I want to prove the following: Suppose A is a unital C*algebra and X is a compact Hausdorff space, then any positive map $\phi: A \to C(X) $ and any positive map $\psi: C(X) \to A$ is completely positive, where C(X), is the set of continuous functions from X to the complex numbers.
For $\phi$ I want to show that $\sum_{i, k} \phi(a_{ik}) (x )f_{k}(x ) \bar{f} _{i } (x )) $ is positive for any x in X and with $a_{ik} $ the elements of a positive matrix in $M_{n} (A) $ and $f_{j} $ some functionals in C(X).
EDIT
I was able to prove that $\phi$ is completely positive, so now I'm trying to find a way to use this property of $\phi$ to prove that $\psi$ is also completely positive.
I know of no argument that lets you use one of the complete positivities to show the other.
The standard argument to show that $\psi$ is cp (due to Stinespring, if I'm not wrong) is not entirely obvious.
First you note that $M_n(C(X))=C(X,M_n(\mathbb C))$, so you consider $A\in C(X,M_n(\mathbb C))^+$ and you want to show that $\psi^{(n)}(A)\geq0$.
Using compactness and a partition of unity, you approximate $A$ with a function of the form $\sum_jg_j(t)A_j$, with $g_j:X\to\mathbb C$ positive, and $A_j\in M_n(\mathbb C)^+$.
You show that, being positive $\psi$ is bounded and so $\psi^{(n)}$ is bounded.
By the above, it is enough to show that $\psi^{(n)}(\sum_jg_j A_j)\geq0$.
As sums of positives are positives, you are reduced to show that $\psi^{(n)}(gA)\geq0$ when $g\geq0$ and $A\geq0$.
The above is obvious since $\psi^{(n)}(gA)=\psi(g)\,A$.