Proving the isomorphism $A \otimes B \cong B\otimes A$ of the tensor products of abelian groups $A,B$ given the definition by the quotient groups. 
For two Abelian groups $A$ and $B$ we define their tensor product
$A\otimes B$ as the quotient of the free Abelian group on the set of
formal generators $\{a \otimes b \mid a \in A; b \in B\}$ by the
subgroup generated by elements of the form $$a_1 \otimes b + a_2
 \otimes b − (a_1 + a_2) \otimes b$$ and $$a\otimes b_1 +a\otimes b_2
 −a\otimes(b_1 +b_2).$$ By abuse of notation we write $a\otimes b$ for
the corresponding element in the quotient $A \otimes B.$

I'd like to prove that $A\otimes B \cong B\otimes A$. Now my first thought was using the map $$a\otimes b \mapsto b \otimes a$$
Now it's obvious this is compatible with the relations. But I don't know how the quotients $A\otimes B$ and $B\otimes A$ become isomorphic. According to the solution, "it descends to the quotients", but I don't know how. Could someone please elaborate how we precisely get the isomorphism??
 A: I'll use more convenient notations: in the free abelian group $\mathbf Z^{(X)}$ generated by a set $X$, I'll denote $[x]$ the element $e_x$, i.e. the map which sends $x$ to $1$ and any $x'\ne x$ to $0$.
This being said, you have a bijective map from $\mathbf Z^{(A\times B)}$ to $\mathbf Z^{(B\times A)}$, which sends $[(a,b)]$ to $[(b,a)]$. This map sends generators of the relations defining the tensor product in the first free group, namely $[(a_1,b)]+[(a_2,b)]-[(a_1+a_2,b)]$ onto the generators of the relations defining the tensor product in the second free group,  $[(b,a_1)]+[(b,a_2)]-[(b,a_1+a_2)]$, hence the subgroup  $R_{A\times B}$ generated by the first set onto the subgroup $R_{B\times A}$ generated by the second group.
Therefore we have a commutative diagram of abelian groups
\begin{alignat}{5}
0\longrightarrow &R_{A\times B}\hookrightarrow&&\mathbf Z^{A\times B}\longrightarrow A\otimes B\longrightarrow 0 \\
&\quad\downarrow&&\enspace\downarrow\\
0\longrightarrow &R_{B\times A}\hookrightarrow&&\mathbf Z^{A\times B}\longrightarrow B\otimes A\longrightarrow 0 
\end{alignat}
which induces a morphism from $A\otimes B$ to $B\otimes A$ by the universal property of kernels. As the the two vertical maps are group isomorphisms, the induced morphism  is an isomorphism too.
A: Let $A\widetilde{\otimes} B$ denote the free abelian group on symbols $a\otimes b$. We have, by the universal property of free abelian groups, a group homomorphism $\tilde{f}:A\widetilde{\otimes} B \to B\widetilde{\otimes} A$, given on generators by $a\otimes b\mapsto b\otimes a$.
Let $I$ be subgroup in $A\widetilde{\otimes} B$ generated by expressions $a_1 \otimes b + a_2
 \otimes b − (a_1 + a_2) \otimes b$ and $a\otimes b_1 +a\otimes b_2
 −a\otimes(b_1 +b_2)$,  with $a,a_i\in A,b\in B$. Define the subgroup $J\subset B\tilde{\otimes} A$ in a similar way.
As you noted, if $x\in I$ then $\tilde{f}(x)\in J$, so $\tilde{f}$ induces  a homomorphism $$f:A\otimes B = A\widetilde{\otimes}B/I\to B\widetilde{\otimes}A/J= B\otimes A,$$
given by $f(x+I)=\tilde{f}(x)+J$.
In a similar way, you have a homomorphism $g$ in the other direction, given by $g(y+J)=\tilde{g}(y)+I$.
It remains to check that $f$ and $g$ are inverses of each other. That's clear, as on generators $$(g\circ f)(a\otimes b +I)=g(b\otimes a +J)=a\otimes b +I.$$
The same for $f\circ g$. This completes the proof.
