Does this result holds regarding unitization of $C^*-$ algebra?

Question

For any $C^*$-algebra $V$, let $V^+$ denotes the unitization of $V$. Let $V$ and $W$ be two $C^*$-algebras then is it true that $$( V \otimes^{max}W)^+\cong V^+ \otimes^ {max} W^+$$

I guess it’s true but I can’t see the proof of this.Any reference or ideas?

Answer 1

This is not true. Consider $V=W=C_0(\mathbb R)$ (since both $V$ and $W$ are nuclear, I'll drop the $max$). Then we have $$(C_0(\mathbb R\otimes C_0(\mathbb R))^+\cong C_0(\mathbb R^2)^+\cong C((\mathbb R^2)^+)\cong C(S^2),$$ while $$C_0(\mathbb R)^+\otimes C_0(\mathbb R)^+\cong C(\mathbb R^+)\otimes C(\mathbb R^+)\cong C(S^1)\otimes C(S^1)\cong C(\mathbb T^2),$$ where $\mathbb T^2$ is the $2$-torus.

Answer 2

Consider $V=W=c_0$. On $(c_0\otimes c_0)^+$, the projections are $\{e_n\}_n$ and $1$. So if $pq\ne0$ for two distinct projections, then one of them is $1$. On $c_0^+\otimes c_0^+$, you have the projections $1\otimes e_1$ and $e_1\otimes 1$, both not $1\otimes 1$ and with nonzero product. So the two algebras cannot be isomorphic.