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Is the number of cut points the same if they are homeomorphic. What I mean, if you had a structure that had only one 4-cut point, and then another topology with five 4-cut points. Can they ever be homeomorphic?

I know they to topologies have to have the same path components and this implies they have say a topology with 10-cut point isn't homeomorphic to one without a 10-cut point. However, I have strong intuition that its even stronger than this and the number of cut points matter. However, I can't prove it.

simplicity
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  • From wikipedia: Cut-points are not necessarily preserved under continuous functions, (example: f: [0, 2π] → R2, given by f(x) = (cos x, sin x)), but are preserved under homeomorphisms. Therefore the circle is not homeomorphic to a line segment, as the circle has no cut-points, but every point of the interval (except the two endpoints) is a cut-point. – Carl Nov 05 '11 at 00:47
  • Circle has 1 cut point. The line segment has 2 cut points. That's not the question. – simplicity Nov 05 '11 at 00:50
  • The important part was "but are preserved under homeomorphisms" – Carl Nov 05 '11 at 01:10
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    @carl I get it now. Thought you had to prove more, but yeah that is sufficient. – simplicity Nov 05 '11 at 01:15

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Let $X$ and $Y$ be topological spaces and let $h$ be a homeomorphism between them. If $x$ is any point of $X$, then $h \restriction X\setminus \{x\}$ is a homeomorphism between $X\setminus\{x\}$ and $Y\setminus\{h(x)\}$: a restriction of a continuous function to a subspace is still continuous, and the same holds for the inverse, while $h$ is still a bijection between these 2 spaces.

This means that if $x$ is a cut point, which means that $X\setminus\{x\}$ is disconnected, then so is $Y\setminus\{h(x)\}$ and hence $h(x)$ is a cut point as well. On the other hand, if $x$ is a non-cutpoint so is $h(x)$, so homeomorphic spaces have the same number of them.

This is often applied to show that the circle (no cut points) and proper intervals (at least one) are not homeomorphic.

Henno Brandsma
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