We will prove first that ${\mathcal{L}}f^{\vee}=({\mathcal{L}}f)^{-1}$. Fix $C\in\Pi(f)$ and set $C^{\prime}={\mathcal{L}}f(C)$. Let $y_{0}\in M_{\mathbb{R}}$ such that $C=C_{y_{0}}$ and let $u_{0}\in\operatorname{ri}(C)$. Hence $u_{0}\in C_{y_{0}}=\partial f^{\vee}(y_{0})$ and so $y_{0}\in\partial f(u_{0})=C^{\prime}$ by Proposition 3.21 and Lemma 3.28. Hence

$${\mathcal{L}}f^{\vee}({\mathcal{L}}f(C))={\mathcal{L}}f^{\vee}(C^{\prime})=\bigcap_{x\in C^{\prime}}\partial f^{\vee}(x)\subset\partial f^{\vee}(y_{0})=C.$$

On the other hand, let $x_{0}\in\operatorname{ri}(C^{\prime})$. In particular, $x_{0}\in\partial f(u_{0})$ and so $u_{0}\in\partial f^{\vee}(x_{0})={\mathcal{L}}f^{\vee}(C^{\prime})$ for all $u_{0}\in C$. It implies

$$C\subset{\mathcal{L}}f^{\vee}(C^{\prime})={\mathcal{L}}f^{\vee}({\mathcal{L}}f(C)).$$

Thus ${\mathcal{L}}f^{\vee}({\mathcal{L}}f(C))=C$ and applying the same argument to $f^{\vee}$ we conclude that ${\mathcal{L}}f^{\vee}=({\mathcal{L}}f)^{-1}$ and that ${\mathcal{L}}f$ is bijective.

Now we have to prove that ${\mathcal{L}}$ is a duality between $\Pi(f)$ and $\Pi(f^{\vee})$. Let $C,D\in\Pi(f)$ such that $C\subset D$. Clearly, ${\mathcal{L}}f(C)\supset{\mathcal{L}}f(D)$. The reciprocal follows by applying the same argument to $f^{\vee}$. The fact that $C$ and ${\mathcal{L}}f(C)$ lie in orthogonal affine spaces has already been shown during the proof of Lemma 3.28 above, see (3.30). ∎