Is $\kappa$ an isomorphism？ Given a morphism of schemes $X\overset{\varphi}{\rightarrow}\operatorname{Spec}A, \forall f\in A$.
We have the following two commutative diagrams.
$$\begin{array}[c]{ccc}
A&\stackrel {\lambda}{\longrightarrow}&\mathcal{O}(X)\\
\downarrow&&\downarrow\\
A_f&\stackrel {\mu}{\longrightarrow}&\mathcal{O}(\varphi^{-1}(D(f)))\\
\end{array}$$
$$\begin{array}[c]{ccc}
A&{\longrightarrow}&A/\ker\lambda\\
\downarrow&&\downarrow\\
A_f&{\longrightarrow}&A_f/\ker\mu
\end{array}$$
$$(A/\ker\lambda)_{f+\ker\lambda}\overset{\kappa}{\longrightarrow} A_f/\ker\mu$$
$$\frac{a+\ker\lambda}{f^n+\ker\lambda}\mapsto \frac{a}{f^n}+\ker\mu$$
$$\because\frac{a+\ker\lambda}{f^n+\ker\lambda}=\frac{b+\ker\lambda}{f^m+\ker\lambda}\Rightarrow f^k(af^m-bf^n)\in \ker\lambda \Rightarrow \frac{a}{f^n}+\ker\mu=\frac{b}{f^m}+\ker\mu,$$
$\therefore \kappa$ is defined well.
Is $\kappa$ an isomorphism?
 A: It's enough to prove that $A_f/I_f\cong (A/I)_{\overline{f}}$ for all ideals $I\subset A$ and all $f\in A$, where $\overline{f}$ denotes the image of $f$ under the natural quotient map $A\to A/I$. Geometrically, both sides corrsepond to intersecting the open subscheme $\operatorname{Spec} A_f$ with the subscheme corresponding to the ideal $I$, so depending on what level of rigor you're after, this could be considered a proof.
Here is a full proof that $\kappa:A_f/I_f\to (A/I)_{\overline{f}}$ is an isomorphism. First, note that we may assume $f\notin I$, as otherwise both sides are the zero ring. Let $\varphi: A\to (A/I)_{\overline{f}}$ be the composition of the natural maps $A\to A/I$ and $A/I \to (A/I)_{\overline{f}}$. Then as $\varphi(f)$ is invertible, $\varphi$ factors through $A_f$, ie it gives $\widetilde{\varphi}: A_f\to (A/I)_{\overline{f}}$ which is a map of $A_f$ modules. Next, note that $I_f$ is exactly the kernel of this map - it is clear that $\widetilde{\varphi}(I_f)=0$ and for the reverse direction $\widetilde{\varphi}(a)=0 \Leftrightarrow \widetilde{\varphi}(f^na)=0$, but for some $n>0$, $f^na\in A\subset A_f$ which implies that $\varphi(f^na)=0$ and thus $f^na \in I \Rightarrow a\in I_f$.
