At first a definition of injective (unital) $C^*$-algebras: A unital $C^*$-algebra $C$ is called injective if for all completely positive contractive (cpc, for short) maps $\varphi:X\to C$, where $X\subseteq A$ is an operator system in a unital $C^*$-algebra $A$, there exists a cpc map $\overline{\varphi}:A\to C$ such that $\overline{\varphi}\circ i_X =\varphi$ (here $i_X$ denotes the inclusion $i_X:X\to C$).

Even though it is a well-known fact that quotiens of nuclear $C^*$-algebras are nuclear, I'm searching for alternative (preferably more elementary) proofs of this fact without using biduals of $C^*$-algebras. More precisely, the setting is: Given a short exact sequence of $C^*$-algebras $$0\to J \xrightarrow{\text{i}} A\xrightarrow{\pi} A/J\to 0$$ with $J$ and $A$ nuclear. I want to proof that $A/J$ is nuclear .

The proof which I know can be found in Blackadar, "operator algebras- theory of $C^*$-algebras and von Neumann algebras", corollary IV.3.1.13 and uses that $A$ is nuclear iff $A^{**}$ is injective and that $A^{**}\cong J^{**}\oplus (A/J)^{**}$. The strategy is then to check that $(A/J)^{**}$ is injective (without details): you have an embedding $i_{A/J}:(A/J)^{**}\to A^{**}$ and given a cpc map $\varphi:X\to (A/J)^{**}$ ($X$ is an operator system in a unital $C^*$-algebra $D$) there exists a cpc map $\gamma:D\to A^{**}$ such that $\gamma\circ i_X=i_{A/J}\circ \varphi$ (here is $i_x:X\to D$ the inclusion). Finally $\gamma$ induces $\overline{\varphi}$ as desired, so that $(A/J)^{**}$ is injective.

Now, I'm searching for a different proof (as I said). For instance I tried to prove it with the five lemma. For this I tried to proof that $$0\to J\otimes_{min}D \xrightarrow{i\otimes id_D} A\otimes_{min}D \xrightarrow{\pi\otimes id_D} A/J\otimes_{min}D \to 0$$ is exact for arbitrary $C^*$-algebras $D$. Hoewever, I don't know how to proof that $\ker(\pi\otimes id_D)\subseteq im(i\otimes id_D)$.

Or if you know a proof that $id:A/J\to A/J$ is nuclear, it would be fine as well.

I appreciate your help, regards.

  • 3
    $\begingroup$ My impression has always been that there are no "very easy" proofs (there are some comments on this in Fillmore's A User's Guide To Operator Algebras). I can't resist the comment that in contrast, it is very easy to show that quotients of amenable algebras are amenable... $\endgroup$
    – Yemon Choi
    Oct 27, 2016 at 15:01


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