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I have a question about covering maps. If $\phi_1: X_1 \rightarrow Y_1$ is a covering map, and $\phi_2: X_2 \rightarrow Y_2$ is a covering map, then is it true that $\phi_1 \times \phi_2: X_1 \times X_2 \rightarrow Y_1 \times Y_2$ a covering map?

Stefan Hamcke
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  • Yes. I believe the evenly covered neighborhood in the product is the product of the evenly covered neighborhoods. – Zach L. Jun 11 '13 at 20:51
  • See here: http://math.stackexchange.com/questions/1405942/is-my-proof-that-the-product-of-covering-spaces-is-a-covering-space-correct – HUO May 03 '16 at 13:56

1 Answers1

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Yes. This is a case where you should just follow the definition. The map $\phi_1\times\phi_2$ is continuous and surjective, and every point of $Y_1\times Y_2$ has a neighborhood that is evenly covered (remember the definition of the product topology).

Zev Chonoles
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