[proofplan] The vector $v$ satisfies exactly the same $\mathfrak b$-relations as the generator $1\in\mathbb C_\lambda$, so sending $1$ to $v$ defines a $U(\mathfrak b)$-module map $\mathbb C_\lambda\to V$. We then use this map to define $\Phi$ on pure tensors by $\Phi(u\otimes c)=u\psi(c)$, and we verify directly that this formula respects the tensor relations over $U(\mathfrak b)$. Finally, $U(\mathfrak g)$-linearity is immediate from left multiplication, and uniqueness follows because $M(\lambda)$ is generated by $m_\lambda$ as a $U(\mathfrak g)$-module. [/proofplan] [step:Send the one-dimensional highest weight module into $V$] Let $\psi:\mathbb C_\lambda\to V$ be the complex-[linear map](/page/Linear%20Map) defined by $\psi(c)=c v$, where $c\in\mathbb C$ is identified with $c\cdot 1\in\mathbb C_\lambda$. We prove that $\psi$ is a $\mathfrak b$-[module homomorphism](/page/Module%20Homomorphism). For $h\in\mathfrak h$ and $c\in\mathbb C_\lambda$, \begin{align*} \psi(hc)=\psi(\lambda(h)c)=\lambda(h)c v=h(c v)=h\psi(c). \end{align*} For $x\in\mathfrak n^+$ and $c\in\mathbb C_\lambda$, \begin{align*} \psi(xc)=\psi(0)=0=x(c v)=x\psi(c). \end{align*} Since $\mathfrak b=\mathfrak h\oplus\mathfrak n^+$, these two identities show that $\psi$ is $\mathfrak b$-linear. Equivalently, after extending the actions from $\mathfrak b$ to $U(\mathfrak b)$, $\psi$ is a $U(\mathfrak b)$-module homomorphism. [guided] The point of the hypotheses on $v$ is that $v$ obeys the same defining relations as the generator of $\mathbb C_\lambda$. Define the complex-linear map $\psi:\mathbb C_\lambda\to V$ by sending the basis vector $1\in\mathbb C_\lambda$ to $v$; explicitly, for $c\in\mathbb C$, identified with $c\cdot 1\in\mathbb C_\lambda$, set $\psi(c)=c v$. We must check compatibility with the $\mathfrak b$-action. Since $\mathfrak b=\mathfrak h\oplus\mathfrak n^+$, it is enough to check elements of $\mathfrak h$ and elements of $\mathfrak n^+$. If $h\in\mathfrak h$, then $h$ acts on $\mathbb C_\lambda$ by the scalar $\lambda(h)$, while the hypothesis says that $h$ acts on $v$ by the same scalar. Therefore, for every $c\in\mathbb C_\lambda$, \begin{align*} \psi(hc)=\psi(\lambda(h)c)=\lambda(h)c v=h(c v)=h\psi(c). \end{align*} If $x\in\mathfrak n^+$, then $x$ acts as $0$ on $\mathbb C_\lambda$, and the hypothesis says that $xv=0$. Thus \begin{align*} \psi(xc)=\psi(0)=0=x(c v)=x\psi(c). \end{align*} These two verifications prove that $\psi$ is $\mathfrak b$-linear. Since a $\mathfrak b$-module is equivalently a compatible $U(\mathfrak b)$-module, the same map $\psi$ is $U(\mathfrak b)$-linear. [/guided] [/step] [step:Define the induced map on the tensor product and check it is well defined] Define a complex-bilinear map $\widetilde{\Phi}:U(\mathfrak g)\times\mathbb C_\lambda\to V$ by \begin{align*} \widetilde{\Phi}(u,c)=u\psi(c). \end{align*} We check that $\widetilde{\Phi}$ is balanced over $U(\mathfrak b)$. Let $u\in U(\mathfrak g)$, let $a\in U(\mathfrak b)$, and let $c\in\mathbb C_\lambda$. Because $\psi$ is $U(\mathfrak b)$-linear, \begin{align*} \widetilde{\Phi}(ua,c)=(ua)\psi(c)=u(a\psi(c))=u\psi(ac)=\widetilde{\Phi}(u,ac). \end{align*} By the universal property of the balanced [tensor product](/page/Tensor%20Product), this balanced complex-bilinear map descends to a well-defined complex-linear map \begin{align*} \Phi:U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda\to V \end{align*} satisfying \begin{align*} \Phi(u\otimes c)=u\psi(c) \end{align*} for all $u\in U(\mathfrak g)$ and $c\in\mathbb C_\lambda$. [guided] We want a map out of the induced module, which is a tensor product over $U(\mathfrak b)$. The natural formula is to let $u\in U(\mathfrak g)$ act on the image of the second factor. Define a complex-bilinear map $\widetilde{\Phi}:U(\mathfrak g)\times\mathbb C_\lambda\to V$ by \begin{align*} \widetilde{\Phi}(u,c)=u\psi(c). \end{align*} Here $u\psi(c)$ means the action of $U(\mathfrak g)$ on the $\mathfrak g$-module $V$. To descend to the tensor product over $U(\mathfrak b)$, the map must be balanced: moving an element $a\in U(\mathfrak b)$ from the right side of $u$ to the left action on $c$ must not change the value. Let $u\in U(\mathfrak g)$, $a\in U(\mathfrak b)$, and $c\in\mathbb C_\lambda$. Since $\psi$ is $U(\mathfrak b)$-linear, $a\psi(c)=\psi(ac)$. Therefore \begin{align*} \widetilde{\Phi}(ua,c)=(ua)\psi(c)=u(a\psi(c))=u\psi(ac)=\widetilde{\Phi}(u,ac). \end{align*} This is exactly the balancing relation defining $U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda$. Hence the universal property of the balanced tensor product gives a well-defined complex-linear map \begin{align*} \Phi:U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda\to V \end{align*} with \begin{align*} \Phi(u\otimes c)=u\psi(c) \end{align*} for all $u\in U(\mathfrak g)$ and $c\in\mathbb C_\lambda$. [/guided] [/step] [step:Verify $\mathfrak g$-linearity and the value on the highest weight generator] Let $y\in\mathfrak g$, $u\in U(\mathfrak g)$, and $c\in\mathbb C_\lambda$. Using the left $U(\mathfrak g)$-action on $M(\lambda)$ and the $U(\mathfrak g)$-action on $V$, \begin{align*} \Phi(y\cdot(u\otimes c))=\Phi((yu)\otimes c)=(yu)\psi(c)=y(u\psi(c))=y\Phi(u\otimes c). \end{align*} Hence $\Phi$ is $\mathfrak g$-linear. Since $m_\lambda=1\otimes 1$ and $\psi(1)=v$, \begin{align*} \Phi(m_\lambda)=\Phi(1\otimes 1)=1\psi(1)=v. \end{align*} [guided] The induced module $M(\lambda)$ carries its $\mathfrak g$-action through left multiplication on the $U(\mathfrak g)$ factor. We verify that the map just constructed respects this action. Let $y\in\mathfrak g$, let $u\in U(\mathfrak g)$, and let $c\in\mathbb C_\lambda$. Then $y$ acts on $u\otimes c$ by sending it to $(yu)\otimes c$, so \begin{align*} \Phi(y\cdot(u\otimes c))=\Phi((yu)\otimes c)=(yu)\psi(c)=y(u\psi(c))=y\Phi(u\otimes c). \end{align*} Thus $\Phi$ is compatible with the action of every $y\in\mathfrak g$, and hence $\Phi$ is a $\mathfrak g$-module homomorphism. It remains in this step to check that the prescribed generator goes to the prescribed vector. The highest weight generator is $m_\lambda=1\otimes 1$, where the first $1$ is the unit of $U(\mathfrak g)$ and the second $1$ is the basis vector of $\mathbb C_\lambda$. Since $\psi(1)=v$, we get \begin{align*} \Phi(m_\lambda)=\Phi(1\otimes 1)=1\psi(1)=v. \end{align*} [/guided] [/step] [step:Use generation by $m_\lambda$ to prove uniqueness] The module $M(\lambda)=U(\mathfrak g)\otimes_{U(\mathfrak b)}\mathbb C_\lambda$ is generated by $m_\lambda=1\otimes 1$ as a $U(\mathfrak g)$-module, because every pure tensor has the form \begin{align*} u\otimes c=u\cdot(1\otimes c)=u\cdot(c(1\otimes 1))=c\,u\cdot m_\lambda. \end{align*} Let $\Theta:M(\lambda)\to V$ be any $\mathfrak g$-module homomorphism satisfying $\Theta(m_\lambda)=v$. Since the $\mathfrak g$-actions on $M(\lambda)$ and $V$ extend to the corresponding $U(\mathfrak g)$-actions, $\Theta$ is $U(\mathfrak g)$-linear. Therefore, for every $u\in U(\mathfrak g)$ and $c\in\mathbb C_\lambda$, \begin{align*} \Theta(u\otimes c)=\Theta(c\,u\cdot m_\lambda)=c\,u\cdot\Theta(m_\lambda)=c\,u v. \end{align*} This is exactly the formula defining $\Phi$. Therefore $\Theta=\Phi$, so the required homomorphism is unique. [guided] Uniqueness comes from the fact that the whole Verma module is generated by the single vector $m_\lambda$. We first verify that generation directly from the tensor-product definition. For every $u\in U(\mathfrak g)$ and $c\in\mathbb C_\lambda$, scalar multiplication in the one-dimensional [vector space](/page/Vector%20Space) $\mathbb C_\lambda$ gives $1\otimes c=c(1\otimes 1)$. Therefore \begin{align*} u\otimes c=u\cdot(1\otimes c)=u\cdot(c(1\otimes 1))=c\,u\cdot m_\lambda. \end{align*} Thus every pure tensor lies in the $U(\mathfrak g)$-submodule generated by $m_\lambda$, and pure tensors span the tensor product as a complex vector space. Hence $M(\lambda)$ is generated by $m_\lambda$ as a $U(\mathfrak g)$-module. Now let $\Theta:M(\lambda)\to V$ be any $\mathfrak g$-module homomorphism with $\Theta(m_\lambda)=v$. Since $\Theta$ is $\mathfrak g$-linear, equivalently $U(\mathfrak g)$-linear for the induced enveloping-algebra action, and since it is complex-linear, for every $u\in U(\mathfrak g)$ and $c\in\mathbb C_\lambda$ we have \begin{align*} \Theta(u\otimes c)=\Theta(c\,u\cdot m_\lambda)=c\,u\cdot\Theta(m_\lambda)=c\,u v. \end{align*} The map $\Phi$ constructed above has the same value on every pure tensor, because $\psi(c)=cv$. Since pure tensors span $M(\lambda)$, $\Theta=\Phi$. Therefore there is at most one $\mathfrak g$-module homomorphism sending $m_\lambda$ to $v$, and the homomorphism constructed above is the required unique one. [/guided] [/step]