Dual of a line bundle is its tensor inverse

Evaluation is a morphism: for each open $U\subseteq X$ the map ${\mathcal{L}}^{\vee }(U)\times \mathcal{L}(U)\to {\mathcal{O}}_X(U)$, $(\phi ,s)\mapsto \phi (s)$, is well defined by [F3], is ${\mathcal{O}}_X(U)$-bilinear by additivity and ${\mathcal{O}}_X$-linearity of homomorphisms of module sheaves, and is compatible with restrictions; by [F4] it induces a morphism of ${\mathcal{O}}_X$-modules $\mathrm{ev}:{\mathcal{L}}^{\vee }{\otimes }_{{\mathcal{O}}_X}\mathcal{L}\to {\mathcal{O}}_X$ with $\mathrm{ev}(\phi \otimes s)=\phi (s)$.

On a chart $U$ with trivialisation $\tau :\mathcal{L}{\mathrm{\mid }}_U\to {\mathcal{O}}_U$, the induced trivialisation of the dual is ${\tau }^{\vee }:{\mathcal{L}}^{\vee }{\mathrm{\mid }}_U\to {\mathcal{O}}_U$, $\psi \mapsto \psi ({\tau }^{-1}(1))$, which is an isomorphism because $\psi \mapsto \psi ({\tau }^{-1}(1))$ corresponds under $\tau$ to the isomorphism ${\mathrm{Hom}}_{{\mathcal{O}}_U}({\mathcal{O}}_U,{\mathcal{O}}_U)\cong {\mathcal{O}}_U$ of [F7]; moreover every automorphism of the trivial bundle ${\mathcal{O}}_U$ is multiplication by a unit of ${\mathcal{O}}_X(U)$, again by [F7].

On such a chart $U$, transport the evaluation morphism along the trivialisations ${\tau }^{\vee }\otimes \tau$: the result is the map ${\mathcal{O}}_U{\otimes }_{{\mathcal{O}}_U}{\mathcal{O}}_U\to {\mathcal{O}}_U$, $a\otimes b\mapsto ab$, which is an isomorphism by [F5], with inverse $c\mapsto c\otimes 1$. By [F1] such trivialising charts cover $X$, so $\mathrm{ev}$ restricts to an isomorphism on every member of a cover of $X$, and is therefore an isomorphism by [F6]; this proves the first claim.

Transition units: let ${\tau }_i,{\tau }_j$ be two trivialisations as in the Statement, and put $u_{ij}=({\tau }_i\circ {\tau }_j^{-1})(1)\in \mathrm{\Gamma }(U_i\cap U_j,{\mathcal{O}}_X)$; by step 1.2 the automorphism ${\tau }_i\circ {\tau }_j^{-1}$ of ${\mathcal{O}}_{U_i\cap U_j}$ is multiplication by $u_{ij}$, which is a unit because the automorphism is invertible. Computing the transition of the duals, for $a\in \mathrm{\Gamma }(U_i\cap U_j,{\mathcal{O}}_X)$ and with ${\phi }_i={\tau }_i^{\vee }{}^{-1}(1),{\phi }_j={\tau }_j^{\vee }{}^{-1}(1)$ one has ${\tau }_i^{\vee }(a{\phi }_j)=a{\phi }_j({\tau }_i^{-1}(1))=a({\tau }_j\circ {\tau }_i^{-1})(1)=a{u}_{ij}^{-1}$, since ${\tau }_j\circ {\tau }_i^{-1}=({\tau }_i\circ {\tau }_j^{-1}{)}^{-1}$ is multiplication by $u_{ij}^{-1}$; so the transition unit of ${\mathcal{L}}^{\vee }$ is $u_{ij}^{-1}$.

Compatibility with evaluation: under the trivialisation ${\tau }_i^{\vee }\otimes {\tau }_i$ the evaluation pairing becomes multiplication ${\mathcal{O}}_{U_i}\otimes {\mathcal{O}}_{U_i}\to {\mathcal{O}}_{U_i}$, $a\otimes b\mapsto ab$, by step 2.1, and under the transitions of step 2.2 both sides transform by $u_{ij}^{-1}$ and $u_{ij}$ respectively, so the transition unit of ${\mathcal{L}}^{\vee }\otimes \mathcal{L}$ is $u_{ij}^{-1}{u}_{ij}=1$ and evaluation is the identity trivialisation on overlaps; all constructions are local and canonical, so no choice principle is used.