Trying to compute $\operatorname{kernel}$ of the map $\pi\otimes \rho: A \otimes^{\text{min}} B \to A/I \otimes^{\text{min}} B/J$

Question

Let $A$ and $B$ be $C^{\ast}$-algebras and $I$ and $J$ be ideals of $A$ and $B$ respectively. Let $\pi: A \to A/I$ and $\rho: B\to B/J$ be quotient maps. Let $\pi\otimes \rho: A \otimes^{\text{min}} B \to A/I \otimes^{\text{min}} B/J$ be $C^{\ast}$-homomorphism. Moreover, assume that $A$ or $B$ is exact, then is the following true?

$\operatorname{ker}(\pi \otimes \rho)= I \otimes^{\text{min}} B+ A \otimes^{\text{min}} J$

Clearly, $I \otimes^{\text{min}} B+ A \otimes^{\text{min}} J \subseteq \operatorname{ker}(\pi \otimes \rho)$. Also, it is clear that the sequence $ I\otimes^{\text{min}} B \to A \otimes^{\text{min}}B \to A/I \otimes^{\text{min}} B$ is exact as $B$ is exact. How to proceed further?

Answer 1

The following is a deep result, recorded as corollary 9.4.6 in Brown Ozawa:

If $A$ is an exact $C^*$-algebra and $B$ is any $C^*$-algebra and $I\subset A\otimes B$ is an ideal, then $$I=\overline{\text{span}}\{a\otimes b: a\otimes b\in I\}.$$

Now in our setting $\ker(\pi\otimes\rho)$ is an ideal. I claim that if $a\in A$ and $b\in B$ are such that $a\otimes b\in\ker(\pi\otimes\rho)$, then $a\otimes b\in (A\otimes J)\cup(I\otimes B)\subset A\otimes J+I\otimes B$. We prove this by contradiction:

Assume that $a\not\in I$ and $b\not\in J$. Then $\pi(a)\ne0$ and $\rho(b)\ne0$. Now norms on the tensor product are cross norms, so $$\|\pi\otimes\rho(a\otimes b)\|=\|\pi(a)\otimes\rho(b)\|=\|\pi(a)\|\cdot\|\rho(b)\|\ne0$$ and thus $\pi\otimes\rho(a\otimes b)\ne0$, i.e. $a\otimes b\not\in\ker(\pi\otimes\rho)$, which is a contradiction.

The above fact then yields that $\ker(\pi\otimes\rho)$ is the closed linear span of elements of $A\otimes J+I\otimes B$, thus $\ker(\pi\otimes\rho)\subset A\otimes J+I\otimes B$ as we wanted.