why is $f\otimes g:A\otimes_{\min}C\to B \otimes_{\min} D$ injective?

Question

If $f:A\to B$, $g:C\to D$ are injective $\ast$-homomorphisms between $C^*$-algebras $A, B, C, D$, is the induced map on the spatial tensor product $$f\otimes g:A\otimes_{\min}C\to B \otimes_{\min} D$$ injective as well?

I tried to prove it, but I'm stuck. The spartial tensor norm on $A\otimes B$ is $$\|\sum\limits_{i=1}^na_i\otimes b_i\|_{\min}=\sup\{\|\pi_A\otimes \pi_B)(\sum\limits_{i=1}^na_i\otimes b_i)\|: \pi_a,\pi_B\;\ast-\text{respresentations}\}.$$

Maybe the fact $\pi_A$ and $\pi_B$ injective $=>$ $\pi_A\otimes \pi_B$ injective will be helpful. But my problem is that it shouln't be enough to check injectivity of $f\otimes g$ on sums $z=\sum\limits_{i=1}^na_i\otimes c_i$.

Could anyone give some help?

Answer 1

The key is that, for any injective representations (i.e., $*$-homormophisms, so in particular $f$ and $g$), $$ \|\sum a_j\otimes b_j\|_\min=\|(f\otimes g)\left(\sum a_j\otimes b_j\right)\| $$ (technically, this might require using a further set of faithful representations to get embeddings $A\hookrightarrow B(H)$, $B\hookrightarrow B(K)$, but it doesn't change the argument).

You have, as the min norm is the spatial norm, $$ \|(f\otimes g)\left(\sum a_j\otimes b_j\right)\|_\min=\|\sum a_j\otimes b_j\|, $$ so $f\otimes g$ is isometric (thus bounded) on a dense subset of $A\otimes_\min B$. So it extends to a $*$-homomorphism $A\otimes_\min B\to C\otimes_\min D$ which will be still be isometric.

Answer 2

Tensor products of c.p. faithful maps with respect to the minimal norm are faithful:

D. Avitzour, Free products of C*-algebras, Trans. Amer. Math. Soc. 271 (1982), 423–435.

Certainly, *-homomorphisms are c.p.