Product (arbitrary) of open functions is open. Let $f_{\alpha}\colon X_{\alpha}\to Y_{\alpha}$ be open, for all $\alpha \in J$. Then $\prod_{\alpha} f_{\alpha}\colon \prod_{\alpha}X_{\alpha} \to \prod_{\alpha}Y_{\alpha}$ is open? Both $ \prod_{\alpha}X_{\alpha}$ and $ \prod_{\alpha}Y_{\alpha}$ have the product topology.
What if every $f_{\alpha}$ is also surjective?
 A: 

2.3.29. Proposition. The Cartesian product $f=\prod_{s\in S} f_s$, where $f_s \colon X_s \to Y_s$ and $X_s\ne\emptyset$ for $s\in S$, is open if and only if all mappings $f_s$  are open and there exists a finite set $S_0\subset S$ such that $f_s(X_S)=Y_S$ for $s\in S\setminus S_0$.
Proof. From 1.4.14 and the equality $f(\prod_{s\in S}W_s)=\prod_{s\in S}f_s(W_s)$ it follows that if the mappings $f_s$ satisfy the above conditions, then $f$ is open.
Conversely, suppose that $f$ is an open mapping . Take an $s_0\in S$ and a non-empty open set $U\subset X_{s_0}$. the set $U\times \prod_{s\in S\setminus\{x_0\}} X_s$  is non-empty and open in $\prod_{s\in S} X_s$, so that the set
  $$p_{s_0} f(U\times\prod_{s\in S\setminus\{s_0\}} X_s)=f_{s_0}(U)$$
  is open in $Y_{s_0}$, because the projection $p_{s_0}$ is an open mapping. This implies that $f_{s_0}$ is open. As $\prod_{s\in S} X_s\ne\emptyset$, the set $f(\prod_{s\in S} X_s) = \prod_{s\in S} f_s(X_s)$ is a non-empty open subset of $\prod_{s\in SY_s}$, so that it contains a set of the form $\prod_{s\in S} W_s$, where $W_s\ne Y_s$ only for $s$ in a finite set $S_0\subset S$; then for $s\in S\setminus S_0$ we have $f_s(X_s)=Y_s$. $\square$


(cited from “General Topology” by Ryszard Engelking. )  

1.4.14. Theorem. A continuous mapping $f\colon X\to Y$ is open if and only if there exists a base $\mathcal B$ for $X$ such that $f(U)$ is open in $Y$ for every $U\in\mathcal B$. $\square$ 

