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$f:A \to B$ a homeomorphic mapping from $A$ to $B$

$=> f:A \backslash U\to B \backslash f(U)$ is also homeomorphic, $U \subseteq A$.

Why does this implication hold?

sonicboom
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  • It seems to hold intuitively but how would you put it formally? – sonicboom Apr 01 '14 at 16:34
  • What is the topology on $A\setminus U$ and on $B \setminus f(U)$? Once you know that, you just have to prove that $f \colon A\setminus U \to B\setminus f(U)$ is open and continuous, which is going to be totally trivial. – Magdiragdag Apr 01 '14 at 16:52

2 Answers2

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Since $f$ is homeomorphic it isn't difficult to show that $\bar{f}$ (the quotient mapping) is bijective. To prove continuity we take an open set $\bar{O}$ in $B\setminus f(U)$, which is of the form $O\cap f(U)$ with $O$ open in $B$. Now $\bar{f}^{-1}(\bar{O}) = f^{-1}(O)\cap A\setminus U$. We know that $f$ is a homeomorphism, so $f^{-1}(O)$ is open in $A$. Then by definition of the quotient topology $\bar{f}^{-1}(\bar{O})$ is open in $A\setminus U$.

Marc
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  • Usually $\setminus$ denotes set difference, not quotient. – Asaf Karagila Apr 01 '14 at 16:43
  • Ah, you're right. The slashes are the wrong way. In that case I don't know how to answer this question, because now we don't know what topology we're dealing with? – Marc Apr 01 '14 at 16:45
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If you meant quotient, an answer was given. If you meant set difference, simply notice that restricting a homeomorphism to a subset U gives a homeomorphism from U to its image (because it's still bijective, restrictions of continuous functions are continuous, and the inverse is also a restriction so it is also continuous.)

This counts here, because we can consider your function as the restriction to the complement of U.

GraduateStudent
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