Let $\mathcal{F}$ and $\mathcal{G}$ be semistable sheaves with the same Hilbert polynomial, i.e. $P(\mathcal{F})=P(\mathcal{G})$ (pay attention that these are Hilbert polynomial and not reduced Hilbert polynomial). If $\mathcal{F}$ or $\mathcal{G}$ is stable, than why is any nonzero morphism $f:\,\mathcal{F}\to\mathcal{G}$ an isomorphism? Alternatively, why every nonzero morphism, that is injective or suriective, an isomorphism?

$\begingroup$ Why do you know that your assertions are true? $\endgroup$ – Stefan Kohl Jul 5 '14 at 17:23

1$\begingroup$ Dear Stephan, the first assertion is proposition 1.2.7 of the book "The Geometry of moduli spaces of Sheaves" by HuybrechtsLehn and the second assertion is their answer to the first one. I don't understand their reason and I am not able to prove the first assertion by myself. $\endgroup$ – User3773 Jul 5 '14 at 17:35
Suppose $f:F \to G$ is injective, and let $C$ be the cokerel, so we have an exact sequence $0 \to F \to G \to C \to 0$. Since the Hilbert polynomial is additive on short exact sequences, and $P(F)=P(G)$, we get that $P(C)$=0. So for all $m>>0$, $h^0(X,C(m))=P(C)(m)=0$. Serre has a theorem that says that for all $m>>0$ $C(m)$ is globally generated. But we just saw that for $m>>0$, $C(m)$ has no global sections besides $0$, and so $C(m)=0$, so $C=0$.
The case when $f$ is surjective is similar; look at the kernel of $f$.