3
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I haven't seen it stated as such, which is why I'm raising it here for confirmation.

In most references I'm seeing a homeomorphism $\phi$ between topological spaces $(W, \mathscr S)$ and $(X, \mathscr T)$ defined as:

  1. a bijection, $\phi: W \to T$
  2. which is continuous, i.e. $\phi^{-1} (T \in \mathscr T) \in \mathscr S$, where $\phi^{-1} (T \in \mathscr T)$ is the pre-image of $T$
  3. whose inverse $\phi^{-1}: T \to W$ is continuous.

It's fairly easy to show that a bijection $\phi: W \to T$ gives a bijection $\phi: \mathscr P (W) \to \mathscr P (T)$ between their powersets (with the inverse defined by the pre-image), so condition 3 becomes $\phi (S \in \mathscr S) \in \mathscr T$, i.e. $\phi $ is an "open map"

Recognizing that a topological space consists of a set and its collection of (open) subsets can't we then just say that a homeomorphism is mapping $\phi: W \to X, \phi: \mathscr S \to \mathscr T$ which is invertible in both cases.

Equivalently, since $\mathscr S , \mathscr T$ are also sets, a homeomorphism is a bijection between the underlying sets and between the corresponding topologies ?.

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  • $\begingroup$ Yes, if I haven't missed something, you statement is correct... $\endgroup$ – user302982 Feb 14 '16 at 13:31
  • $\begingroup$ Then you are talking about one map having two domains (codomains). I wouldn't plead for that. $\endgroup$ – drhab Feb 14 '16 at 14:06
  • 2
    $\begingroup$ Note that a continuous map is a map $\phi \colon W \to X$, wich also acts as a map $\phi^* \colon \mathscr T \to \mathscr S$ (note the direction). Indeed a homeomorphism is a map for which both $\phi$ and $\phi^*$ are bijections. $\endgroup$ – martini Feb 14 '16 at 14:10

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