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From the negative answer to this question we know that C$^*$-algebras that are isomorphic after tensoring with $M_n$ for all $n\geq 2$ need not be isomorphic. So what happens when we strengthen this?

In the following, if $\mathcal A\subset B(\mathcal H)$ and $\mathcal B\subset B(\mathcal K)$ then the minimum tensor product $\mathcal A\otimes_\textrm{min} \mathcal B$ is the closure of the algebraic tensor product $\mathcal A\odot \mathcal B$ in the $B(\mathcal H\otimes \mathcal K)$ norm.

Question 1: Let $\mathcal A,\mathcal B$ be C$^*$-algebras such that for all C$^*$-algebras $\mathcal C$ non-isomorphic to $\mathbb C$
$$\mathcal A\otimes_\textrm{min} \mathcal C \simeq \mathcal B\otimes_\textrm{min} \mathcal C $$ Does this imply $\mathcal A\simeq \mathcal B$?

Secondly, can one accomplish this with a single C$^*$-algebra.

Question 2: Does there exist a C$^*$-algebra $\mathcal C$ not isomorphic to $\mathbb C$ such that for all C$^*$-algebras $A,B$ we have $$ \mathcal A \otimes_\textrm{min} \mathcal C \simeq \mathcal B\otimes_\textrm{min} \mathcal C \ \Rightarrow \ \mathcal A \simeq\mathcal B\ ? $$

One can also ask these questions with the max tensor product.

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    $\begingroup$ That "set" is in fact a proper class because there are $C^\ast$-algebras of arbitrarily large cardinality. You could introduce a bound on the cardinality to work around that. $\endgroup$ – Johannes Hahn Aug 22 '15 at 20:39
  • $\begingroup$ @JohannesHahn Of course you are right, I'll fix that. $\endgroup$ – Chris Ramsey Aug 22 '15 at 22:57
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I think the solution could be $\mathcal{C}= \mathbb{C} \oplus \mathbb{C}$. Then you should show $A \oplus A \cong B \oplus B$ ($=B \otimes \mathcal{C}$) implies $A \cong B$.

Let $\pi:A \oplus A \rightarrow B \otimes B$ be the isomorphism. Consider the 4 projections $p_1:= \pi(1 \oplus 0)$, $p_2:=\pi(0 \oplus 1)$, $q_1 := 1 \oplus 0$, $q_2 := 0 \oplus 1$ in the center of (the multiplier algebra of) $B \oplus B$.

Note that $\pi(A \oplus 0) = p_1 \cdot (B \oplus B)$. By refining those 4 projections you see that $$B \oplus B = (W \oplus X) \oplus (Y \oplus Z)$$ where $A \cong X\oplus Y \cong W \oplus Z$, the two copies of $A$.

Hence $W \oplus X = Y \oplus Z$. Even if the sitiuation is as in the beginning, by refining you could proceed further.

It appears to me that at the end $A \cong B$, as the double of $A$ is the double of $B$ means that $A$ has the same components as $B$ under the infinite refinements.

ED:

Even it is wrong, let me still give an essentially same but more transparent argument:

Write $A$ as an (infinite) direct sum $A = \bigoplus_i A_i$, where each $A_i$ has no further direct summands (=inessential ideals). Similarly $B=\bigoplus_j B_j$. Then it is clear that $A$ and $B$ share the same components (=summands).

Just as I said above.

ED:ED:

It is clear that it is not possible to write $A$ as a direct sum as claimed in general.

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    $\begingroup$ I don't think this works. According to the answer to this question there exist non-homeomorphic compact Hausdorff spaces $X$ and $Y$ such that $X\coprod X \cong Y\coprod Y$. $\endgroup$ – Nik Weaver May 18 '18 at 18:48
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    $\begingroup$ See also this question. $\endgroup$ – Nik Weaver May 18 '18 at 19:01

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