Quotients of the maximal tensor product

Question

Let $A$ and $B$ be C*-algebras and let $\gamma$ be any C*-norm on the algebraic tensor product $A\odot B$. Why is $A\otimes_\gamma B$ a quotient of $A\otimes_{{\rm max}}B$, where $\otimes_{{\rm max}}$ stands for the maximal tensor product of C*-algebras? How does this quotient map look like?

Answer 1

By definition of the max norm, you have $$ \|x\|_\gamma\leq\|x\|_\max $$ for all $x$ in $A\odot B$. This tells you that the identity map $A\odot B\to A\odot B$ is a continuous function when you consider $\|\cdot\|_\max$ in the domain, and $\|\cdot\|_\gamma$ in the codomain. This map, call it $\pi$, can be extended to the completion, since given a Cauchy sequence $\{x_n\}$ we have $$ \|\pi x_n-\pi x_m\|=\|\pi(x_n-x_m)\|\leq \|\pi\|\,\|x_n-x_m\|, $$ so $\{\pi(x_n)\}$ is also Cauchy. The extension $\pi$ is also a $*$-homomorphism, since it is on a dense subset of $A\otimes_\max B$. So we have obtained a canonical $*$-homomorphism $$ \pi:A\otimes_\max B\to A\otimes_\gamma B. $$ By the First Isomorphism Theorem, $A\otimes_\gamma B$ is isomorphic to a quotient of $A\otimes_\max B$ (namely, its quotient by the kernel of $\pi$).

An explicit form of the quotient map would require having explicit forms of the completions $A\otimes_\max B$ and $A\otimes_\gamma B$ of $A\odot B$. I don't think such a thing exists in any meaninful example, and it is what it makes dealing with tensor products an art.