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Note The following question was asked by @Tian Vlašić, who then closed it after several interesting comments had been made. I was interested enough that I wanted to revive it.

Let us say that a property $P$ of topological spaces is transferred by bijective continuous maps if for every pair $X,Y$ of topological spaces, the statements $P(X)$ and there exists a bijective continuous map $f: X \rightarrow Y$ implies the statement $P(Y)$.

It is clear that every property of topological spaces that is transferred by bijective continuous maps is a topological property.

I have noticed that the following well-known topological properties are in fact transferred by bijective continuous maps:

  • the cardinality of the underlaying set of the topological space

  • the cardinality of the topology of the topological space

  • Hausdorffness of the topological space

I was quite surprised that Hausdorffness of the topological space is a property of topological spaces transferred by bijective continuous maps (note that this is a stronger statement than the statement that Hausdorffness of the topological space is a topological property).

My question is the following. What are some other well-known topological properties that are transferred by bijective continuous maps?

Note Some comments (from F. Shrike) on this question already noted the following:

  • Cardinality is pretty trivially transferred
  • Compactness and (path-)connectivity are transported (even if you remove injectivity as an assumption). A curious question is whether or not arc-connectivity is transported
  • Compactness and connectivity are easy from surjectivity, they have extremely well known proofs (and duplicates on this site). As for arc-connectivity, I have no idea if this is true. It's true if the domain is Hausdorff by a difficult theorem I don't know the proof of + your claim that Hausdorffness is transported.
FShrike
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John Hughes
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    The Hausdorff property isn't transferred, is it? Let $X$ be any Hausdorff space with at least two points, and let $Y$ be the same set with the trivial (indiscrete) topology. The identity map $f : X \to Y$ is a continuous bijection. Or have I misunderstood a definition? – Nate Eldredge Aug 22 '23 at 14:05
  • Separability is transferred, first countability is not. – Nate Eldredge Aug 22 '23 at 14:11
  • So if P is to be a transferred property, then for any set $X$ of a given cardinality, one of the following must be true: (1) every topology on X has property P; (2) no topology on X has property P; (3) the trivial topology on X has property P and the discrete topology does not. – Nate Eldredge Aug 22 '23 at 14:13
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    Being "transferred by bijective continuous maps" is equivalent to being "transferred by coarser topologies" in the sense that if property $P$ holds for a topological space $(X,\tau)$, it also holds for $(X,\tau')$ for any coarser topology $\tau' \subseteq \tau$. – Geoffrey Trang Aug 22 '23 at 14:30
  • It was good of you to revive this. I'm not sure why Tian deleted it – FShrike Aug 22 '23 at 16:32
  • On reflection, we should note that Tian was incorrect in claiming that Hausdorffness is transferred (and therefore my comment about arc-connectivity might be false); take any non trivial Hausdorff space $(X;\mathcal{T})$ and $\mathrm{Id}:(X;\mathcal{T})\to(X;\text{indiscrete topology})$ which is a continuous bijection – FShrike Aug 22 '23 at 16:35
  • Thanks, folks, for the revived interest. I'm pleased that others found this dark corner of the subject a worthwhile place to shed a little light. – John Hughes Aug 22 '23 at 17:14
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    Arcwise connectedness (defined as "any two points are connected by an embedded copy of $[0,1]$") does not transfer, since the trivial topology is not arcwise connected (it doesn't contain any embedded copy of $[0,1]$ at all). See https://mathworld.wolfram.com/Arcwise-Connected.html – Nate Eldredge Aug 22 '23 at 17:42

1 Answers1

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This is merely a partial answer:

While path-connectedness is preserved by such maps, dis-connectedness and path-disconnectedness are not. If $X$ is the topologist's sine curve $$ \{(0,0)\} \cup \{(x, \sin(1/x)) \mid 0 < x \le 1\} $$ then it is path-disconnected. But $f: X \to Y : (x, y) \mapsto x$ is a bijective continuous map, and its image is the unit interval, which is both connected and path connected. Of course, the inverse map is not continuous.

FShrike
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John Hughes
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    The topologist's sine curve is connected. But a simpler counterexample for disconnectedness would be a two-point set with the discrete and the trivial topology. – Nate Eldredge Aug 22 '23 at 14:33
  • D'oh! Of course it is. It's the classic example of "connected but not path-connected." I guess this shows what being a non-mathematician for 25 years will do to your brain. Thanks for correcting this, and for the better example. – John Hughes Aug 22 '23 at 17:12
  • Thanks, FShrike, for editing to remove my error. – John Hughes Aug 22 '23 at 21:19