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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.

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Mr.Topology
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  • Yes, here's a hint: What does $U$ look like in terms of the topologies for $X_1$ and $X_2$ (remember, $X_1\times X_2$ is given the product topology from the topologies on $X_1$ and on $X_2$). – Justin Benfield Jun 12 '16 at 19:00
  • Yes, I am mistaken, or it is correct until now? I try to follow your hint. – Mr.Topology Jun 12 '16 at 19:04
  • Hint: If you use the letter $U$ for everything, you'll just get messed up. You're working at a point $(x_1, x_2)$. Call the fundamental nhd for $p_1$ around $x_1$ by the name $V$, covered by sets $V_i$ in the $Y_1$. Call the similar nhd for $p_2$ by the name $W$, covered by sets $W_j$. Then can you think of an open set $U$ containing $(x_1, x_2)$ that might be a good candidate for a fundamental neighborhood? – John Hughes Jun 12 '16 at 19:04

1 Answers1

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Hint: Prove (or recall) the following statements to arrive at a complete proof of your desired result:

A product of homeomorphisms is a homeomorphism of the product spaces.

For all continuous $f:A\to C$ and $g:B\to D$ and all $S\subseteq A$ and $T\subseteq B$ we have the equality $$(f\times g)|_{S\times T}=f|_S\times g|_T$$

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 $(x_1,x_2)$.

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User12345
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  • The seconds statement is known. I want to prove, that the product of homeomorphisms is a homeomorphism. So, let $f:X_1\to Y_1$ and $g:X_2\to Y_2$ be homeomorphisms. I want to show, that $(f\times g): X_1\times X_2\to Y_1\times Y_2$ is a homeomorphism. Therefore I need to show, that $(f\times g)$ is a bijection. This is trivial. Now I need to show, that $(f\times g)$ is continuous. I show first, that $(f\times g)^{-1}(Y_1\times Y_2)=f^{-1}(Y_1)\times g^{-1}(Y_2)$. – Mr.Topology Jun 13 '16 at 17:25
  • $(f\times g)^{-1}(Y_1\times Y_2)={(x_1, x_2)\in X_1\times X_2| (f\times g)(x_1, x_2)\in Y_1\times Y_2}$. Let $(x_1,x_2)\in (f\times g)^{-1}(Y_1\times Y_2)$ be random, then $(f\times g)(x_1,x_2)\in Y_1\times Y_2$. Hence $f(x_1)\in Y_1$ and $g(x_2)\in Y_2$, therefor $x_1\in f^{-1}(Y_1)$ and $x_2\in g^{-1}(Y_2)$. Hence $(x_1,x_2)\in f^{-1}(Y_1)\times g^{-1}(Y_2)$. You can show "$\supseteq$" in the same way. – Mr.Topology Jun 13 '16 at 17:29
  • To show, that $(f\times g)$ is continuous. I show, that the preimage of an open set, is open. Let $U\times V\subseteq Y_1\times Y_2$ be open. By definition of the product topology $U\subseteq Y_1$ and $V\subseteq Y_2$ are open sets. Now $(f\times g)^{-1}(U\times V)=\underbrace{f^{-1}(U)}{open}\times \underbrace{g^{-1}(V)}{open}$. Hence open, by definition of the product topology. Therefor the preimage of an open set, is open, and $(f\times g)$ is continuous. – Mr.Topology Jun 13 '16 at 17:32
  • To show, that the inverse function is continuous, should be similar. – Mr.Topology Jun 13 '16 at 17:33