I was trying to compute the fundamental group of torus by forcing myself of using the Seifert Van Kampen theorem. One knows that the answer is the direct product of $ \mathbb{Z} $ and $ \mathbb{Z} $.
But from the way I tried to solve it something goes wrong and I can't find out what it is.
Here's what I do:
First I remove the Circle $AB$ from the torus which gives me an open set $X_1$.
For the second one I remove the circle $BC$ from the torus instead of the other one which gives me another open set $X_2$.
Now $X_0:=X_1 \cap X_2$ is the torus without those two circles, which is a simply connected subset of torus.
Now if applying the Seifert Van Kampen theorem to these path connected subsets of the torus you'll get
$\pi_1(\mathbb{T}) = \mathbb{Z}*\mathbb{Z}$
while writing this I realized something that solves the fundamental group problem I asked. However I'm still interested to know the answer of the question below:
Is there a free product with amalgamation of $\Bbb Z*\Bbb Z$ isomorphic to the direct product of $ \mathbb{Z} $ and $ \mathbb{Z} $?
Best Answer
As mentioned by OP in comments, the question asks: Is there a group $G$ which is an amalgam of the form $$ A\star_{C} B, $$ with the factors $A\cong B\cong {\mathbb Z}$, such that $G$ is isomorphic to ${\mathbb Z}^2$?
This question has a negative answer.
Proof. Suppose, to the contrary, that there exists an isomorphism $$ \phi: A\star_{C} B\to {\mathbb Z}^2. $$
There are two cases to consider.
a. $C$ is a nontrivial group. Then the images $\phi(A), \phi(B)$ are infinite cyclic subgroups of ${\mathbb Z}^2$ which have nontrivial intersection $\phi(C)$. But any two cyclic subgroups with nontrivial intersection in ${\mathbb Z}^n$ generate a cyclic subgroup. (I leave it as an exercise.)
This would imply that ${\mathbb Z}^2$ is cyclic (since $\phi$ is surjective). The latter is obviously false. (Again, a linear algebra exercise.)
b. $C$ is the trivial subgroup. Then $G\cong A\star B\cong F_2$, the free group on two generators. But a free group on two generators (say, $a, b$) is nonabelian, for instance, since the word $aba^{-1}b^{-1}$ is reduced and nonempty, hence, represents a nontrivial element of $F_2$ (by one of the equivalent definitions of $F_2$). qed
Remark. Incidentally, an amalgam of the form $A\star_{C} B$ with $A, B, C$ infinite cyclic, can contain ${\mathbb Z}^2$ as an index two subgroup. I leave this as an exercise in algebraic topology of surfaces.