Let $0\ne\mathcal{B}\subset\mathcal{A}$ be a $C^*$-subalgebra and assume that $\mathcal{A}$ is unital. Suppose that there exists a conditional expectation $$\mathbb{E}_{\mathcal{B}}:\mathcal{A}\to\mathcal{B}.$$
Further assume that every state on $\mathcal{B}$ extends uniquely to a state on $\mathcal{A}$ (e.g., when $\mathcal{B}$ is a hereditary subalgebra). Is it the case then that every state on $\mathcal{A}$ factorizes through $\mathbb{E}_{\mathcal{B}}$?
For any state $\varphi$ on $\mathcal{B}$, let us denote its extension to $\mathcal{A}$ by $\bar{\varphi}$. Say that $\psi$ is a state on $\mathcal{A}$. Then, $\psi|_{\mathcal{B}}$ is a state on $\mathcal{B}$.
By assumption of unique extension, $\bar{\psi_{\mathcal{B}}}=\psi_{\mathcal{B}}\circ\mathbb{E}_{\mathcal{B}}$.
On the other hand, $\psi$ itself is an extension of $\psi_{\mathcal{B}}$. So, $$\psi=\bar{\psi_{\mathcal{B}}}=\psi_{\mathcal{B}}\circ\mathbb{E}_{\mathcal{B}}$$
Is this alright? The question comes down to whether $\psi$ itself can be considered as an extension of $\psi|_{\mathcal{B}}$ or not?
This is where I am confused at. Thank you for your time.