I want to prove the following result:
Theorem: Let $\mathfrak{g}$ be a semi-simple Lie algebra, $\mathfrak{h}$ be a Cartan subalgebra, $\mu \in \mathfrak{h}$, and $W_{\mu} = \mathfrak{U}_{\mathfrak{g}}/I_{\mu}$ be the Verma module with highest weight $\mu$. Suppose that $\sigma_{\mu}: \mathfrak{g} \rightarrow \mathfrak{gl} \left( W_{\mu} \right)$ be a highest weight cyclic representation with highest weight $\mu$. Then, there is a surjective intertwining map $\phi: V_{\mu} \rightarrow W_{\mu}$.
Here, $\mathfrak{U}_{\mathfrak{g}}$ is the universal enveloping algebra of $\mathfrak{g}$, and $I_{\mu}$ is a left-ideal in $\mathfrak{U}_{\mathfrak{g}}$ generated by elements of the form $H - \langle \mu, H \rangle \cdot 1$, where $H \in \mathfrak{h}$, and $X \in \mathfrak{g}_{\alpha}$, where $\alpha$ is a positive root. Also, $V_{\mu} = W_{\mu}/U_{\mu}$, where $U_{\mu}$ is the space of all elements $v$ of $W_{\mu}$ which do not have a $v_0$-component (where $v_0$ is a highest weight vector for $W_{\mu}$), and for any collection $X_1, X_2, \cdots, X_l$ of root vectors with positive roots, the vector $\pi_{\mu} \left( X_1 \right) \pi_{\mu} \left( X_2 \right) \cdots \pi_{\mu} \left( X_l \right)v$ also does not have a $v_0$-component.
The representation $\pi_{\mu}: \mathfrak{U}_{\mathfrak{g}} \rightarrow W_{\mu}$ is given by $\pi_{\mu} \left( t \right) \left( \left[ v \right] \right) = \left[ tv \right]$, where $\left[ \cdot \right]$ denotes the equivalence class of an element in $W_{\mu}$. This representation can be restricted to $\mathfrak{g}$ (through the inclusion) to give a representation of $\mathfrak{g}$.
To prove the result, I have tried the following:
Firstly, using the universal property of universal enveloping algebra $\mathfrak{U}_{\mathfrak{g}}$, we get an associative algebra homomorphism $\tilde{\sigma}_{\mu}: \mathfrak{U}_{\mathfrak{g}} \rightarrow \mathfrak{gl} \left( W_{\mu} \right)$ which satisfies $\tilde{\sigma}_{\mu} \left( 1 \right) = I$ and $\tilde{\sigma}_{\mu} \left( Z \right) = \sigma{\mu} \left( Z \right)$, for $Z \in \mathfrak{g}$.
Then, we define $\psi: \mathfrak{U}_{\mathfrak{g}} \rightarrow W_{\mu}$ as
$$\psi \left( t \right) = \tilde{\sigma}_{\mu} \left( t \right) w_0,$$
where $w_0$ is a highest weight vector for the representation $\sigma_{\mu}$ of $\mathfrak{g}$. Since $\tilde{\sigma}_{\mu}$ is an associative algebra homomorphism, $\psi$ is a linear map between vector spaces. Now, for $H \in \mathfrak{h}$, we see that
$$\psi \left( H - \langle \mu, H \rangle \cdot 1 \right) = \tilde{\sigma}_{\mu} \left( H - \langle \mu, H \rangle \cdot 1 \right) w_0 = \sigma_{\mu} \left( H - \langle \mu, H \rangle \cdot 1 \right) w_0 = 0.$$
Also, for any $X \in \mathfrak{g}_{\alpha}$, where $\alpha$ is a positive root, we have
$$\psi \left( X \right) = \tilde{\sigma}_{\mu} \left( x \right) w_0 = \sigma_{\mu} \left( X \right) w_0 = 0,$$
since $w_0$ is a highest weight vector for $\sigma_{\mu}$. That is, we see that $I_{\mu} \subseteq \ker \psi$. Hence, this map descends to a linear map $\tilde{\psi}: W_{\mu} = \mathfrak{U}_{\mathfrak{g}}/I_{\mu} \rightarrow W_{\mu}$.
Now, I want to prove that $U_{\mu} \subseteq \ker \tilde{\psi}$ so that it can descend to a map $\phi: V_{\mu} \rightarrow W_{\mu}$. However, I cannot prove it! If we start with any $v \in U_{\mu}$, then it does not have a $v_0$-component. We want that $\tilde{\psi} \left( v \right) = 0$. This would mean that $\tilde{\sigma}_{\mu} \left( v \right) w_0 = 0$. However, I cannot see why this should happen?
Is there something wrong in this attempt? Any help will be appreciated!