Let $\mathcal{A}$ and $\mathcal{B}$ be ${\rm C}^*$-algebras, $\psi: \mathcal{A} \to \mathcal{B}$ a linear positive map (i.e. $a \ge 0 \Rightarrow \psi(a) \ge 0$) and $p \in A$ projection such that $\psi$ is strictly positive (i.e. $a > 0 \Rightarrow \psi(a) > 0$ ) on $p\mathcal{A}p$ and $(1-p)\mathcal{A}(1-p)$.
Question: Is $\psi$ strictly positive on $\mathcal{A}$?
[Else, is it true for $\mathcal{A} = M_n(\mathbb{C})$? for $n=2$?]
No.
Take $\mathcal{A} =\mathcal{B} = M_{2}(\mathbb{C})$, $p=\begin{pmatrix} 1 & 0 \newline 0 & 0 \end{pmatrix}$ and $q=\begin{pmatrix} 1/2 & 1/2 \newline 1/2 & 1/2 \end{pmatrix}$.
Then $\psi(X) = qXq$ defines a positive linear map, which is strictly positive on $p\mathcal{A}p$ and $(1-p)\mathcal{A}(1-p)$ because $qpq = q(1-p)q = \begin{pmatrix} 1/4 & 1/4 \newline 1/4 & 1/4 \end{pmatrix}$, but $\psi$ is not strictly positive on $\mathcal{A}$ because $q(1-q)q = 0$.