I'm studying Seifert Van Kampen theorem and I evaluate the fundamental group of torus seen as a quotient space in particular $$T = [0,1]\times[0,1]/\sim\\ (x,0)\sim(x,1) \\ (0,y)\sim(1,y)$$ With $\pi:[0,1]\times[0,1]\to [0,1]\times[0,1]/\sim$ the projection on the quotient. I choose $A= \pi([0,1]\times[0,1]-\{p\})$ where $p\in [0,1]\times[0,1]$ and $B$ is the image of the interior of $[0,1]\times[0,1]$ trough $\pi$. All the hypothesis of SVK are respected and I think that in the end I'm convinced saying that $$\pi_1(T) =\langle [a],[b]|aba^{-1}b^{-1}\rangle$$ where $[a],[b]$ are the generator of $A$ since $B$ is the trivial group. Here is where I get confused, it's been told me that $$\pi(T) \cong\mathbb{Z}\times\mathbb{Z}$$ Why this isomorphism holds? It's because the only abelian free group generated by two elements is the cartesian product? Because the relation I write indicates that the operation is component-wise? There are others ideas that can help understand this isomorphism and lastly is it possible to write down this isomorphism explicitly? Sorry for what maybe it's a dumb question. Thank you for the help!
P.S. I found a question that talks about this topic and mention abelianization of a free group, I'm not very familiar with this concept, is this the key point to understand what I ask above?
You need to understand what it means for a group $G$ to have a presentation $\langle S|R\rangle$, where $S$ is a set of generators and $R$ is a set of relations. We first need to construct the free group $F$ generated by $S$. Then $R$ is an arbitrary subset of $F$, and $G$ is isomorphic to the quotient of $F$ by the smallest normal subgroup $N$ of $F$ containing $R$. In particular this means that $G$ is generated by the image $S$ under this isomorphism. We write $S$ to denote the image of $S$ in $G$ (this, of course, is an abuse of lenguaje).
Lets explain this in terms of universal properties: Each element of $R$ is of the form $s_1^{a_1}\cdots s_n^{a_n}$ for some $s_j\in S$ and $a_j\in\mathbb{Z}$. Then the above means that for every group $H$ and every map $f:S\to H$ such that $$ f(s_1)^{a_1}\cdots f(s_n)^{a_n} $$ for every relation in $R$, there exists an unique group homomorphism $f':G\to H$ such that $f'(s)=f(s)$ (recall that we identify the elements of $S$ with its image in $G$ under the isomorphism $F/N\to G$).
Now, in the present case, we have two generators $a$ and $b$, and one relation $aba^{-1}b^{-1}$. This means that for every group $H$ and every function $f:\{a,b\}\to H$, such that $$ f(a)f(b)f(a)^{-1}f(b)^{-1} = 1 $$ then $f$ extends uniquely to a group homomorphism $f':\pi_1(T)\to H$.
Take $H=\mathbb{Z}\times \mathbb{Z}$ and the map $f:\{a,b\}\to H$ given by $$ f(a)=(1,0), \quad f(b)=(0,1), $$ then the condition $f(a)f(b)f(a)^{-1}f(b)^{-1} = 1$ is clearly satisfied and hence we obtain a map $f':\pi_1(T)\to \mathbb{Z}\times \mathbb{Z}$ such that $f'(a)=(1,0)$ and $f'(b)=(0,1)$.
Conversely, let $g:\mathbb{Z}\times \mathbb{Z}\to \pi_1(T)$ be given by $$ g(p,q) = a^p b^q. $$ $g$ is a group homomorphism (you can check it, you will need to use the fact that $aba^{-1}b^{-1}=1$, which is the same as to say that $ab=ba$). Then you can verify that $g$ is an inverse for $f'$ and consequently $f'$ is a group isomorphism.