How to show the following morphism is independent of the choice of neighborhoods?

Question

In Görtz and Wedhorn's AGI (3.4)

Let $X$ be a scheme. Let $x ∈ X$, and let $U \subset X$ be an affine open neighborhood of $x$, say $U = \mathop{Spec} A$. Denote by $p \subset A$ the prime ideal of $A$ corresponding to $x$. Then $\mathcal{O}_{X,x} = \mathcal{O}_{U,x} = A_p$, and the natural homomorphism $A \to A_p$ gives us a morphism $$j_x \colon \mathop{Spec} \mathcal{O}_{X,x} = \mathop{Spec} A_p \to \mathop{Spec}A = U \subset X $$ of schemes. By Proposition 3.2 (2) this morphism is independent of the choice of $U$.

The Proposition 3.2 (2) is the following:

Let $X$ be a scheme. The affine open subscheme are a basis of the topology.

My question: Why is this morphism independent of the choice of $U$?

Answer 1

Let's start with the Affine Communication Lemma

Let $X$ be a scheme. $U, \ V$ are open affine subschemes of $X$. Then given $x \in U \cap V \ \ \exists W \subset U \cap V$ such that $x \in W$ is distinguished open in both $U$ and $V$.

So in your case you get two morphisms $$ j_1 : Spec \ \mathcal O_{X,x} \rightarrow U$$ and $$ j_2 : Spec \ \mathcal O_{X,x} \rightarrow V $$

Since $ W$ is a distinguished open affine in $U$ the first morphism factors through $$ Spec \ \mathcal O_{X,x} \rightarrow W \to U$$ and the second factors through $$ Spec \ \mathcal O_{X,x} \rightarrow W \to V$$.

The first arrow in both the factors is the same morphism by commutativity while the second is just inclusion in both cases. So the composition gives you the same morphism.