I want to prove this statement:
Let $p_i: Y_i\to X_i$ with $i=1,2$ be covering spaces.
Show, that $p_1\times p_2: Y_1\times Y_2\to X_1\times X_2$ is a covering space.
Therefore I want to show, that for every $(x_1, x_2)\in X_1\times X_2$ exists a neighborhood $U$, such that $(p_1\times p_2)^{-1}(U)$ is a disjoint union of open sets, which are mapped homeomorphically onto $U$ by $p_1\times p_2$.
First of all, I have to find a convenient neighborhood $U$ for every $(x_1,x_2)$. Since $p_i$ is a covering space, there exists neighborhoods $U_i$ for every $x_i$. Therefore I would try $U:=U_1\times U_2$.
Now I have to show, that $(p_1\times p_2)^{-1}(U)$ is the disjoint union of open sets, which are mapped homeomorphically onto $U$ by $p_1\times p_2$.
$(p_1\times p_2)^{-1}(U)=\{(y_1,y_2)\in Y_1\times Y_2|(x_1, x_2)\in U\}$
$U_1$ and $U_2$ both hold these properties, because $p_i$ is a covering space.
Hence $U_1=\bigcup_{l\in L}$ and $U_2=\bigcup_{j\in J}$ are both disjoint unions of open sets.
Then $U=\bigcup_{(l,j)\in L\times J} U_l\times U_j$
And now I have to show, that $(p_1\times p_2)(U_l\times U_j)\sim U$ for every $(l,j)\in L\times J$
Is this correct? Or am I mistaken?
Thanks in advance for your comments.
Best Answer
Hint: Prove (or recall) the following statements to arrive at a complete proof of your desired result:
Now given a point $(x_1,x_2)\in X_1\times X_2$ and "evenly covered" neighborhoods $U_1$ and $U_2$ of $x_1$ and $x_2$ respectively, using the statements above and your rewrite of $(p_1\times p_2)^{-1}(U_1\times U_2)$ in the question (by the way, you might consider a different notation for neighborhoods in $Y_1$ and $Y_2$) show that $U_1\times U_2$ is an evenly covered open neighborhood for $(x_1,x_2)$.