Direct sum of standard Hilbert A-modules

Question

For any $C^\ast$-algebra $A$, don't we have the isomorphism $\mathcal{H}_A \oplus \mathcal{H}_A \cong \mathcal{H}_A$? In some places, it is only mentioned when $A$ is $\sigma$-unital. But $\mathcal{H}_A \cong A \otimes l^2$, and since tensor product commutes with direct sums, we should be getting $(A \otimes l^2) \oplus (A \otimes l^2) \cong A \otimes (l^2 \oplus l^2) \cong A \otimes l^2$. What's wrong here?

Answer 1

There is no mistake. The result holds for general $C^*$-algebras, as you proved.

Answer 2

Nothing wrong and your proof shows that the hypothesis $A$ is $\sigma$-unital is useless (you could complete your algebraic proof whith a little argument justifying why the isomorphism goes over to the completions).

You will find a related theorem, the "Stabilization or Absorption Theorem", in Bruce Blackadar's K-Theory for Operator Algebras: (for any $C^*$-algebra $B$),

"The theorem in this form is due to Kasparov [...]

Theorem 13.6.2. If $E$ is a countably generated Hilbert $B$-module, then $E\oplus H_B\cong H_B.$"

and

"Exercise 13.7.1 (a) Use the stabilization theorem to prove that if $B$ is $\sigma$-unital and $E$ is a countably generated full Hilbert $B$-module, then $E^\infty\cong H_B.$" ($E^\infty:=E\otimes l^2$, and "full" means $\langle E,E\rangle=B$.)

Enlightning comment below by @MaoWao: "I think the stabilization theorem is the reason why some authors state this result [$\mathcal{H}_A \oplus \mathcal{H}_A \cong \mathcal{H}_A$] only for $σ$-unital $C^∗$-algebras, because only then it is a direct consequence."