If A is an associative k-algebra, and $A^{\text{op}}$ represents the opposite k-algebra (i.e. $a*_{A^{\text{op}}} b := b\cdot a)$.
In the following we consider ${\cal A_1}=A$ as a right $A\otimes A^{op}$ bimodule and ${\cal A_2}=A$ as a left $A\otimes A^{op}$ bimodule. I want to prove the following isomorphism of $k-k$ bimodules (aka vector spaces): $$\large {\cal A_1} \otimes_{A \otimes A^{\text{op}}} {\cal A_2} \cong A/[A,A].$$
Reference and context: Page 17 of this article. I am reading articles on TQFT with the aim of understanding extended TQFTs. I hope I have presented the question faithfully from the linked article.
My thoughts: I am generally not comfortable with $A^{op}$ stuff. I know that a left $A \otimes A^{op}$ module is $A-A$ bimodule. And similarly right $A \otimes A^{\text{op}}$ module is a left $(A \otimes A^{op})^{op} = A^{\text{op}} \otimes A$ module.
The members of the bimodule ${\cal A_1} \otimes_{A \otimes A^{\text{op}}} {\cal A_2} =: \mathfrak{A}$ are generated by elements of the form $\alpha \otimes \beta$ with the condition $$\alpha \otimes (a\cdot \beta \cdot b) = (b \cdot \alpha \cdot a) \otimes \beta$$ for all $a,b,\alpha, \beta \in A$.
I see that the constraints force $$a \otimes b = 11a \otimes b = 1 \otimes ab1 = 1 \otimes 1ab = b \otimes a.$$
In order to prove the isomorphism I could think of two maps:
1) Consider the map $A \to \mathfrak{A}$ given by:
$$a \mapsto a \otimes 1.$$
2) Consider another map $\mathfrak{A} \to A/[A,A]$ given by:
$$a \otimes b \mapsto [ab].$$ Here $[x]$ represents the coset of an element $x (\in A)$ in the quotient.
I don't know how to proceed. Please help.
You have the right idea; to prove that $A/[A,A]$ and $\mathfrak{A}$ are isomorphic, it suffices to define homomorphisms $f : A/[A,A]\to\mathfrak{A}$ and $g : \mathfrak{A}\to A/[A,A]$ such that $gf = \operatorname{id}_{A/[A,A]}$ and $fg = \operatorname{id}_{\mathfrak{A}}.$
To get the map $f : A/[A,A]\to\mathfrak{A},$ it suffices to check that $\tilde{f}$ factors through the canonical projection $\pi : A\to A/[A,A].$ To do this, we may show that $\tilde{f}$ vanishes on generators of $[A,A].$ To that end, let $a,b\in A.$ Then \begin{align*} \tilde{f}(ab - ba)&= (ab - ba)\otimes 1\\ &= (ab)\otimes 1 - (ba)\otimes 1\\ &= (ab\cdot 1)\otimes 1 - (ba\cdot 1)\otimes 1\\ &= b\otimes (1\cdot 1\cdot a) - a\otimes (1\cdot 1\cdot b)\\ &= b\otimes a - a\otimes b\\ &= 0, \end{align*} so that $\tilde{f} : A\to\mathfrak{A}$ indeed factors as $A\xrightarrow{\pi} A/[A,A]\xrightarrow{f}\mathfrak{A}.$
The induced map $f : A/[A,A]\to\mathfrak{A}$ is given by \begin{align*} f : A/[A,A]&\to\mathfrak{A}\\ [a]&\mapsto a\otimes 1, \end{align*} so we readily check that the composition $A/[A,A]\xrightarrow{f}\mathfrak{A}\xrightarrow{g} A/[A,A]$ is the identity, where $g$ is the map you defined in the question.
It remains to be shown that the composition $\mathfrak{A}\xrightarrow{g} A/[A,A]\xrightarrow{f}\mathfrak{A}$ is also the identity. It suffices to show that this holds for simple tensors. This composition is given on simple tensors by \begin{align*} fg(a\otimes b) &= f([ab])\\ &= ab\otimes 1\\ &= b\otimes a\\ &= a\otimes b, \end{align*} so that $fg = \operatorname{id}_{\mathfrak{A}},$ as desired. Thus, $\mathfrak{A}\cong A/[A,A].$