Induced Map on Spectra by Integral Ring Homomorphism I do not want the solution, but this is rather a question about clearing a misunderstanding.
Atiyah 's 5.1 asks to show that given an integral ring homomorphism $f:A \to B$, the induced map $f^*:Spec(B) \to Spec(A)$ is closed map.
I simply showed that $$f^*(V(J))=V(f^{-1}(J))$$ which is a closed set in $Spec(A)$. I am surprised that I didn't use integral homomorphism. Where exactly am I wrong?
 A: $\newcommand{\pp}{\mathfrak{p}}\newcommand{\qq}{\mathfrak{q}}$I'll copy your argument (with slight modifications) from the comments:

I was able to show one inclusion. Now I want to prove that $V(f^{-1}(J)) \subset f^{\ast}(V(J))$. For that I am choosing a prime ideal $\pp \in V(f^{-1}(J))$. Then it must happen that $\pp$ contains $f^{-1}(J)$. Now by lifting property, I have a $\color{red}{\text{prime}}$ ideal $\pp' \subset B$ such that $f^{-1}(\pp')=\pp$. But now all I need is to prove that $J \subset \pp'$ but $f^{-1}(J) \subset f^{-1}(\pp')$ doesn't imply $J \subset \pp'$.

You are almost there. Note that you have already used the integrality to invoke the lifting property (for example, in the non-integral extension $\Bbb Z \hookrightarrow \Bbb Q$, you cannot find a lift of $2 \Bbb Z$).
As you have noted, it is not direct that $J \subset \pp'$. The way to combat this is to look at quotients instead, where your prime ideals will manifestly contain the desired ideals.
To elaborate, let $I = f^{-1}J$ and consider the extension $A/I \hookrightarrow B/J$. This extension is integral (Proposition 5.6). Moreover, $\pp/I$ is a prime ideal in $A/I$ (where $\pp$ is as before). Now, using the lifting theorem gives you an ideal $\pp'/J$ in $B/J$ that contracts to $\pp/I$. This corresponds to a prime ideal $\pp' \subset B$ containing $J$ that contracts to $\pp$. (Check this if you're unsure.)
This finishes the proof.
