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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!

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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 $\mathbb C$.

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  • $\begingroup$ There was a natural surjective * homomorphism from $A\otimes_{max} B$ to $A⊗_{\alpha}B$,can we deduce that $A\otimes_{max} B$ is larger than $A⊗_{\alpha}B$? $\endgroup$ – math112358 May 13 '18 at 9:19
  • $\begingroup$ You would have to tell me what "larger" means. $\endgroup$ – Martin Argerami May 13 '18 at 13:56
  • $\begingroup$ I mean $ A\otimes_\gamma B$ is the subset of $A\otimes_{max} B$? I prove that $A\otimes_{max} B$ is the subset of $ A\otimes_\gamma B$ above,Can you point out my mistake?Thanks! $\endgroup$ – math112358 May 13 '18 at 14:17
  • $\begingroup$ I cannot point a mistake because I see no proof. But it is just not true, as the answer to the question you quoted shows. A Cauchy sequence for $\|\cdot\|_\alpha$ needs not be Cauchy for $\|\cdot\|_\max$. $\endgroup$ – Martin Argerami May 13 '18 at 14:25
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    $\begingroup$ In an algebra where both $\|\cdot\|_\alpha$ and $\|\cdot\|_\infty$ are defined, yes. $\endgroup$ – Martin Argerami May 17 '18 at 12:21

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