The following question arose while thinking about a step in the proof of Huybrechts-Lehn, Theorem 1.3.1 (the Harder-Narasimhan filtration for the projective line $\mathbb{P}^{1}$):

**Setup**: Let $k$ be a field, let $X$ be a projective $k$-scheme, let $\mathcal{O}_{X}(1)$ be a fixed very ample line bundle on $X$, let $E$ be a vector bundle on $X$. For an integer $b \in \mathbb{Z}$, let $\mathrm{ev}_{E(b)} : \mathrm{H}^{0}(X,E(b)) \otimes_{k} \mathcal{O}_{X} \to E(b)$ be the evaluation map and let $E_{b} \subseteq E$ be the image of the twist $\mathrm{ev}_{E(b)}(-b) : \mathrm{H}^{0}(X,E(b)) \otimes_{k} \mathcal{O}_{X}(-b) \to E$.

**Question**: Why is $E_{b} \subset E_{b+1}$?

(And does this inclusion depend on a choice of global section $s \in \Gamma(X,\mathcal{O}_{X}(1))$?)

*Remark*: Note that $\mathrm{Hom}_{\mathcal{O}_{X}}(\mathcal{O}_{X}(-b),\mathcal{O}_{X}(-b-1)) = 0$ so there is no map $\varphi : \mathrm{H}^{0}(X,E(b)) \otimes_{k} \mathcal{O}_{X}(-b) \to \mathrm{H}^{0}(X,E(b+1)) \otimes_{k} \mathcal{O}_{X}(-b-1)$ satisfying $\mathrm{ev}_{E(b+1)}(-b-1) \circ \varphi = \mathrm{ev}_{E(b)}(-b)$.