Isomorphism of rings induced from schemes

Question

I'm new to schemes and reading Hartshorne. This is a detail in the proof of Proposition 3.2 chapter 2. I think it should be obvious but I don't see why.

Say we have $X=\operatorname{Spec}(A)$ is an affine scheme. Assume there is an open set $U=\operatorname{Spec}(B)$ of $X$. Then for some $f\in A$, $D(f)\subseteq U$. This is true because $D(f)$ form a base for the topology.

Now let $g$ be the image of $f$ in $B$. Hartshorne claims that $A_f \cong B_g$. I don't understand this detail. Sure there is such a map induced from $A\to B$. But I don't see any details that indicate this is an isomorphism

Answer 1

consider $D(f)=Spec\,A_f\to U=Spec\,B\to X=Spec A$ any map between affine schemes come from a map on rings so the natural map $A\to A_f$ factors through $B$, clearly $g$ goes to $f$ in $A_f$ so it is invertible in $A_f$, By universal property you get a map $B_g\to A_f$ which is easy to check that this is the inverse of natural map $A_f\to B_g$