If $\cal{A,B}$ are unital $C^\ast$-algebras and $\varphi:\cal{A\to B}$ is an unital $\ast$-homomorphism. Then it is clear that $\rho\circ\varphi$ is a state on $\cal{A}$ for any state $\rho$ on $\cal{B}$. But if $\cal{A}$ and $\cal{B}$ are both non-unital, is it true?
It is not difficult to show that $\rho\circ\varphi$ is positive and $\|\rho\circ\varphi\|\leq1$, but I do not know how to prove $\|\rho\circ\varphi\|\geq1$. I try to prove it by approximate identity, but not get it.
Thnks a lot.
No, it is not true. Fix a state $f$ on $K(H)$. Let $\mathcal A=\mathcal B=K(H)\oplus K(H)$, $\varphi(a,b)=(a,0)$, $\rho(a,b)=(f(a)+f(b))/2$. Then $\|\rho\circ\varphi\|=1/2$.
The key point is that you are not replacing the unital condition on the homomorphism for something that makes it "almost" unital. Note that in the example above you can replace $K(H)$ with $\mathbb C$: you get unital C$^*$-algebras, but the homomorphism not being unital still can diminish the norm.