I have a question regarding the exercise 3.29 in Atiyah-MacDonald. In the previous exercises it was introduced the constructible topology. Basically the space $\DeclareMathOperator{\Spec}{Spec}\Spec(A)$ for a ring $A$ is endowed with a topology such that the closed set are images of $f^\ast(\Spec(B))$ where $f:A\to B$ is a ring homomorphism. It was shown that this approach provides indeed a topology and is finer than the Zarisky topology. Now the question: Let $f:A\to B$ a ring homomorphism is $f^\ast:\Spec(B)\to \Spec(A)$ continuous if $\Spec(A)$ and $\Spec(B)$ are endowed with the constructible topology?
My attempt: Let $g:A\to C$ a ring homomorphism then \begin{equation*} f^{\ast-1}(g^\ast(\Spec(C)))=f^{\ast-1}(g^\ast(\Spec(C))\cap f^\ast(\Spec(B)))= f^{\ast-1}(h^\ast(\Spec(C\otimes_A B))), \end{equation*} where $h:A\to C\otimes_A B$ is the ring homomorphism $h(x)=1_C\otimes f(x)=g(x)\otimes 1_B$. If we now use the mapping $\alpha:B\to C\otimes_A B, b\mapsto 1_C\otimes b$ then one has $h=\alpha\circ f$ hence $h^\ast=f^\ast\circ \alpha^\ast$, but from this follows only $\alpha^\ast(\Spec(C\otimes_A B))\subset f^{\ast-1}(g^\ast(\Spec(C)))$. Is there a way to show equality or is the continuity trivial and I don't see it? Or is the map $f^\ast$ not continuous in general with respect to the constructible topology?
Have you done exercise 3.28 yet? If not you should do it first, if so then what you've shown is that the $X_f$ and $V(f)$ sets give a sub-basis for the constructable topology, so to show that $f^\ast$ is continuous it suffices to show that sets of the form $f^{\ast -1}(X_f)$ and $f^{\ast -1}(V(g))$ are closed. That fact is $(i)$ and $(ii)$ of exercise 1.21.