Given two C*-algebras $A$ and $B$, the maximal tensor product $A\otimes_{max}B$ is *bigger* than the minimal tensor product $A\otimes_{min}B$ in the sense that there exists an epimorphism $$A\otimes_{max}B \to A\otimes_{min}B,$$ which is the identity map on the respective copies of the algebraic tensor product $A\otimes_{alg}B$.

The *free product* $A{*}B$ is even bigger than $A\otimes_{max}B$ according to the exact same criteria described above (notice that $A{*}B$ contains a cannonical copy of the *vector space* $A\otimes_{alg}B$).  This is of course due to the fact that, within the free product, the elements of $A$ are not required to commute with those of $B$.

My question is whether $A\otimes_{max}B$ is still maximal if we weaken the condition of commutativity of *elements* by replacing it with the commutativity of *sets*.

> Question:  Suppose that $C$ is a C*-algebra containing copies of $A$ and $B$, such that $C$ coincides with the closed linear span of $$AB = \{ab:a\in A,\ b\in B\}$$ as well as that of $BA$.  In symbols $C=\overline{\hbox{span}}\,AB=\overline{\hbox{span}}\,BA$.  Suppose moreover that there exists a *-homomorphism $$\varphi:C\to A\otimes_{max}B$$ such that $\varphi(ab) = a\otimes b$, for all $a$ in $A$, and all $b$ in $B$.  Is $\varphi$ necessarily an isomorphism?