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Suppose we have a map $ f: X \rightarrow Y $ where $ X $ and $ Y $ are topological spaces. Also suppose that $ U $ and $ V $ are open subsets of $ X. $

Suppose that $ V \supset U $ if and only if $ f(V) \supset f(U). $ Is this another way of saying that $ f $ is a homeomorphism? If so, why?

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    $\begingroup$ It’s not. Counterexamples aren’t hard to come by. $\endgroup$ – Randall Feb 24 at 16:29
  • $\begingroup$ Does this property imply anything about the map? $\endgroup$ – Math N00b Feb 24 at 16:30
  • $\begingroup$ I don't think this implies anything particularly nice without further assumptions on your spaces (i.e. separation axioms) $\endgroup$ – Aweygan Feb 24 at 16:36
  • $\begingroup$ It may imply that the function is an open map. $\endgroup$ – Randall Feb 24 at 16:55
  • $\begingroup$ This might be it. I now see that the map in question is bijective, continuous and open. The author is attempting to show that it is also a homeomorphism, and he bases that claim on this property. $\endgroup$ – Math N00b Feb 24 at 17:08
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Define $f$ to be the identity map from $\mathbb{R}$ with the standard topology to $\mathbb{R}$ with the discrete topology. Then for any two sets (even if they are not open) $U,V$ we have $U\subset V$ if and only if $f(U)\subset f(V)$. But obviously $f$ is not a homeomorphism. It is not even continuous.

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