Take a unital cp map $f:B\to A$ between unital $C^\*$ algebras. Given a state $\psi:B\to \mathbb{C}$ what conditions are necessary for there to exist a state $\phi:A\to \mathbb{C}$ so that $\phi\circ f=\psi$? I am sure that the answer must be known, and apologise for my ignorance. 


In the case where $f$ is a unital $\*$-algebra hom, and we consider the image of $f$ as a subalgebra, suitable conditions are given in -- Joel Anderson, Extensions, restrictions, and representations of states on $C^\*$-algebras, Transactions of the American Mathematical Society 249(2):303-329, 1979. 


This comes from considering a possibility of defining the degree of a cp map at a pure state. The classical theory requires taking the inverse image of points, which translates to the current question. The states may be assumed pure if it helps (as in the paper above). 


As Nik points out below, the Hahn Banach theorem proves this if we have the inequality |ψ(b)|≤‖f(b)‖ for all b∈B.



direct images of states in $C*$ algebras
This is an expansion of Nik Weaver's comments. I'm using [Effros+Ruan, "Operator Spaces"](http://www.ams.org/mathscinet-getitem?mr=1793753) as a reference, but this is no doubt a little overkill.


A unital CP map is self-adjoint, so $f(B) \subseteq A$ is a (non-closed) *operator system*, namely a linear subspace which is unital and self-adjoint. (Most results in this area assume a closed subspace, but this is unnecessary). We have the following result:



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> A unital linear map between operator systems is completely positive if and only if it is completely contractive.
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This implies that a unital *functional* from an operator system to $\mathbb C$ is positive if and only if it is contractive.


So give a state $\psi:B\rightarrow\mathbb C$, when does there exist a state $\phi:A\rightarrow\mathbb C$ with $\phi\circ f = \psi$?



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> **Well-defined:** If and only if $f(x)=0 \implies \psi(x)=0$.
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Given this condition, there is a linear map $\phi:f(B)\rightarrow\mathbb C$ with $\phi f = \psi$. Notice that $\phi(1) = \phi(f(1)) = \psi(1) = 1$.



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> **Positive:** $f(x)\geq 0 \implies \psi(x)\geq 0$.
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Then if $x\in f(B)$ and $x\geq 0$ there is $y$ with $f(y)=x$ so $\psi(y)\geq 0$ so $\phi(x)\geq 0$. Thus $\phi:f(B)\rightarrow\mathbb C$ is a positive unital functional, and so contractive. Hahn-Banach it to a contractive, unital functional on all of $A$, which must be a state, as required.


*Equivalently* (but not additionally) you could impose a different condition:



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> **Contractive:** $|\psi(x)| \leq \|f(x)\|$
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Then with $y=f(x)$ we have $|\phi(y)| = |\psi(x)| \leq \|f(x)\| = \|y\|$ so $\phi$ is a unital contraction (hence positive). We may again apply Hahn-Banach.


