How does $F:Spec \ A \rightarrow Spec \ A'$ induce $F':A' \rightarrow A?$

Question

I'm reading Algebraic Geometry by Ravi Vakil, and I'm trying to do exercise 4.3.A (the newest edition, I think June 2013). Show there is a bijection of the isomorphisms $\pi :Spec \ A \rightarrow Spec \ A'$ and $\pi ' : A' \rightarrow A.$ If $ f:A' \rightarrow A$ is an isomorphism of rings, we can induce $f':Spec \ A' \rightarrow Spec \ A$ by defining, for any prime ideal $P \in Spec \ A',$ $f'(P)=f^{-1} (P).$ But if we have a isomorphisms between sets of prime ideals, $\pi :Spec \ A \rightarrow Spec \ A',$ how does that induce an isomorphism $A' \rightarrow A?$

Answer 1

$\pi$ is not just a morphism of prime ideals it is a morphism of sheaves. And if $X=spec A$ then $A=\Gamma(X,\mathcal{O})$. So the map you are looking for is the induced map on global sections.