ are traveled), to see that (modulo homotopy), $c_{1}=b$, $c_{2}=a^{-1}$, $c_{3}=b^{-1}$, $c_{4}=a$, so $c=ba^{-1}b^{-1}a$.
