1

In one of the book of Munkres, he claims that a map $p$ is open because:

$p^{-1}(p(U))$ with U open is open, and so $p$ is open.

I don't understand why this is true, can you help me? More precisely, I don't see why that fact show that $p$ maps open sets to open sets. Thanks !

EDIT: In this context, $p$ is indeed a quotient map, sorry for the omission, it was not clear in the proof that this property was useful.

matyce
  • 67
  • Please post the quote in full context – D_S May 12 '22 at 13:58
  • Presumably one has extra information regarding the map $p$. Not all maps are open, after all. Please edit your post to include the relevant context. – lulu May 12 '22 at 14:00
  • Notice that if $p$ is continuous and $p(U)$ is open, then $p^{-1}(p(U))$ is open. However, withouth further context the question has no sense, which map is $p$? – Marcos May 12 '22 at 14:33
  • It is likely (but just a guess on my part) that $p$ is a quotient map. In this case the definition of $A$ being open is that $p^{-1}(A)$ is open. Munkres (if I am right) is applying this definition to $A=p(U)$. Not all quotient maps are open, so he must have more conditions. – Jamie Radcliffe May 12 '22 at 14:54
  • $p$ must at least be assumed to be a quotient map. Otherwise this isn't true. – freakish May 12 '22 at 15:47

0 Answers0