1

Let $p,q$ be a covering maps, $p:\tilde X \rightarrow X$ and $q:\tilde Y \rightarrow X$ and let $Z=\lbrace(\tilde x, \tilde y)| p(\tilde x)=q(\tilde y) \rbrace$, I want to prove that $f:\tilde Z\rightarrow X$ definided by $f(\tilde x, \tilde y)=p(\tilde x)$ is a covering map too, I try putting for any $x$ in $X$ the same open set $U_x$ given by the covering map $p$ and applying the properties of this open set $U_x$ in $f$, by inverse image of $f$, but at the and this don´t work, some help to prove this is well. Thanks.

InsideOut
  • 6,883
sti9111
  • 1,559
  • Is $X=Y$? Otherwise the expression $p(\tilde{x})=q(\tilde{y})$ does not make sense. – archipelago May 20 '15 at 07:31
  • If $X=Y$ it is even enough to assume that only one of the two maps is a covering. $Z$ is the so-called pullback of those maps and pullbacks of covering spaces are covering spaces as asked in the following question: http://math.stackexchange.com/questions/494024/pullback-of-a-covering-map – archipelago May 20 '15 at 07:32
  • In that case, your question is answered by my second comment. – archipelago May 21 '15 at 14:01

0 Answers0