At page 69，chapter 3.4 in Görtz & Wedhorn 's "Algebraic Geometry 1" book, there is the following text:
Let $X$ be a scheme. Let $x \in X$, and let $U \subset X$ be an affine open neighborhood of x, say $U = \operatorname{Spec}A$. Denote by $p \in \operatorname{Spec}A$ the prime ideal of $A$ corresponding to $x$. Then $\mathscr O_{X,x} =\mathscr O_{U,x} = A_p$, and the natural homomorphism $A \rightarrow A_p$ gives us a morphism of scheme
$j_x :\operatorname{Spec} \mathscr O_{X,x} = \operatorname{Spec} A_p \rightarrow \operatorname{Spec}A = U \subset X$
By Proposition $3.2(2)$ this morphism is independent of the choice of $U$.
Proposition $3.2(2)$: Let $X$ be a scheme. The affine open subschemes are a basis of the topology.
I just don't understand why $j_x$ is independent of the choice of $U$... Let's choose a different $U' \subset A$, then the $j'_x$ with respect to $U'$ is on longer the same as the original $j_x$ since they have a different codomain!(Although the domains are the same.)
My question is: In what sense does $j_x$ consider themselves the same?
Let $U = \operatorname{Spec}(A), V = \operatorname{Spec}(B)$ be affine neighbourhoods of $x$ with $x$ corresponding to primes $\mathfrak{p}, \mathfrak{q}$ in $A, B$ respectively. The map $A\to A_{\mathfrak{p}}$ is essentially the map from the sections on $U$ to the stalk, i.e., sending a section $s$ to the associated germ $(U, s)$.
Let $\operatorname{Spec}(A_f)$ be an affine neighbourhood of $x$ in $U\cap V$ (note here that $A_f$ is inverting $f\in A$). The inclusion $\operatorname{Spec}(A_f)\hookrightarrow V$ induces a map (of rings) $B\to A_f$.
Because everything glues well(the following goes back and forth between the locally ringed structure on $X$ and the affine schemes $U, V$), the following square commutes (the simpler case is when $U = V$)
$$\require{AMScd} \begin{CD} B @>>> A_f\\ @VVV @VVV \\ B_{\mathfrak{q}} @= A_{\mathfrak{p}} \end{CD}$$ (where the equality is an isomorphism through a hidden $\mathcal{O}_{X, x}$). To see that it commutes you have to go from $B$ to the sections over $V$, restrict to the sections over $\operatorname{Spec}(A_f)$ etc.
Now, because it commutes, the map $\operatorname{Spec}(\mathcal{O}_{X, x})\to \operatorname{Spec}(B)$ factors through $\operatorname{Spec}(A_f)$. Similarly, the one to $\operatorname{Spec}(A)$ factors through this same map.
So, if you take a prime in $\mathcal{O}_{X, x}$ it maps to a point in $U$ and one in $V$ and both of these are the same element living in the intersection.
When you include the codomain open sets in $X$, what this says is that the maps $\operatorname{Spec}(\mathcal{O}_{X, x})\to X$, induced as above, factor through every (affine) open neighbourhood of $x$ and is independent of the choice of such a neighbourhood (because given two neighbourhoods, it factors through an even smaller neighbourhood). Intuitively, the points of $\operatorname{Spec}(\mathcal{O}_{X, x})$ are points in every neighbourhood of $x$ and in the affine case, they are primes "below" $\mathfrak{p}$.