Let $p_1:\tilde X_1 \rightarrow X_1$ and $p_2:\tilde X_2 \rightarrow X_2$ be two covering spaces. Prove: $p = p_1 \times p_2:\tilde X_1 \times \tilde X_2 \rightarrow X_1 \times X_2$ is a covering space.
Clearly $p$ is surjective and continuous, and given $x \in X_1 \times X_2$ and an open set $U$ s.t. $x \in U$, $p^{-1}(U) = p_1^{-1}(U_{x_1}) \times p_2^{-1}(U_{x_2}) $, which is a product of a union of disjoint open sets and therefore is a union of disjoint and open sets.
I'm pretty sure my proof is wrong, I think that the actual proof has something to do with the fundamental groups of the spaces but I can't quite understand how the fundamental group relates to the covering space.
Is this proof wrong?
Best Answer
This proof is indeed wrong.
For a counterexample, consider the case where $\tilde X_1 = X_1 = \mathbb{R}$ and $\tilde X_2 = X_2 = \mathbb{R}$ and $p_1,p_2,p$ are all simply identity maps. In that case, you would be saying that for any open subset $U \subset \mathbb{R} \times \mathbb{R}$, letting $V = p^{-1}(U)=U$, the set $V$ can be written as a product $V = A \times B$ for some sets $A \subset \mathbb{R}$ and $B \subset \mathbb{R}$.
But this is not possible: not every open subset $V \subset \mathbb{R} \times \mathbb{R}$ can be written in the form $V = A \times B$. Think, for example, of the open ball in $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ of radius $1$ centered on the origin $O = (0,0)$: $$V = B(O,1) = \{(x,y) \in \mathbb{R}^2 \bigm| x^2 + y^2 = 1\} \subset \mathbb{R}^2 = \mathbb{R} \times \mathbb{R} $$ This open ball is not equal to any subset of $\mathbb{R}^2$ of the form $A \times B$, for any $A \subset \mathbb{R}$ and $B \subset \mathbb{R}$.
To summarize, your proof uses the statement that every open subset of the Cartesian $\tilde X_1 \times \tilde X_2$ is a Cartesian product of an open subset of $\tilde X_1$ times an open subset of $\tilde X_2$. This statement is false.