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In page 104 of this book, there is the following exercise and solution:

S.123: Show that if $X$ is a compact space, and $f: X \rightarrow Y$ is a condensation, then $f$ is a homeomorphism.

Proof: We must only prove that f^{-1} is continuous. Given a closed $F \subset X$, the set $(f^{-1})^{-1}(F) = f(F)$ is closed in $Y$. (This is because $X$ is compact implies $F$ is compact implies $f(F)$ is compact so it is closed.) so $f$ is a closed map which implies $f^{-1}$ is continuous.

My question is: where is the fact that $f$ is "a condensation" used?

Thank you!

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  • $\begingroup$ What is a "condensation"? $\endgroup$ – Daniel Fischer Jun 1 '14 at 10:56
  • $\begingroup$ Also, does the author maybe use "compact" to denote quasicompact spaces? $\endgroup$ – Daniel Fischer Jun 1 '14 at 10:57
  • $\begingroup$ You didn't show that $f$ is bijective. $\endgroup$ – user99914 Jun 1 '14 at 10:58
  • $\begingroup$ Don't we need to assume $Y$ is Hausdorff? $\endgroup$ – David Mitra Jun 1 '14 at 11:06
  • $\begingroup$ We absolutely do, @David. $\endgroup$ – Daniel Fischer Jun 1 '14 at 11:13
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A condensation is a 1-1 onto continuous function. You use 1-1 and onto implicitly, because you assume that the inverse function exists. And of course a homeomorphism is continuous, so you use continuity of $f$ as well.

In other words: A homeomorphism is a bijection that is continuous and whose inverse is also continuous. The proof shows the latter, because the first 2 are the definition of condensation.

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  • $\begingroup$ I see. Somhow I thought that by a condensation, the author means an accumulation point.. Thank you! $\endgroup$ – topsi Jun 1 '14 at 11:04
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    $\begingroup$ @ShirSivroni There is the notion of a "condensation point". $x$ is a condensation point of $A$ if every neighbourhood of $A$ intersects $A$ in an uncountable set. But that's a totally different notion. But I can see it might be confusing. Especially Russian topologists like the term condensation for a special continuous function. So they can see that $X$ "condenses to" $Y$, when there is a condensation between $X$ and $Y$. $\endgroup$ – Henno Brandsma Jun 1 '14 at 11:07
  • $\begingroup$ I see. Thank you for clarifying this! $\endgroup$ – topsi Jun 1 '14 at 11:21

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