Decomposing a representation of a C$^{*}$-algebra into a direct sum of irreducible ones Suppose we have C$^{*}$-algebras $A$ and $B$ and a $*$-homomorphism $\phi\colon A\to B$. I came across a paper, where it was claimed that every irreducible representation of $B$ decomposes into a direct sum of a unique set of irreducible representations of $A$. 
I am confused as to how we actually do this. If $\pi$ is a representation of $B$, then, clearly, $\pi\circ\phi$ is a representation of $A$. I am aware that every non-degenerate representation decomposes as a direct sum of cyclic representations, but this is not the same thing.
Any help would be much appreciated. 
 A: As stated, I don't think the assertion makes sense. Consider for instance $A=C[0,1]$, $B=C[0,1]\oplus M_2(\mathbb C)$, and 
$$
\phi(x)=(x,x(0)I),\ \ \ \ \pi(x,M)=M. 
$$
Then $\pi $ is irreducible, but there is no natural way to see it as a sum of representations of $A$ (which are all point evaluations). 
What does hold is that $\pi\circ\phi$ is a representation of $A$ and, at least in the separable case, every representation is a direct sum of irreducibles. This is a nontrivial result, see for instance Corollary II.5.9 in Davidson's book. I think that the general version follows from work by Hadwin, but I don't have a reference. 
A: The answer to your question is no.
Consider the case where $A = B$ and $\phi$ is the identity. In this case one can see that the answer to your question is no, for the reason that not every *-representation is the direct sum of irreducible representations. For an example, consider the identity representation of the $C^*$-algebra generated by the multiplication operator $M_x$ on $L^2[0,1]$; this representation has no irreducible summand, and so in particular it is not a direct sum of irreducible representations.
