Kernel of $F\to F^{**}$ has rank $0$ I want to show that every coherent sheaf over a curve is a direct sum of a torsion sheaf and a locally free sheaf via the following:
Let $X$ be a smooth projective $1$-dimensional scheme. Let $F$ be a coherent sheaf on $X$ and $T$ the kernel of the canonical map $F \to F^{**}$.
Then

*

*$T$ has rank 0

*$F = T \oplus $(locally free sheaf).

Why is this the case? Note that I do not assume that $F$ is torsion-free (=locally free).

I'll do the affine version of this, which is commutative algebra.
$\def\Ext{\operatorname{Ext}}$Let $A$ be commutative integral domain  of global dimension $1$ (for example, the coordinate algebra of an affine smooth curve) and let $M$ be finitely presented $A$ module. Let us write $(-)^*$ for the functor $\hom(-,A)$.
Consider a finite projective resolution $0\to P_1 \xrightarrow{d} P_0 \to M\to 0$. Applying $(-)^*$ we get an exact sequence $$0\to M^*\to P_0^*\to P_1^*\to\Ext^1(M,A)\to 0$$ The modules $P_i^*$ are finitely generated projectives: since $A$ has global dimension $1$, we see that the finitely generated module $M^*$ is projective, and what we have is a projective resolution of $\Ext^1(M,A)$, of length $2$. We apply $(-)^*$ to it, and obtain what is now a complex: $$P_1^{**}\to P_0^{**}\to M^{**}\to 0$$ that has $P_1^{**}$ in degree zero and differentials going up: its cohomology is $\Ext^\bullet(\Ext^1(M,A),A)$. As $A$ has global dimension $1$, the map $P_0^{**}\to M^{**}$ is surjective. Since the map $P_1^{**}\to P_0^{**}$ is just $P_1\xrightarrow{d}P_0$, because the $P_i$ are reflexive, its cokernel is just $M$. We thus get a map $M\to M^{**}$, and looking at it hard shows that it is the canonical one. We also get a description of the kernel of this map, and, putting everything together, an exact sequence $$0\to\Ext^1(\Ext^1(M,A),A)\to M\to M^{**}\to 0$$ As $M^*$ is f.g. projective, so is $M^{**}$, and this sequence splits and there is an iso $$M\cong M^{**}\oplus\Ext^1(\Ext^1(M,A),A).$$
Tensoring the exact sequence with the fraction field of $A$ gives an exact sequence, and the map obtained from $M\to M^{**}$ is an iso: it follows from this that $\Ext^1(\Ext^1(M,A),A)$ is torsion.
