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 $\mathrm{ev}\_{E(b+1)}(-b-1) \circ \varphi = \mathrm{ev}\_{E(b)}(-b)$.



The evaluation map on twists of a vector bundle and an induced filtration
There is a commutative diagram
$$\require{AMScd}
\begin{CD}
H^0(E(b)) \otimes H^0(\mathcal{O}(1)) \otimes \mathcal{O}(-b-1) 
@>{\mathrm{ev}\_{\mathcal{O}(1)}}>> 
H^0(E(b)) \otimes \mathcal{O}(-b) 
\\
@VVV @V{\mathrm{ev}\_{E(b)}}VV
\\
H^0(E(b+1)) \otimes \mathcal{O}(-b-1) 
@>{\mathrm{ev}\_{E(b+1)}}>> 
E
\end{CD}
$$
where the left vertical arrow is the multiplication of global sections. The top arrow is surjective (because $\mathcal{O}(1)$ is very ample), hence the composition of the top and right arrows has image $E\_b$. By commutativitity of the diagram it is contained (canonically) in the image $E\_{b+1}$ of the bottom arrow.


