In Huybrechts and Lehn it says that for two semistable sheaves $\mathcal{F},\mathcal{G}$ and a sheaf homomorphism $f : \mathcal{F} \to \mathcal{G}$, writing $$\mathcal{E} := \text{im}(f)$$ yields the inequality $$p(\mathcal{F}) \leq p(\mathcal{E}) \leq p(\mathcal{G}),$$ where $p(-)$ denotes taking the Hilbert polynomial. I understand why $p(\mathcal{E}) \leq p(\mathcal{G})$; it is because $\mathcal{G}$ is semistable and $\mathcal{E}$ is a subsheaf of it.
Yet, I do not see why $p(\mathcal{F}) \leq p(\mathcal{E})$. In fact, I am so clueless that I see the opposite! For considering the exact sequence $$0 \to \text{ker}(f) \to \mathcal{F} \to \mathcal{E} \to 0$$ we get $p(\mathcal{F}) = p(\text{ker}(f))+ p(\mathcal{E})$ and so $p(\mathcal{E}) \leq p(\mathcal{F})$ by positivity of the coefficients of the Hilbert Polynomial.