[proofplan]
We first regard the vector $v\in V$ as defining a map from the one-dimensional $\mathfrak b$-module $\mathbb C_\lambda$ into the restriction of $V$ to $\mathfrak b$. The highest weight relations on $v$ are exactly the condition that this map is $\mathfrak b$-linear. We then use the induced-module construction of $M(\lambda)=U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda$ to extend the map by the formula $u\otimes c\mapsto u\cdot\theta(c)$. Finally, uniqueness follows because $M(\lambda)$ is generated as a $U(\mathfrak g)$-module by $v_\lambda$.
[/proofplan]
[step:Construct the $\mathfrak b$-module map determined by the highest weight vector]
Let $c_\lambda\in\mathbb C_\lambda$ denote the basis vector satisfying $h c_\lambda=\lambda(h)c_\lambda$ for every $h\in\mathfrak h$ and $x c_\lambda=0$ for every $x\in\mathfrak n^+$. Define the complex-[linear map](/page/Linear%20Map)
\begin{align*}
\theta:\mathbb C_\lambda\to V
\end{align*}
by $\theta(c_\lambda)=v$.
We prove that $\theta$ is a $\mathfrak b$-[module homomorphism](/page/Module%20Homomorphism). Since $\mathfrak b=\mathfrak h\oplus\mathfrak n^+$, it is enough to check the action of elements of $\mathfrak h$ and $\mathfrak n^+$. For $h\in\mathfrak h$,
\begin{align*}
\theta(hc_\lambda)=\theta(\lambda(h)c_\lambda)=\lambda(h)v=h v=h\theta(c_\lambda).
\end{align*}
For $x\in\mathfrak n^+$,
\begin{align*}
\theta(xc_\lambda)=\theta(0)=0=xv=x\theta(c_\lambda).
\end{align*}
Thus $\theta$ is $\mathfrak b$-linear, and therefore also $U(\mathfrak b)$-linear.
[guided]
The first point is to translate the hypotheses on $v$ into a module map. Let $c_\lambda\in\mathbb C_\lambda$ be the chosen basis vector of the one-dimensional $\mathfrak b$-module $\mathbb C_\lambda$. By definition of $\mathbb C_\lambda$, every $h\in\mathfrak h$ acts on $c_\lambda$ by the scalar $\lambda(h)$, and every $x\in\mathfrak n^+$ acts by zero:
\begin{align*}
h c_\lambda=\lambda(h)c_\lambda,\qquad x c_\lambda=0.
\end{align*}
Define the complex-linear map
\begin{align*}
\theta:\mathbb C_\lambda\to V
\end{align*}
by $\theta(c_\lambda)=v$.
We need to verify that $\theta$ respects the $\mathfrak b$-action. Since $\mathfrak b=\mathfrak h\oplus\mathfrak n^+$, checking elements of $\mathfrak h$ and elements of $\mathfrak n^+$ checks all [Lie algebra](/page/Lie%20Algebra) generators of $\mathfrak b$. If $h\in\mathfrak h$, then the definition of $\mathbb C_\lambda$ and the assumed weight relation for $v$ give
\begin{align*}
\theta(hc_\lambda)=\theta(\lambda(h)c_\lambda)=\lambda(h)v=h v=h\theta(c_\lambda).
\end{align*}
If $x\in\mathfrak n^+$, then the definition of $\mathbb C_\lambda$ and the assumed highest weight relation for $v$ give
\begin{align*}
\theta(xc_\lambda)=\theta(0)=0=xv=x\theta(c_\lambda).
\end{align*}
Hence $\theta$ is a $\mathfrak b$-module homomorphism. Because a Lie algebra module structure is equivalently a compatible module structure over the universal enveloping algebra, this also means that $\theta$ is $U(\mathfrak b)$-linear.
[/guided]
[/step]
[step:Extend the map across the induced tensor product]
Define
\begin{align*}
\varphi:M(\lambda)\to V
\end{align*}
on pure tensors by
\begin{align*}
\varphi(u\otimes c)=u\cdot\theta(c),
\end{align*}
where $u\in U(\mathfrak g)$ and $c\in\mathbb C_\lambda$.
This formula is well-defined on the [tensor product](/page/Tensor%20Product) over $U(\mathfrak b)$. Indeed, for $a\in U(\mathfrak b)$,
\begin{align*}
\varphi(ua\otimes c)=(ua)\cdot\theta(c)=u\cdot(a\cdot\theta(c)).
\end{align*}
Since $\theta$ is $U(\mathfrak b)$-linear, $a\cdot\theta(c)=\theta(a\cdot c)$, so
\begin{align*}
\varphi(ua\otimes c)=u\cdot\theta(a\cdot c)=\varphi(u\otimes a\cdot c).
\end{align*}
This is exactly the balancing relation defining $U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda$.
The map $\varphi$ is a $\mathfrak g$-module homomorphism. If $y\in\mathfrak g$, then for every $u\in U(\mathfrak g)$ and $c\in\mathbb C_\lambda$,
\begin{align*}
\varphi(y\cdot(u\otimes c))=\varphi(yu\otimes c)=(yu)\cdot\theta(c)=y\cdot(u\cdot\theta(c))=y\cdot\varphi(u\otimes c).
\end{align*}
Thus $\varphi$ is $\mathfrak g$-linear.
[/step]
[step:Verify that the extension sends the highest weight vector to $v$]
By definition, the highest weight vector of the Verma module is
\begin{align*}
v_\lambda=1\otimes c_\lambda\in U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda.
\end{align*}
Therefore
\begin{align*}
\varphi(v_\lambda)=\varphi(1\otimes c_\lambda)=1\cdot\theta(c_\lambda)=v.
\end{align*}
So the constructed homomorphism has the required value on $v_\lambda$.
[/step]
[step:Use generation by $v_\lambda$ to prove uniqueness]
Let
\begin{align*}
\psi:M(\lambda)\to V
\end{align*}
be any $\mathfrak g$-module homomorphism satisfying $\psi(v_\lambda)=v$. Since $M(\lambda)=U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda$ is generated by $v_\lambda=1\otimes c_\lambda$ as a $U(\mathfrak g)$-module, every element of $M(\lambda)$ is a finite sum of elements of the form $u\cdot v_\lambda$ with $u\in U(\mathfrak g)$. For such an element,
\begin{align*}
\psi(u\cdot v_\lambda)=u\cdot\psi(v_\lambda)=u\cdot v.
\end{align*}
The constructed map $\varphi$ satisfies the same formula:
\begin{align*}
\varphi(u\cdot v_\lambda)=\varphi(u\otimes c_\lambda)=u\cdot v.
\end{align*}
Hence $\psi$ and $\varphi$ agree on a spanning set of $M(\lambda)$, so $\psi=\varphi$. This proves uniqueness and completes the proof.
[/step]
