Open maps in the product and box topology

Question

I am new to topology and I have some trouble with the following exercise in my topology class:
Let I be an infinite index set and for each $i \in I$ we are given a non-empty topological space $X_i$.
If X is another topological space and $f_i:X \rightarrow X_i$ are open maps for each $i \in I$, then $f:X \rightarrow \prod_{i\in I}X_i$, $f(x)=(f_i(x))_{i \in I}$, is an open map aswell.
I have to decide and prove/give a counterexample if the statement is true when $\prod_{i\in I}X_i$ is equipped with (a) the product topology and (b) the box topology.

I know a function $f:X \rightarrow Y$ is called open, if for all open $U \subseteq X$ the image $f(U) \subseteq Y$ is open.
I also know that if I is finite then box top. = prod. top. (but in this problem I is infinite).

Answer 1

$\newcommand{\box}{\prod_i U_i \ \text{for $U_i$ open in $X_i$}}$ $\newcommand{\pro}{ \{\ p^{-1}(U_i), U_i \text{ open in } X_i \}\ }$ For box topology remember the definition of the box topology the basis consists of $\box$

By definition of the box topology the basis for box topology is $\box$ and since each $f_i$ is open and the fact that the Cartesian product of open sets is open we reach our conclusion

For the product topology recall the definition of the product topology, here $p$ represents the projection map Definition of product topology: $\pro$

Use the definition $\pro$ in a similar manner to arrive at the conclusion. Note how $p^{-1} \circ f$ is $ X \rightarrow X_i \rightarrow \prod X_i$ Since both $p^{-1}$ and $f$ are open (and continuous for continuity see what is famously known as "pasting Lemma") we have the desired result