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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.

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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 5 '11 at 0:47
Circle has 1 cut point. The line segment has 2 cut points. That's not the question. – simplicity Nov 5 '11 at 0:50
The important part was "but are preserved under homeomorphisms" – Carl Nov 5 '11 at 1:10
@carl I get it now. Thought you had to prove more, but yeah that is sufficient. – simplicity Nov 5 '11 at 1:15
up vote 1 down vote accepted

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.

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