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Let $A$ a C*-algebra (unital or not). Its suspension is defined to be: $$ S(A) \equiv A\otimes C_0((0,1);\,\mathbb{C}) $$where $C_0$ denotes all continuous functions which vanish at infinity.

We know that if $X$ is locally compact and if $X^{+}$ is its one-point compactification, then $$ \widetilde{C_0(X;\,\mathbb{C})}\cong C(X^{+};\,\mathbb{C})$$

My question is:

1) Given two C*-algebras $A$ and $B$ (unital or not), is it true that $$ \widetilde{A\otimes B} \cong \widetilde{A}\otimes\widetilde{B} $$?

2) Does it then follow that $$ \widetilde{S(A)}\cong \widetilde{A}\otimes C(S^1;\,\mathbb{C})$$?

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  • $\begingroup$ This seems to be false. See Wegge-Olsen pp. 136. $\endgroup$ – PPR May 29 '16 at 10:50
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As noted in the comments, this is false. Intuitively, on the left-hand side in 1) you add one point to both spaces and then take the cartesian product whereas on the right-hand side you first take the product and then add one point.

For example let $A=B=C_0((0,1))$. Then $\widetilde{A\otimes B}=\widetilde{C_0((0,1)\times (0,1))}=C(S^2)$ and $\tilde A\otimes \tilde B=C(S^1)\otimes C(S^1)=C(\mathbb{T}^2)$.

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