Let $T\to S$ be scheme morphism, and let $\mathcal{L}$ be a globally generated line bundle on $T$. By choosing $n$ global sections of $\mathcal{L}$ that generate at each point, we obtain a morphism $T\to \mathbb{P}^n_S$. With this generality, is there a coordinate-free way to view this morphism? What if $T$ and $S$ are of finite type over a field?
For example, I believe that in the particular case that $S=\text{Spec}(k)$ for $k$ a field, and the $n$ sections are linearly independent over $k$, then we can identify $\mathbb{P}^n=(\mathbb{P}H^0(T,\mathcal{L}))^*$ and the map can be taken to be
$$x\mapsto\{s\in H^0(T,\mathcal{L}):s_x\in\frak{m}_x\mathcal{L}_x\}.$$
Best Answer
I don't believe you can find a totally coordinate-free way to phrase things, unless you're just tautologically hiding the coordinates. The choice of $n+1$ global sections of $\mathcal{L}$ which generate are the coordinates on $\Bbb{P}^n_S$, so you're implicitly starting with a set of coordinates. Moreover, the line bundle $\mathcal{L}$ does not uniquely determine a collection of global sections which give a unique morphism $T\to\Bbb{P}^n_S$ -- this choice is essential, and can give you different morphisms.
Of course, you can rephrase your starting assumptions. Let $f : T\to S$ be the original morphism. Your sections are equivalent to the data of a surjection $f^\ast\mathcal{O}_S^{n+1} \cong\mathcal{O}_T^{n+1}\to\mathcal{L},$ and such a surjection is precisely the data of an $S$-morphism $T\to\Bbb{P}(\mathcal{O}_S^{n+1}) \cong \Bbb{P}^n_S,$ as if $\mathcal{E}$ is a locally free sheaf (of finite rank) on $S,$ then $$\operatorname{Hom}_S(T,\Bbb{P}(\mathcal{E}))\cong\{\textrm{invertible quotients of }f^\ast\mathcal{E}\}.$$
But there are many morphisms $T\to\Bbb{P}^n_S$ in general, and they will depend on the coordinates you pick, which your example does not seem to.
As I described in my comment, if we don't require $n$ such that $n+1$ is the minimum number of global sections required to generate $\mathcal{L}$, then there really isn't a nice "coordinate-free" way to describe the morphism. Let $T = S = \operatorname{Spec}k,$ and let $\mathcal{L} = \mathcal{O}_T,$ and let $n=1.$ Then the choice of two global sections is the choice of a point $[a_0 : a_1]\in\Bbb{P}_k^1,$ but this choice will give you completely different points as you change what global sections you pick. You could say that this morphism simply picks out a point in $\Bbb{P}(\mathcal{E}),$ where $\mathcal{E}$ is free of rank $2,$ but you already knew this because $T = \operatorname{Spec}k.$
Even if we force $n$ to be such that $n+1$ is the minimum number of global sections required to generate $\mathcal{L},$ we cannot say too much. let $T =\Bbb{P}^1_k$, $S = \operatorname{Spec}k,$ and let $\mathcal{L} = \mathcal{O}(1).$ Then we need $n+1$ global sections to generate $\mathcal{L},$ but we have many choices of what these may be. We might choose $x_0,\dots, x_n,$ but we might also choose $x_0 + x_1 + \dots + x_n, x_1 + \dots + x_n,\dots, x_{n-1} + x_n, x_n.$ The corresponding maps are distinct maps to projective space, as the two invertible quotients are not equivalent: there is no isomorphism $i : \mathcal{L}\to\mathcal{L}$ making the diagram $$\require{AMScd} \begin{CD} \mathcal{O}_T^{n+1} @>(x_0,x_1,\dots,x_n)>> \mathcal{L} \\ @V{\operatorname{id}}VV @VViV\\ \mathcal{O}_T^{n+1} @>>(x_0 + \dots + x_n, \dots, x_n)> \mathcal{L} \end{CD} $$ commute (recall that $\operatorname{Isom}(\mathcal{L},\mathcal{L})\cong\mathcal{O}(T)^\times$ for any line bundle $\mathcal{L}$ on $T$). That is, there is no isomorphism of invertible quotients between $(x_0,\dots, x_n) : \mathcal{O}_T^{n+1}\to\mathcal{L}$ and $(x_0 + \dots + x_n,\dots, x_n) : \mathcal{O}_T^{n+1}\to\mathcal{L}.$
Perhaps you have something particular in mind when you say coordinate-free, or a particular type of description you're looking for. If this is the case and my answer does not address those, it would be helpful to narrow down exactly what you're looking for and requiring, and add that to the question.