I'm trying to get my head round the following calculation of the fundamental group of the torus, using Seifert Van-Kampen (I know it's easier to do this by considering covering spaces, but I'm trying to learn the Seifert Van-Kampen method).
Consider the torus $T$ as the unit square in $\mathbb R^2$ with opposite edges identified. Let $A$ be an open disc (say of radius $1/4$ about the origin), and $\overline{A}$ be it's closure. Let $B$ be complement of $A$, so that $A \cap B = S^1$. Then we have $T = A \cup B$ where $A$ and $B$ are a neighbourhood deformation retract pair, so we can use Van Kampen. It is clear that $\pi_1(A) = \{e \}$ and $\pi_1(A \cap B) = \mathbb Z$. So we need to find $\pi_1(B)$. Observe that $B$ deformation retracts onto the boundary of the square which, after identification, is homeomorphic to $S^1 \vee S^1$; in this way we see that $B$ and $S^1 \vee S^1$ are homotopy equivalent (inclusion in one direction, the retraction in another), and so $\pi(S^1 \vee S^1) = F_2$ (which can be calculated by Van Kampen, or by using the theory of covering spaces). So, by Van Kampen, $\pi_1(T) \cong F_2 *_{\mathbb Z} \{e\}$. This is the quotient of $F_2$ by the normal closure of the image of $l_* : \pi_1(A \cap B) \to \pi_1(B)$ (where $l_*$ is the homomorphism induced from the inclusion map $l : A \cap B \hookrightarrow B$). The following is causing me confusion: The generator of $\pi_1(A \cap B)$ is homotopic in $B$ to a loop of the form $u.v.\overline{u}.\overline{v}$, where $u$ and $v$ represent the two generators of $F_2$. So $\pi_1(T) \cong \langle a,b ; aba^{-1}b^{-1} \rangle$, the free abelian group on two generators.
Any explanation or geometric intuition would be really appreciated. Thanks