Let $A$ and $B$ be $C^*$-algebras and let $\alpha$ be any $C^*$-norm on the algebraic tensor product $A⊙B$. Why is $A⊗_{\alpha}B$ is the subalgeba of $A\otimes_{max} B$?
In many reference books,they mentioned that there was a natural surjective * homomorphism from $A\otimes_{max} B$ to $A⊗_{\alpha}B$ ,which reveals that $A⊗_{\alpha}B$ is the subalgeba of $A\otimes_{max} B$.
According to the definition,$A\otimes_{max} B$ denote the $||.||_{max}$, $A⊗_{\alpha}B$ is the completion of $A⊙B$ with respect to $\alpha$,$||.||_{max}$ is the maximal $C^*$-norm,I prove that $A\otimes_{max} B$ is the subalgebra of $A⊗_{\alpha}B$(refer to Completion with respect to stronger norm is no subset?) .
Can anyone point out my mistake?Thanks in advance!
It isn't. The existence of a $*$-epimorphism $\phi:A\to B$ does not tell you that $B$ can be seen as a subalgebra of $A$; only as quotient, which is most often not a subalgebra.
To see an easy example, let $A=C_0(\mathbb R)$, $B=\mathbb C$. Then you have for instance the $*$-epimorphism $\phi:A\to B$ given by $\phi(f)=f(0)$, but no subalgebra of $C_0(\mathbb R)$ is isomorphic to $\mathbb C$.