Let $C$ be regular curve. Consider a finite and locally free morphism $f: C \to \mathbb{P}^1$. (the latter mean that $f_* \mathcal{O}_{C}$ is a free $\mathcal{O}_{P^1}$ module)
Let $\mathcal{F}$ be a quasicoherent sheaf of finite type on $C$.
Assume that we know that the pushforward $f_*\mathcal{F}$ is torsion-free. How to deduce that then $\mathcal{F}$ is also torsion-free?
My ideas: the problem is local and $f$ is affine (because $f$ finite). Therefore $f$ reduces to a morphism $\phi:= f^{\#}: R \to A$ of Dedekind rings $R = \Gamma(U, \mathcal{O}_{P^1}),R = \Gamma(f^{-1}(U), \mathcal{O}_{C}) $.
In this setting $\mathcal{F}= \widetilde{M}$ for $A$-module $M$.
The pushforward is $f_*\mathcal{F}= f_*\widetilde{M}= \widetilde{M \vert_R}$.
By assumption classification theorem for finitely generated modules over Dedekind rings $M \vert_R$ is free $R$-module since it has no torsion.
So it suffice to show that $M$ is also free as $A$-module.
But how? The $A$-module structure of $M$ is fixed at the beginning. The naive approach by tensoring $M \vert_R$ by $A$ induces a (free)$A$-module structure on $M \vert_R$ but it doesn't "bring" the previous $A$-module structure back.