In their book "Quantum Groups and Knot Invariants", Kassel, Rosso, and Turaev prove that $U_q\mathfrak{sl}(N+1)$ has a PBW basis. I'm having trouble following the last step, though.
In corollary $3.9$ on page $41$ they write (emphasis mine)
Corollary. The ordered monomials $E_{\beta_1}^{m_1} \cdots E_{\beta_s}^{m_s}$ (resp. $F_{\beta_1}^{m_1} \cdots F_{\beta_s}^{m_s}$) with $(m_1, \ldots, m_s) \in \mathbb{N}^s$ form a basis of $V_+$ (resp. of $V_-$)
Proof. The ordered monomials $F_{\beta_1}^{m_1} \cdots F_{\beta_s}^{m_s}$ span $V_-$. This is proved using a version of Proposition $3.2$ for the root vectors $F_\alpha$. Proposition $3.8$ shows that $(E_{\beta_1}^{m_1} \cdots E_{\beta_s}^{m_s})_{\beta \in \mathbb{N}^s}$ is in duality with a generating set, which implies that it is linearly independent. Reversing the argument, we see that $(F_{\beta_1}^{m_1} \cdots F_{\beta_s}^{m_s})_{\beta \in \mathbb{N}^s}$ is linearly independent as well. $\square$
In proposition $3.8$ the authors compute $\varphi \left (K E_{\beta_1}^{m_1} \cdots E_{\beta_s}^{m_s} \ , \ K' F_{\beta_1}^{m_1'} \cdots F_{\beta_s}^{m_s'} \right )$ for a bilinear map $\varphi : V_+ \times V_- \to k$.
I understand the proof that the monomials $E_{\beta_1}^{m_1} \cdots E_{\beta_s}^{m_s}$s span $V_+$, (resp. $F_{\beta_1}^{m_1} \cdots F_{\beta_s}^{m_s}$ span $V_-$) but I don't see how "duality with a generating set" shows that the $E_{\beta_1}^{m_1} \cdots E_{\beta_s}^{m_s}$s are linearly independent. Presumably this "duality" has something to do with the bilinear pairing $\varphi$, but I still don't follow the argument.
Any help is appreciated!
I don't have a copy of KRT with me, but I think they must mean something along the lines of the following:
Suppose that we have a nondegenerate pairing $b\colon V\times W \to \mathsf k$ (where $V$ and $W$ are $\mathsf k$-vector spaces for some field $\mathsf k$) so that $\forall v \in V$ there is some $w \in W$ with $b(v,w)\neq 0$, and similarly, for all $w \in W$ there is some $v\in V$ with $b(v,w)\neq 0$. Then
Thus combining 1. and 2. we see that $\dim(V)=\dim(W)$ and $\theta_V$ and $\theta_W$ are isomorphisms.
Now suppose that $S_W\subset W$ spans $W$, say $S_W = \{w_i:i \in I\}$ and that we have a subset $S_V\subseteq V$ where $S_V=\{v_i: i \in I\}$ with the property that $b(v_i,w_j) = \delta_{ij}$ for any $i,j \in I$. Then we say that $S_V$ and $S_W$ are in duality. Any two sets in duality in this way must be both be linearly independent: For example, to see that the set $S_W$ is linearly independent, suppose that $J\subseteq I$ is a finite subset and we have $\sum_{i \in J} \lambda_i w_i=0$ for some $\lambda_i \in \mathsf k$, ($i \in J$). Then for a given $i_0 \in J$, $$ 0=b(\sum_{i \in J} \lambda_i w_i,v_{i_0}) =\sum_{i \in J} \lambda_ib(w_i,v_{i_0}) = \lambda_{i_0}. $$ Since this holds for any $i_0\in J$ it follows $\lambda_i=0$ for all $i \in J$ and hence $S_W$ is linearly independent as claimed. A symmetric argument shows that $S_V$ is linearly independent. But if we also know $S_W$ spans $W$, then it follows that $S_W$ is a basis of $W$, and then $S_V$ must be a basis of $V$ since it is dual to the basis $\theta_W(S_W)$ of $V^*$.
Note that here the argument assumes that $V$ and $W$ are finite-dimensional, but it applies also to $V$ and $W$ when $V= \bigoplus_{n \in \mathbb N} V_n, W = \bigoplus_{m \in \mathbb N} W_m$ with each $V_n,W_n$ finite dimensional equipped with bilinear pairing $b_n\colon V_n\times W_n\to \mathsf k$. $V_{\pm}$ in KRT can be viewed as graded vector spaces of this sort by giving $E_{\alpha}$ degree $\text{ht}(\alpha)$ and similarly for $F_\alpha$.