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).
 A: 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.
A: Thanks to KReiser for pointing out this only works for $X = \mathbb P^1$.
Essentially I use the Structure theorem for finite modules over PIDs. Locally, $X = \mathbb P^1$ is $\DeclareMathOperator{\Spec}{Spec} \Spec A$ for a PID $A$. Then $F$ has the form
$$\tag{1}F|_{\Spec A} \cong A/(a_1) \oplus \dotsm \oplus A/(a_r) \oplus A^n.$$
Then $\operatorname{Ker}(F \to F^{**})$ is exactly the torsion part $A/(a_1) \oplus \dotsm \oplus A/(a_r)$. This is true because any morphism
$$\varphi: A / (a_i) \to A$$
extends to a morphism $\psi: A \to A$ by precomposing with $A \to A / (a_i)$.
Then $\psi$ is given by multiplication with some element $b$. On the other hand, $b a_i = \psi(a_i) = 0$. Since $A$ is an integral domain, $b = 0$, and so $\varphi = 0$.
To patch the affine situation together on the scheme, one defines the torsion subsheaf
$$T = \operatorname{Tors}(F)$$
over an open $U \subset X$ to consist of all torsion elements, i.e. $T(U) = \{\, a \in F(U) \,|\, \exists b \in \mathcal O_X(U): ab = 0 \,\}$. Then under the isomorphism (1) you get
$$T(\Spec A) \cong A/(a_1) \oplus \dotsm \oplus A/(a_r).$$

For the general situation, one might still consider the torsion subsheaf $\DeclareMathOperator{\Tors}{Tors} \Tors(F) \subset F$, which is the maximal zero-dimensional subsheaf of $F$. For affine opens $U \subset X$ this is exactly the torsion submodule $\{s \,|\, as = 0 \text{ for some } a \in \mathcal O(U)\,\}$, and on stalks we have
$$\Tors(F)_x = \Tors(F_x).$$
So the structure theorem implies that the quotient $E = F / \Tors(F)$ is locally free, because all local rings $\mathcal O_{X,x}$ are PIDs by the smoothness assumption.
From this description it is straight forward to see that $\Tors F = \operatorname{Ker}(F \to F^{**})$: both are subsheaves of $F$ which agree on all stalks.
Right now I don't have a really good argument why the sequence
$$0 \to \Tors(F) \to F \to E \to 0$$
splits.
