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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.

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.

If $\mathfrak S$ denotes the set of all non-zero C$^*$-algebras (up to $*$-isomorphism) of some bounded cardinality, for instance separable, then $(\mathfrak S, \otimes_\textrm{min})$ and $(\mathfrak S, \otimes_\textrm{max})$ are abelian semigroups with identity element $\mathbb C$ (or commutative monoids if you prefer that language) since both tensor products are commutative and associative.

Question 1: Is there any cancellation in these semigroups? Does there exist $\mathcal C\in \mathfrak S$ not isomorphic to $\mathbb C$ such that for all $A,B\in \mathfrak S$ we have $$ \mathcal A \otimes_\textrm{min} \mathcal C \simeq \mathcal B\otimes_\textrm{min} \mathcal C \Rightarrow \mathcal A \simeq \mathcal B $$ or the similar statement with $\otimes_\textrm{max}$?

One could start by asking the following:

Question 2: 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 $\mathcal A\simeq \mathcal B$?

This second question is an extension of this question.

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.

If $\mathfrak S$ denotes the set of all non-zero C$^*$-algebras up to $*$-isomorphism, then $(\mathfrak S, \otimes_\textrm{min})$ and $(\mathfrak S, \otimes_\textrm{max})$ are abelian semigroups with identity element $\mathbb C$ (or commutative monoids if you prefer that language) since both tensor products are commutative and associative.

Question 1: Is there any cancellation in these semigroups? Does there exist $\mathcal C\in \mathfrak S$ not isomorphic to $\mathbb C$ such that for all $A,B\in \mathfrak S$ we have $$ \mathcal A \otimes_\textrm{min} \mathcal C \simeq \mathcal B\otimes_\textrm{min} \mathcal C \Rightarrow \mathcal A \simeq \mathcal B $$ or the similar statement with $\otimes_\textrm{max}$?

One could start by asking the following:

Question 2: Let $\mathcal A,\mathcal B\in \mathfrak S$ such that for all $\mathcal C\in \mathfrak S$ non-isomorphic to $\mathbb C$
$$\mathcal A\otimes_\textrm{min} \mathcal C \simeq \mathcal B\otimes_\textrm{min} \mathcal C $$ Does this imply $\mathcal A\simeq \mathcal B$?

This second question is an extension of this question.

If $\mathfrak S$ denotes the set of all non-zero C$^*$-algebras (up to $*$-isomorphism) of some bounded cardinality, for instance separable, then $(\mathfrak S, \otimes_\textrm{min})$ and $(\mathfrak S, \otimes_\textrm{max})$ are abelian semigroups with identity element $\mathbb C$ (or commutative monoids if you prefer that language) since both tensor products are commutative and associative.

Question 1: Is there any cancellation in these semigroups? Does there exist $\mathcal C\in \mathfrak S$ not isomorphic to $\mathbb C$ such that for all $A,B\in \mathfrak S$ we have $$ \mathcal A \otimes_\textrm{min} \mathcal C \simeq \mathcal B\otimes_\textrm{min} \mathcal C \Rightarrow \mathcal A \simeq \mathcal B $$ or the similar statement with $\otimes_\textrm{max}$?

One could start by asking the following:

Question 2: 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 $\mathcal A\simeq \mathcal B$?

This second question is an extension of this question.