Corestriction of completely positive maps

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Let $A$ and $B$ be $C^*$-algebras. A linear map $\varphi: A \to B$ is called completely positive (= c.p.) if all inflation maps $$\varphi_n: M_n(A) \to M_n(B) : [a_{i,j}]\mapsto [\varphi(a_{i,j})]$$ are positive, i.e. map positive matrices of $M_n(A)$ to positive matrices in $M_n(B)$.

I want to prove the following (if it is true at all):

Let $C$ be a $C^*$-subalgebra of $B$ with $\varphi(A) \subseteq C \subseteq B$ and $\overline{\varphi}: A \to C: a \mapsto \varphi(a)$ be the co-extended map. I want to show that

$$\varphi \mathrm{\ is \ completely \ positive }\iff \overline{\varphi} \mathrm{\ is \ completely \ positive }$$

Attempt (edited): I will use the following fact: If we have a $C^*$-subalgebra $D \subseteq A$, then $D^+ = A^+ \cap D$, i.e. the positive elements of $D$ are the positive elements of $A$ that live in $D$.

Let $[a_{i,j}]\in M_n(A)$. Then $$\varphi_n([a_{i,j}]) = [\varphi(a_{i,j})] = [\overline{\varphi}(a_{i,j})] = \overline{\varphi_n}([a_{i,j}])$$

If $\varphi$ is c.p., then $\overline{\varphi_n}([a_{i,j}])= \varphi_n([a_{i,j}]) \in M_n(B)^+\cap M_n(C) = M_n(C)^+$ when $[a_{i,j}] \in M_n(A)^+$, so $\overline{\varphi}$ is c.p.

If $\overline{\varphi}$ is c.p., then $\varphi_n([a_{i,j}]) = \overline{\varphi_n}([a_{i,j}]) \in M_n(C)^+ \subseteq M_n(B)^+$ when $[a_{i,j}] \in M_n(A)^+$, so $\varphi$ is c.p. $\quad \square$

Is the above proof correct?

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Modulo a few (what I assume are) typos ($D^*=A^+\cap D$ should be $D^+=A^+\cap D$, $M_n(B)^+\subset M_n(C)^+$ should be $M_n(C)^+\subset M_n(B)^+$, and $M_n(C)^+ \cap M_n(B) = M_n(B)^+$ should be $M_n(C)^+ \cap M_n(B) \subset M_n(B)^+$), this is correct.

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The crucial fact upon which the OP' proof depends is that if $C$ is a closed $*$-subalgebra of a $C^*$-algebra $B$, then for every element $c$ in $C$ one has that $$\sigma_C(c)=\sigma_B(c).$$

This is called spectral permanence and it's proof strongly depends on the properties of $C^*$-algebras. In fact there are counter examples for general Banach algebras.

Since positive elements can be characterized as the self adjoint elements whose spectrum lies in $[0,\infty)$, positivity does not depend on the ambient algebra considered.