Why is $B\odot A\cap B\otimes_{min}J=B\odot J$?

Question

Let $A, B$ $C^*$-algebras, $J$ a closed ideal in $A$. Let $B\otimes_{min}J=\overline{B\odot J}^{\|\enspace \|_{min}}$, where $B\odot J$ denotes the tensor product as $*$-algebras and $\|\enspace \|_{min}$ is the spatial tensor norm on $B\odot J$.

My question: Is it true that

1. $B\odot A\cap B\otimes_{min}J=B\odot J$,

and the same for the maximal tensor norm

2. $B\odot A\cap B\otimes_{max}J=B\odot J$?

At first I only want to discuss 1., maybe 2. is similar and then to discuss 1. will be enough for me. The inclusion $B\odot A\cap B\otimes_{min}J\supseteq B\odot J$ is clear. But I'm stuck to proof $\subseteq$.

The beginning of the proof should be: If $x\in B\odot A\cap B\otimes_{max}J$, then $x$ is a sum of elementary tensors, i.e. $x=\sum\limits_{i=1}^n b_i\otimes a_i$, and there is a sequence $(z_m)_m\subseteq B\odot J$ with $\|x-z_m\|_{min}\to 0$ for $m\to\infty$.

Maybe the following calculations/ideas can be used for a proof?:

For $n\in\mathbb{N}$, $z_n$ looks like $z_m=\sum\limits_{l=1}^k b_l^{(m)}\otimes j_l^{(m)}$ and we can set $k=n$ and $b_l^{(m)}=b_l$ for all $m\in \mathbb{N}$ .

If $\pi:A\to B(H_{\pi})$ and $\sigma:B\to B(H_{\sigma})$ are two faithful $*$-representations, then $$\|x-z_m\|_{min}=\|\sum\limits_{l=1}^n \sigma(b_l)\otimes \pi (a_l-j_l^{(m)})\|_{op}\le \sum\limits_{l=1}^n \|\sigma(b_l)\|_{op}\|a_l-j_l^{(m)}\|_A$$ and $a_l\in J$ because $J$ is closed in $A$.

But I'm sure that this is not the proof, because I don't use $\|x-z_m\|_{min}\to 0$ explicitly and I don't know, if $\sum\limits_{l=1}^n \|\sigma(b_l)\|_{op}\|a_l-j_l^{(m)}\|_A\to 0$ for $m\to \infty$. I just wanted to inform you about my experiments. So, my question is

How to prove $\subseteq $ correctly?

Edit: I found a similar statement in the book "C$^*$-algebras and finite-dimensional approximations" by Nathe Brown and Ozawa, Lemma 3.7.7. It's for the minimal tensor product and I only get that $x\in B\otimes_{min}J$ if I see it correctly that it's possible to prove 1 similary as in the book.

Is it correct that I can conclude $x\in B\otimes_{min}J$ as in the book?

I'm not sure if I can conclude $x\in B\odot J$ with the argumentation that $x$ is a sum of elementary tensors $\sum\limits_{i=1}^n b_i\otimes a_i$.

Answer 1

Well, the second equation is true and follows from the universal property of the tensor products for algebras, $\ast$-algebras and the maximal tensor product of $C^\ast$-algebras: $$Hom(A \odot B, X) \cong \{ (f,g) \mid f: A\to X, g: B\to X, \forall a,b: [f(a),g(b)] = 0 \}$$ (explicitely the natural isomorphism maps $\phi: A\odot B\to X$ to $f(a):=\phi(a\otimes 1)$ and $g(b):=\phi(1\otimes b)$ and the pair $(f,g)$ to $\phi(a\otimes b):=f(a)g(b)$).

Dealing with ideals means dealing with kernels of morphisms. So the first step is to recognize that $J\otimes B$ is the kernel of $A\otimes B\to A/J\otimes B$ which follows from the universal property of the tensor product combined with the universal property of quotients. You just have to realise that the canonical map $A\otimes B\to A/J\otimes B$ corresponds via the universal property to the pair of morphisms consisting of the quotient map $A\to A/J$ and $id_B$.

Once one knows that we need to show that the kernel of $A\otimes B\to A/J\otimes J$ restricted to $A \odot B$ is exactly $J\odot B$. But using the universal property we realise that the restricted map also corresponds to the quotient map $A\to A/J$ and $id_B$, which means that it is basically the same map as $A\odot B\to A/J\odot B$ (in fact it is the same as that map composed with the inclusion $A/J\odot B\to A/J\otimes B$) and therefore is has the same kernel, namely $J\odot B$.

Looking at this argument closely, we have actually shown more. Related equations like $A\odot B_0 \cap J\otimes B = J\odot B_0$ for subalgebras $B_0\subseteq B$ are also true by the same line of reasoning.