Lemma. If $\underline{w}_l\neq 0$, there is a $f\neq\sigma(l)$ such that \begin{equation*} \underline{w}_l+\bar\pi_f=v_{lf}, \end{equation*} and similarly for $\underline{\pi}_f$.

Proof. $(\underline w_l,\bar\pi_f)_{lf\in\ll\cup\ff}$ is a dual-optimal payoff. Suppose the claim is false. Then we have $\underline w_l+\bar\pi_f\geq v_{lf}+\epsilon$ for some $\epsilon>0$ and all $f\neq\sigma(l)$. Modifying the payoff by letting $w'_l=\underline w_l-\epsilon'$ and $\pi'_{\sigma(l)}=\pi_{\sigma(l)}+\epsilon'$ is feasible for any $0<\epsilon^\prime<\epsilon$, the payoff has the same value, so it too solves the dual, contradicting the minimality of $\underline w_l$. ∎

Call this constraint the opportunity constraint for $l$. There is also an opportunity constraint for matching $f$ with $\sigma^{-1}(f)$.