Showing a morphism $f : X \to \mathbb{P}^n_A$ is uniquely determined by an invertible sheaf $\mathcal{L}$ on a scheme $X$ generated by global sections So it is a well-known theorem that given an invertible sheaf $\mathcal{L}$ on a scheme $X$ over a ring $A$, such that $\mathcal{L}$ is generated by global sections $s_0, \dots, s_n$, then there exists a unique morphism $f : X \to \mathbb{P}^n_A$ such that $f^*x_i = s_i$ for all $i$.
I want to prove the uniqueness part of this theorem. The proof I've seen of this starts out more or less in the following way:
We first suppose a morphism $f : X \to \mathbb{P}^n_A$ such that $f^*x_i = s_i$ for all $i$ exists. Then we see that $X$ is covered by the open sets $X_{s_i} = \{x \in X \mid s_i(x) \neq 0\}$ and we can regard $f$ as being glued by the morphisms $$f_i = f|_{X_{s_i}} : X_{s_i} \to D_+(x_i) = \operatorname{Spec} A[x_0, \dots, x_n]_{(x_i)}$$ and this in turn corresponds to a morphism of $A$-algebras
$$f^\sharp : A[x_0, \dots, x_n]_{(x_i)} \to \Gamma(X_{s_i}, \mathcal{O}_X).$$
Then we notice that $x_i$ generates $\mathcal{O}(1)$ on $D_+(x_i)$ and that $x_j = \frac{x_j}{x_i}$ in $A[x_0, \dots, x_n]_{(x_i)}$ for $0 \leq j \leq n$.
And now this is where the argument gets a bit sketchy for me. The proof I've seen then goes on to say that similarly, pulling back via $f^\sharp$ gives $$s_j = f^*_i(x_j) = f_i^\sharp\left(\frac{x_j}{x_i}x_i\right) = f_i^\sharp\left(\frac{x_j}{x_i}\right)s_i$$ where we interpret $\frac{x_j}{x_i}$ as a fraction of $\Gamma(D_+(x_i), \mathcal{O}_{\mathbb{P}^n_A})$.
Then the proof claims that from each morphism $f : X \to \mathbb{P}^n_A$ we get $n+1$ distinguished sections $s_0, \dots, s_n$ from which we can determine the morphism $f_i$. (Call this statement (1))
And then finally the  proof concludes that $f$ is uniquely determined by the data $L, s_0, \dots, s_n$. (Call this statement (2))

Now I don't understand how statements 1 and 2 follow from what is done in the proof up until that point. In particular it is not clear to me how the sections $s_0, \dots, s_n$ can determine the morphism $f_i$, or what it even means for these sections to "determine the morphism" in this context.
 A: First, let us be careful with the notations: if $A$ is a ring, the $i$-th affine patch $D_+(x_i)$ in $\mathbb P^n_A=\operatorname{Proj}A[x_0,...,x_n]$ is given by $\operatorname{Spec}(A[(x_j/x_i)_{j\neq i}])$, not $\operatorname{Spec}(A[(x_j)_{0\leq j \leq n}]_{(x_i)})$.
Let $\mathcal L$ be an invertible sheaf on a scheme $X$, and let $s_0,...,s_n$ be global sections that generate $\mathcal L$ at every point. We want to use the $s_i$ in a natural way to define a morphism $X \to \mathbb P^n_X=\operatorname{Proj}(\mathcal O_X[x_0,...,x_n])$. (Then, by composition, any map $X \to \operatorname{Spec} A$ gives you a morphism $X \to \mathbb P^n_A$.) Intuitively speaking, we want this morphism to "send $x_i$ to $s_i$". Since the $s_i$ are not local sections of $\mathcal O_X$ per se, the formal way to do it is the following.
For each $i$, call $U_i$ the maximal open of $X$ on which $s_i$ generates $\mathcal L$. The $U_i$ cover $X$ by hypothesis. For each $i$ we have a natural morphism of $\mathcal O_X$-modules $f_i\colon \mathcal O_X\to \mathcal L$ mapping $1$ to $s_i$, and the restriction of $f_i$ to $U_i$ is an isomorphism. Let's call $s_j^{(i)}$ the preimage of $s_j$ in $O_X(U_i)$ under the isomorphism $f_i|_{U_i}$.
By the sheaf property, all we need to do is exhibit morphisms $U_i \to \mathbb P^n_X$ which coincide on the pairwise intersections $U_i \cap U_j$. Define a map from $U_i$ to the patch $D_+(x_i)$ of $\mathbb P^n_X$ by mapping $x_j/x_i$ to $s_j^{(i)}/s_i^{(i)}$. These maps glue accordingly since on $U_i \cap U_j$ we have $(s_i^{(j)}/s_j^{(j)})^{-1}=s_j^{(i)}/s_i^{(i)}$, so you have defined a morphism $X \to \mathbb P^n_X$.
This morphism is "determined by the $s_i$" in the sense that follows. Pick another set of global sections $(t_0,...,t_n)$ generating $\mathcal L$, and assume that the resulting morphisms $X \to \mathbb P^n$ are the same. Then it follows immediately from our definition of the morphism that for each $i$, the vanishing sets of $t_i$ and $s_i$ are equal, i.e. there is a unit $\lambda_i\in\mathcal O_X^\times(X)$ such that $s_i=\lambda_i t_i$. By the gluing conditions, we find that for each $i,j$ we have $\lambda_i=\lambda_j$ on $U_i \cap U_j$. Therefore, there exists a global section $\lambda$ of $\mathcal O_X^\times(X)$ such that for each $i$, we have $s_i=\lambda t_i$. This corresponds to the intuitive idea that $(s_0:...:s_n)=(t_0:...:t_n)$ in the projective space.
