Confusion in Lie algebra notes I'm self-studying through these notes, and I ran into a roadblock on the page 38, chapter $sl(2)$ and its irreducible representations.
Right after defining $U(sl(2)) \otimes_{U(b^+)} \mathbb C$ author writes:

It is a “universal” such
  module in the sense that any other cyclic module with cyclic vector satisfying
  (2.4) is a homomorphic image of the one we just constructed.

I don't see how to show that.
It seems that I should combine universal properties of $U(sl(2))$ (every representation of $sl(2)$ is a homomorphic image of $U(sl(2))$) and tensor product, but it's not really clear to me how to do it.

Second question, a little bit softer... I learned about tensor product of modules, but its use in this proof seems like magic to me. I understand why it's useful to construct the universal cyclic module with cyclic vector $v$ satisfying $h.v = \lambda v, e.v = 0$, but I don't understand how could one think of tensoring over $U(b^+)$ as a right tool for that.
 A: $\def\sl{\mathfrak{sl}}$
$\def\b{\mathfrak{b}}$
Suppose $V$ is a cyclic representation of $\sl_{2}$ and let $v$ be a cyclic vector satisfying 2.4. Then, $v$ is an eigenvector for $\b$. Hence, if $N$ is a nontrivial strictly lower triangular element, we can define a map of associative algebras
$$U(\sl_{2}) \otimes_{U(\b)} \mathbb{C} \rightarrow V$$
by sending $N^{m} \otimes 1 \mapsto N^{m}(v).$ This completely defines our morphism because by the Poincare Birkhoff Witt Theorem, the set $\{N^{m} \otimes 1\}_{m \in \mathbb{N}}$ is a basis for $U(\sl_{2}) \otimes_{U(\b)} \mathbb{C}.$ However, even if you don't want to use the PBW theorem, what we are actually doing is sending
$$X \otimes \alpha \mapsto \alpha X(v)$$
where $X \in \sl_{2}, \alpha \in \mathbb{C}$. Now, because $V$ is cyclic, this map is surjective. This is what we needed.
Edit: If you're also confused about how to make $\mathbb{C}$ into a left $U(\b)$-module so that the tensor product is well-defined, note that $\mathbb{C}v$ actually gives a one dimensional representation of $\b$, which is good enough.
