This is a rather basic question. I was reading some notes on geometric representation theory by Gaitsgory and his defition of Verma module is the following: Let $ \lambda $ be a weight of $ \mathfrak{g}$. The Verma module $M_\lambda \in \mathfrak{g} - mod$ is defined such that for any object $ M \in \mathfrak{g} -mod$ we have : $Hom_\mathfrak{g} (M_\lambda , M) = Hom_\mathfrak{b} (\mathbb{C}^\lambda , M)$. Where $\mathfrak{b}$ is the borel subalgebra and $\mathbb{C}^\lambda $ is the one dimensional $\mathfrak{b}$ module.
The definition of Verma module that I am used to is that we define $M_\lambda := U(\mathfrak{g}) \otimes_{U ( \mathfrak{b} )} \mathbb{C}^\lambda$.
Gaitsgory says that his defintion implies this one but gives no proof and I cant see how it should be. My first thought was that maybe I should play around with some adjoint functors but I don't really see how to proceed. An answer would be nice but maybe its a lot to write so I would love even a hint to get started. Thanks.
Let me add on to pre-kidney's answer, which to me is just explaining the tensor-hom adjunction by words; algebraically, it can be expressed in the following lines. (Note that I identify $\mathfrak{g}-mod$ with $U(\mathfrak{g})-mod$, and consider Hom-spaces in the later category)
Let $M(\lambda):= U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\mathbb{C}_\lambda$, we have $$ \begin{array}{rcl} Hom_{U(\mathfrak{g})}(M(\lambda),M) &=& Hom_{U(\mathfrak{g})}(U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\mathbb{C}_\lambda,M)\\ &=& Hom_{U(\mathfrak{b})}(\mathbb{C}_\lambda, Hom_{U(\mathfrak{g})}(U(\mathfrak{g}),M)) \\ &=& Hom_{U(\mathfrak{b})}(\mathbb{C}_\lambda, M) \end{array} $$ Conversely, in Gaitsgory's definition, the equality has a hidden "restrict $M$ to a $U(\mathfrak{b})$-module" on the right hand side, which is algebraically given by $Hom_{U(\mathfrak{g})}(U(\mathfrak{g}),M)$ (since restricting is the adjoint of inducing), so reading the above three line backwards shows that $M_\lambda$ is indeed $M(\lambda)$.