$\underline{Background}$:Let,$X$ be a scheme(for simplicity we can assume it to be noetherian).Let $\mathcal F$ be a locally free sheaf of rank $r$ on $X$.Let, us also assume that {$\operatorname{Spec}A_i:i\in I$} is its trivializing open sets,i.e
we have $\forall i\in I$ , $\mathcal F_{\big|\operatorname{Spec}cA_i}\cong{\mathcal O_{\operatorname{Spec}cA_i^{\oplus r}}}\cong{\oplus_{\substack r}}\tilde{A_i}$
$\underline{Aim}$:I am trying to give a $O_X$ module morphism from $\operatorname{End}(\mathcal F)=\mathcal Hom(\mathcal F,\mathcal F)$ to $O_X$.
$\underline{Attempt}$:$1)$ we can give maps $f(\operatorname{Spec}A_i):\mathcal Hom (F_{\big|\operatorname{Spec}A_i},\mathcal F_{\big|\operatorname{Spec}A_i}) \to \mathcal O(\operatorname{Spec}A_i)=A_i$,
which is equivalent to give a morphism $f(\operatorname{Spec}A_i):\mathcal Hom(\oplus_{\substack r}\tilde{A_i},\oplus_{\substack r}\tilde{A_i})\to A_i$
Let,$g\in \mathcal Hom(\oplus_{\substack r}\tilde{A_i},\oplus_{\substack r}\tilde{A_i})$
$\Rightarrow g(\operatorname{Spec}A_i):\oplus_{\substack r}{A_i}\to \oplus_{\substack r}{A_i}$ is an $A_i$ module morphism.
Since $\oplus_{\substack r}{A_i}$ is a free $A_i$ module of rank r, let {$e_1,…..e_r$} be a basis(its canonical basis say).
Then let the matrix(w.r.to that basis) corresponding to $g(\operatorname{Spec}A_i)$ is $\mathcal M_g$.
Then we define $f(\operatorname{Spec}A_i)(g)=\operatorname{trace}(\mathcal M_g)\in{A_i}$
$\underline{Question}$: Even if we assume that $f(\operatorname{Spec}A_i)$s are valid $A_i$-module morphism how can we say that $\exists \phi :\operatorname{End}(\mathcal F)\to {\mathcal O_X}$ such that $\phi(\operatorname{Spec}A_i)$ is same as $f(\operatorname{Spec}A_i)$ ?
Because we know that we can do that if we have {$\operatorname{Spec}A_i:i\in I$} forms a base of $X$ and $f(\operatorname{Spec}A_i)$s are morphisms of sheaves on that base.but we only have {$\operatorname{Spec}A_i:i\in I$} forms a cover of $X$ ,not necessarily a base.
$\underline{Attempt(2)}$:So to avoid the base problem if we define maps at all affine opens {$\operatorname{Spec}A$}, then we have $g(\operatorname{Spec}A):\mathcal F(\operatorname{Spec}A)\to{\mathcal F(\operatorname{Spec}A)}$.
Now ,we have $\mathcal F$ is coherent on $X$,so $ \mathcal F_{\big|\operatorname{Spec}A}$s are coherent on $\operatorname{Spec}A$.But that does not necessarily mean that $\mathcal F(\operatorname{Spec}A)$s are finitely generated $A$ module,so we do not necessarily have corresponding matrix and hence its trace.
$\underline{Question}$:How do we define $f(\operatorname{Spec}A)$ in this case?
$\underline{Question}$(not directly related to previous discussion)}:For $\mathcal F$ a rank $r$ locally free sheaf is $\operatorname{End}(\mathcal F)$ always a locally free sheaf of rank r?
Any help from anyone is welcome.
Best Answer
For your first question : no you don't need to define a morphism of sheaves $f:\mathcal{F}\to\mathcal{G}$ on a basis. It is enough to define it on a cover $\{U_i\}$. However, you need to check that this agree on the two-fold intersections $U_i\cap U_j$. With this in mind, your attempt 1) can work, you just need to check that the morphisms you defined agree on the intersection of two open subset of the cover.
For your second question and second attempt : yes a coherent sheaf is finitely generated on any affine subset. But you still need to check that the morphisms you defined are compatible with restrictions.
For your last question. $\mathcal{F}$ is locally isomorphic to $\mathcal{O}_X^r$, so $End(\mathcal{F})$ is locally isomorphic to $End(\mathcal{O}_X^r)$ which is the sheaf of $r\times r$-matrices. Hence $End(\mathcal{F})$ is locally free of rank $r^2$.