If $p_i:E_i\rightarrow B_i, \ i\in I$ are covering maps, then is it true that $$\prod_{i\in I} p_i:\prod_{i\in I}E_i \rightarrow \prod_{i\in I}B_i$$ is a covering map ?
It is true if $\mbox{card}\ I <\infty$, but I don't have any idea to prove it if $I$ is infinite ( or find counterexample ).
If $I$ is finite it's true thanks to this question
If $I$ is infinite it's not necessarily true. Let $p : \mathbb{R} \to S^1$ and consider $$\prod_{n \in \mathbb{N}} p : \prod_{n \in \mathbb{N}} \mathbb{R} \to \prod_{n \in \mathbb{N}} S^1.$$
If it were a covering map it would be a local homeomorphism, in particular some neighborhood $U$ of $(0,0\dots)$ would be mapped homeomorphically to some neighborhood $V$ of $(1,1\dots)$. By restricting we can assume that $U = (a_1, b_1) \times \dots \times (a_k, b_k) \times \mathbb{R} \times \mathbb{R} \times \dots$. In particular $U$ is contractible. However, by working with the definition of the product topology again, no neighborhood is simply connected in $\prod_{n \in \mathbb{N}} S^1$ (there will always be nontrivial $S^1$ factors). Contradiction.
