Morphism into Section Well Defined My question concerns the map $\Gamma_*(X, \mathcal{L})_{(s)} \to \Gamma(X_s, \mathcal{O}_X)$ as given in https://stacks.math.columbia.edu/tag/01PW.
Here $X$ is a scheme, $\mathcal{L}$ an invertible sheaf and $\Gamma_*(X, \mathcal{L}) = \bigoplus _n \Gamma(X, \mathcal{L} ^{\otimes n})$ the induced canonical graduated ring. Futhermore $X_s = \{x \in X \mid s \not \in \mathfrak m_x\mathcal{L}_x\}$ for $s \in \Gamma(X, \mathcal{L}) $.
This map is explicitly given by $a/s^n \to (a \mid _{X_s}) \otimes (s^{-n} \mid _{X_s})$, where $grad(a) = n \ grad(s)$.
My question is why this map is well defined, so why $(a \mid _{X_s}) \otimes (s^{-n} \mid _{X_s}) \in \Gamma(X_s, \mathcal{O}_X)$?
And especially what't going wrong if we DON'T restrict the tensor product to $X_s$?
Here the regarding excerpt:

 A: Let $X$ be a scheme, let $\mathcal{L}$ be an invertible $\mathcal{O}_{X}$-module, fix an element $s \in \Gamma(X,\mathcal{L})$. Let $\xi_{s} : \mathcal{O}_{X} \to \mathcal{L}$ be the $\mathcal{O}_{X}$-linear map sending $1 \mapsto s$. Then $X_{s}$ is the largest open subset of $X$ such that the restriction $\xi_{s}|_{X_{s}} : \mathcal{O}_{X_{s}} \to \mathcal{L}|_{X_{s}}$ is an isomorphism, and there exists some $s' \in \Gamma(X_{s} , \mathcal{L}^{\otimes -1}|_{X_{s}})$ such that the map $$ \Gamma(X_{s} , \mathcal{L}^{\otimes n}|_{X_{s}}) \otimes_{\Gamma(X_{s} , \mathcal{O}_{X_{s}})} \Gamma(X_{s},\mathcal{L}^{\otimes -n}|_{X_{s}}) \to \Gamma(X_{s},\mathcal{L} \otimes_{\mathcal{O}_{X}} \mathcal{L}^{\otimes -1}) \stackrel{\sim}{\to} \Gamma(X_{s},\mathcal{O}_{X_{s}}) $$ sends $s^{\otimes n}|_{X_{s}} \otimes (s')^{\otimes n} \mapsto 1$ for any $n \ge 0$. (In 01CY this is stated for $n=1$ but it holds for all $n \ge 1$.)
Then the map $$ \Gamma_{\ast}(X,\mathcal{L})_{(s)} \to \Gamma(X_{s} , \mathcal{O}_{X}) $$ is defined by sending $a/s^{n} \mapsto a|_{X_{s}} \otimes (s')^{\otimes n}$, where $a \in \Gamma(X,\mathcal{L}^{\otimes n})$ for any $n \ge 0$.
In other words the order of "restriction" and "taking inverse" matters, namely "$(s|_{X_{s}})^{-n}$" makes sense while "$(s^{-n})|_{X_{s}}$" does not.
