This is regarding Theorem $18.4$ and Theorem $19.6$ from Topology by Munkres.
Theorem 18.4: Let $$f:A\rightarrow X\times Y$$ be given by the equation $$f(a)=(f_1(a),f_2(a)).$$ Then $f$ is continuous iff $f_1$ and $f_2$ are continuous.
Theorem 19.6: Let $$f:A\rightarrow \prod_{\alpha \in J} X_{\alpha}$$ br given by $$f(a)=(f_{\alpha}(a)))_{\alpha \in J}.$$ Where $f_{\alpha}:A\rightarrow X_{\alpha}$ for each $\alpha.$ Then $f$ is continuous iff $f_{\alpha}$ is continuous for each $\alpha$ provided the product space $\prod_{\alpha \in J} X_{\alpha}$ is given product topology.
Now , $18.4$ clearly shows that result holds irrespective of the type of topology when the product is finite.While $19.6$ shows that the results holds only for product topology and not for box topology as illustrated later by example $2$ , when we are taking arbitrary product.
Now, for $19.6$ , let us take an arbitrary open set from $\prod_{\alpha \in J} X_{\alpha}$ , say $U=\prod_{\alpha\in J} U_{\alpha}$ where each $U_{\alpha}$ is open in $X_{\alpha}.$ Now,$a\in f^{-1}(U) \iff a\in \bigcap_{\alpha \in J} \{f_{\alpha}^{-1}(U_{\alpha})\}.$ Now by continuity of each $f_\alpha$ we know each of $f_\alpha^{-1}(U_{\alpha})$ is open in $A$. But since arbitrary intersection of open sets may or may not be open so so the proof similar to $18.4$ does not work here. Later in the example we see that indeed the intersection ends up becoming $\Phi$ for box topology and thus unable to contain any point.
But , had we taken a closed set say, $C=\prod _{\alpha\in J}C_\alpha$ from $\prod_{\alpha \in J} X_{\alpha}$ s.t each $C_{\alpha}$ is closed in $X_{\alpha}$ containing a point $x=(x_{\alpha})_{\alpha}$, then since arbitrary intersection of closed set is again a closed set we would have $f^{-1}(C)=\bigcap_{\alpha \in J} \{f_{\alpha}^{-1}(C_{\alpha})\}$ will be another closed set in $A$.But inverse of a closed set being closed set implies continuity. But we know our function is not continuous here. So what happened?
In fact , Continuity $\iff$ inverse image of every open set is open $\iff$ inverse of every closed set is closed.
So I'm misunderstanding some point here. Please clarify.
Thank you.
There are a lot more closed sets in $\prod_\alpha X_\alpha$ than those that can be written as $\prod_\alpha C_\alpha$ with $C_\alpha \subset X_\alpha$ closed. This even holds for finitely many spaces: The diagonal is closed in $X \times X$ (when $X$ is Hausdorff) but not of this form.
You've only shown $f^{-1}[C]$ closed for a (too limited) subfamily of closed subsets $C$ of the product.