Let $\mathcal{A}$ and $\mathcal{B}$ be C*-algebras. Let $\phi:\mathcal{A} \rightarrow M_2\mathcal{(B)}$ be *-homomorphism and $$\phi=\begin{bmatrix} \phi_{11} & \phi_{12}\\ \phi_{21} & \phi_{22} \end{bmatrix}$$ If $\phi_{11}:\mathcal{A} \rightarrow M(\mathcal{B})$ is a *-homomorphism is it true that $\phi_{12}=\phi_{21}=0$ and $\phi_{22}$ is a *-homomorphism?
Using $\phi(ab)=\phi(a)\phi(b)$ I got
$\phi_{11}(a)\phi_{21}(b)=0$
$\phi_{12}(ab)=\phi_{11}(a)\phi_{12}(b)+\phi_{12}(a)\phi_{22}(b)$
$\phi_{21}(ab)=\phi_{21}(a)\phi_{11}(b)+\phi_{22}(a)\phi_{21}(b)$
$\phi_{22}(ab)=\phi_{21}(a)\phi_{12}(b)+\phi_{22}(a)\phi_{22}(b)$
I could not conclude anything. Thanks in advance.
If you look at the 1,1 entries of $\phi(a^*a)$ and $\phi(a)^*\phi(a)$, you get (after using the $\phi_{11}$ is a $*$-homomorphism) that $\phi_{21}(a)^*\phi_{21}(a)=0$. So $\phi_{21}(a)=0$ for all $a$.
Now use that $\phi$ preserves adjoints to get that $\phi_{12}=0$. And now your last equation gives you that $\phi_{22}$ is multiplicative.