# Semi-stable locally free sheaves on a curve of constant slope form an abelian category

I have some difficulty understanding the above proof.

Here $C(\mu)$ denotes the category of semi-stable vector bundles (locally free sheaf) of slope $\mu$. Here the sheaves are considered over a smooth connected algebraic curve $X$.

A locally free sheaf $F$ is called semi-stable if for every subsheaf $G$, the slope of $G$, $\mu(G)$, is less than equal to slope of $F$, $\mu(F)$. The slope of a locally free sheaf is defined as $deg(F)/rank(F)$.

I've two questions about the above proof

1. It says that the $ker(f)$ and $coker(f)$ are locally-free sheaves. Well, the $ker(f)$ is a subsheaf of $E$, which is a locally-free sheaf, so $ker(f)$ is also a locally free sheaf ( because these are sheaves on a smooth curve and therefore stalks are pid, and a submodule of a free module over a pid is again a free module). But I've absolutely no idea how it says $coker(f)$ is also a locally-free sheaf. Why is this true?

2. In the proof of $ker(f)$ is semistable, it says if it's not semistable it would have a sub-bundle of slope of slope greater than $\mu$. So, somehow there are assuming slope of $ker(f)$ is also $\mu$. So, I guess since slope of $E$ and $F$ is $\mu$ and they are semistable then it is forcing the slope of both $ker(f)$ and $coker(f)$ to be $\mu$. Is this correct?

1) If $Coker(f)$ is not locally free, it has a summand $T$ which is a torsion sheaf. The kernel $G$ of the composition $F \to Coker(f) \to T$ is a locally free sheaf, whose slope is less than $\mu$. The image of $f$ is contained in $G$, so, we obtain a morphism $E \to G$ from a semistable sheaf to a sheaf of smaller slope, which is impossible.
2) Consider the image $G = Im(f)$. We have an epimorphism $E \to G$, hence all quotients of $G$ have slope greater or equal than $\mu$. On the other hand, we have a monomorphism $G \to F$, hence all subsheaves in $G$ have slope less or equal than $\mu$. It follows that in the Harder-Narasimhan filtration of $G$ all factors have slope $\mu$, hence $G$ is semistable of slope $\mu$. Therefore $Ker(f) = Ker(E \to G)$ and $Coker(f) = Coker(G \to F)$ also have slope $\mu$.
• Thank you very much for the answer! I don't understand why the slope of $G$ should be less than $\mu$. Why can't it be equal to $\mu$? – grok Feb 10 '18 at 17:22
• @grok: Because its rank equals the rank of $F$, and its degree is equal to the degree of $F$ minus the length of $T$, i.e., less than the degree of $F$. – Sasha Feb 11 '18 at 8:15
• Thanks for the reply! What exactly do you mean by the term length of $T$? – grok Feb 12 '18 at 4:31