Let $C$ be curve $f:C \to \mathbb{P}^1$ a finite morphism. This implies that $C= Spec(\mathcal{A})$ where $\mathcal{A}= f_*(\mathcal{O}_C)$ is the induced coherent relative $\mathcal{O}_{\mathbb{P}^1}$-algebra.
Let consider it's dual, the $\mathcal{O}_{\mathbb{P}^1}$-module $\underline{Hom}_{\mathbb{P}^1}(\mathcal{A}, \mathcal{O}_{\mathbb{P}^1})$. By the locally canonical $\mathcal{A}$-action via $s \times h \mapsto (a \mapsto h(sa))$ the $\mathcal{O}_{\mathbb{P}^1}$-module $\underline{Hom}_{\mathcal{O}_{\mathbb{P}^1}}(\mathcal{A})$ becomes an $\mathcal{A}$-module.
My question is why $\underline{Hom}_{\mathcal{O}_{\mathbb{P}^1}}(\mathcal{A}, \mathcal{O}_{\mathbb{P}^1})$ is quasi coherent?
My ideas: The problem is local. So choosing an affine open set $U = Spec(R) \subset \mathbb{P}^1$ we get $\mathcal{A}|_{f^{-1}(U)} = Spec(A)$ for finite presented $R$-module $A= \Gamma(U, \mathcal{A})$, so $A$ is the cokernel of the map $$R^m \to R^n$$.
Quasi coherence is a local property so is whould suffiece to show that $Hom_R(A,R)$ is a cokernel of a map $$A^I \to A^J$$ where $I,J$ are arbitrary index sets, or equivalently the sequence $$0 \to A^I \to A^J \to Hom_R(A,R) \to 0$$ is exact... but from here I don't know what to do (so how to use the finite presentation of $A$)