Let $J$ be a given (countably or uncountably infinite) index set. Let $\{\ X_\alpha \ \colon \ \alpha \in J \ \}$ and $\{\ Y_\alpha \ \colon \ \alpha \in J \ \}$ be collections of topological spaces.
Let the sets $X$ and $Y$ be given by $$X \colon= \Pi_{\alpha \in J} X_\alpha, $$ and $$Y \colon= \Pi_{\alpha \in J} Y_\alpha, $$ respectively.
For each $\alpha \in J$, let $f_\alpha \colon X_\alpha \to Y_\alpha$ be a continuous map.
Let $f \colon X \to Y$ be defined by $$f\left( (x_\alpha)_{\alpha \in J} \right) \colon= \left( f_\alpha(x_\alpha) \right)_{\alpha \in J} \ \ \ \mbox{ for } \ (x_\alpha)_{\alpha \in J} \in X.$$
Then is $f$ continuous if
(i) both $X$ and $Y$ are given the product topologies?
(ii) $X$ is given the product topology and $Y$ is given the box topology?
(iii) $X$ is given the box topology and $Y$ is given the product topology?
(iv) both $X$ and $Y$ are given the box topologyies?
If $J$ is finite, then (i), (ii), (iii), and (iv) reduce to just one case and then $f$ is continuous. Am I right?
(i) Yes. This is well-known fact, which holds because a map $f:Z\to\prod Y_\alpha$ is continuous iff each composition map $\pi_\beta\circ f$ is continuous, where $\pi_\beta: \prod Y_\alpha\to Y_\beta$ is the projection onto $\beta$-th factor.
(ii) No. For a counterexample it suffices to put as $J$ an arbirary infinite set, $X_\alpha=Y_\alpha$ for each $\alpha\in J$ and $f:X_\alpha\to Y_\alpha$ be the identity map. If the box topology on $\prod Y_\alpha$ does not coincide with the product topology, we shall have a counterexmple.
(iii) Yes. This follows from (i), because the box topology on $X$ is stronger than the product topology.
(iv) Yes. If follows since the inverse image of an element of a canonical base of the space $Y$ is open in the space $X$.