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I am self-studying topology from Munkres. One exercise asks, in part, to show that the spaces $(0,1)$ and $(0,1]$ are not homeomorphic. An apparent solution is as follows: If you remove a point, $x$, from $(0,1)$, you get a disconnected space; however, you can remove the point $\{1\}$ from the space $(0,1]$ and the space will still be connected. This apparently means the spaces aren't homeomorphic. I don't quite see the connection to the definition of homeomorphic spaces (that two spaces are homeomorphic if there is a bicontinuous function between them). So, what is the connection between the "bicontinuous" definition of homeomorphisms and the "removing a point" procedure? Sorry for missing something obvious.

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  • $\begingroup$ It's the matter of connected-ness of homeomorphic spaces. $\endgroup$ Dec 17, 2014 at 17:25

6 Answers 6

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What you're missing is the following:

If $f : X \to Y$ is a homeomorphism, then $f|_U : U \to f(U)$ is a homeomorphism for any $U \subseteq X$.

In particular, for any $x \in X$, $f|_{X\setminus\{x\}} : X\setminus\{x\} \to Y\setminus\{f(x)\}$ is a homeomorphism. Also note that the argument you mentioned uses the fact that if two spaces are homeomorphic, then they are either both connected or both disconnected.

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    $\begingroup$ I see. The restricted function is obviously still a bijection, and then continuous maps on topological spaces are still continuous after domain restrictions (Munkres Thm 18.2d), which we apply to both $f$ and $f^{-1}$. $\endgroup$
    – ashman
    Dec 17, 2014 at 21:06
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You should fill out the details of the proof, which will be instructive. (It may also have been discussed in the book, and you missed it. You should check carefully.) There are two steps.

First, observe that if $X$ and $Y$ are homeomorphic, the $X$ is connected if and only if $Y$ is connected: a connected set cannot be homeomorphic to a disconnected set. The reason is essentially this: connectedness is a property phrased completely in terms of open sets, so any argument that $X$ is, or is not, connected, must reduce to an argument about the open sets of $X$, and then the homeomorphism allows one to translate this argument about open sets of $X$ directly to the open sets of $Y$, where it goes through in the same way.

Now similarly, suppose $X$ and $Y$ are homeomorphic; say the homeomorphism is $h$. And suppose there is a point $x\in X$ which can be deleted from $X$, so that $X\setminus\{x\}$ is connected. Then $Y\setminus\{h(x)\}$ must also be connected, by the argument of the previous paragraph.

But in this case $X=(0,1]$ and $Y=(0,1)$, and you claim we have a homeomorphism $h$. We can take $x=1$; then $(0,1]\setminus\{1\}$ is connected, and $(0,1)\setminus\{h(1)\}$ must also be connected, because the two sets are homeomorphic. But $(0,1)\setminus\{h(1)\}$ cannot be connected, so something is wrong; one of our assumptions must be discarded. And the only thing we can discard is that $h$ was a homeomorphism.

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If $f: X \to Y$ is a homeomorphism, then for any $x \in X$, the restricted map $$f\vert_{X - \{ x \}}: X - \{ x \} \to Y - \{ f(x) \}$$ is a homeomorphism, where $X - \{ x \}$ and $Y - \{ f(x) \}$ are endowed with the subspace topologies induced respectively by $X$ and $Y$.

In particular, if $X$ and $Y$ are topological spaces and there are no points $x \in X, y \in Y$ such that $X - \{ x \}$ and $Y - \{ y \}$ are homeomorphic, $X$ and $Y$ are themselves not homeomorphic.

Remark This line of ideas can be used to show that the union of two distinct, intersecting lines in, e.g., $\mathbb{R}^2$ is not a manifold, which is a standard early counterexample in the theory of topological manifolds.

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If two topological spaces, say $X$ and $Y$ are homeomorphic, by definition you have a bicontinous bijection $f:X\rightarrow Y$, so you can consider a restriction of $f$ to the topological subspace $X-\{p\}$, where $p\in X$. This restriction is a bijection between $X-\{p\}$ and $Y-{f(p)}$ and it is trivially bicontinous, so it is still an homeomorphism. This can be done for all $p\in X$, so if you can find a point where this doesn't hold, it implies that $f$ isn't an homeomorphism.

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Let $X$ be a path-connected topological space. We say $x\in X$ is a cut-point of $X$ if the space $X\setminus\{x\}$ (with the subspace topology) is not path connected.

The most basic property of cut-points is the following

Theorem. If $X$ and $Y$ are homeomorphic with a homeomorphism $f\colon X\to Y$, then if $x\in X$ is a cut-point, we must also have $f(x)\in Y$ is a cut point.

There is a surprising amount you can do with cut-points and similar 'removable points which change some property of the space'. For instance, if $X$ and $Y$ are homeomorphic, then if every point of $X$ is a cut-point, so is every point of $Y$ (prove this using the above theorem). The contrapositive of this statement is the one used by Munkres.

The cardinality of the cut-point set of a space is also a homeomorphism invariant. Also the number of components in $X\setminus\{x\}$ is a homeomorphism invariant. So if $X$ has a single cut point which cuts the space into $n$ components, and $Y$ has a single cut-point which cuts the space into $m\neq n$ components, then we know that $X$ and $Y$ are not homeomorphic. You should try to prove these statements.

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Homeomorphisms preserve topological properties. If one space has a certain toplogical property, and another space does not, then the spaces cannot be homeomorphic.

Note this technique is most useful in showing spaces are NOT homeomorphic. It not as useful if you are trying to show they ARE homeomorphic (unless the property actually characterizes a space -- having a certain property may be necessary but not sufficient to ensure two spaces are homeomorphic).

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