Let $(V,R)$ be a not necessarily reduced root system, and $(V^{\vee},R^{\vee})$ the dual root system. A chamber of $R$ is a connected component of the complement in $V$ of the hyperplanes $H_{\alpha} = \{ x \in V : \langle x, \alpha^{\vee} \rangle =0\}$. If $C$ is a chamber, the corresponding base $B = B(C)$ is the unique vector space basis $\alpha_1, ... , \alpha_l$ of roots such that $$C = \{ x \in V : \langle x, \alpha_i^{\vee} \rangle > 0 \textrm{ for } 1 \leq i \leq l \}$$
The elements of $B$ are usually called simple roots. I thought that the set $B^{\vee}=\{\alpha_1^{\vee}, ... , \alpha_l^{\vee} \rangle$ of "simple coroots" was always a base of the dual root system $R^{\vee}$. Bourbaki, Lie Groups and Lie Algebras, Chapter VI, Section 10, seems to indicate otherwise:
Is it true that $B^{\vee}$ is in general not a base of the root system $(V^{\vee},R^{\vee})$?
Let $(-,-)$ be a symmetric, positive definite Weyl group invariant form on $V$, so that $V$ identifies with $V^{\vee}$ via $v \mapsto (-,v)$. Then the coroot $\alpha^{\vee}$ identifies with $\frac{2}{(\alpha,\alpha)}\alpha$. It follows that if $\alpha \in B$ is such that $2\alpha$ is a root, then $\alpha^{\vee}$ cannot be an element of a base of $R^{\vee}$, since it is equal to $2 (2\alpha)^{\vee}$:
$$(2\alpha)^{\vee} = \frac{2}{(2\alpha,2\alpha)} \cdot 2\alpha = \frac{\alpha}{(\alpha,\alpha)} = \frac{1}{2} \alpha^{\vee}$$