Given two distinct points $a,b\in U$ where $U$ is an open subset of $\mathbb R^n$, can one always find a homeomorphism $f:U\to U$ with $f(a)=b$ or does one need any further conditions on $ U $ or the points?
Thank you.
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What open set $U$ are you talking about? Is $U ⊂ ℝ^n$ where $ℝ^n$ is given euclidean topology? – k.stm Jan 17 '15 at 10:33
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Do you want to work with open subsets of $\mathbb R$ or can it be an arbitrary topological spaces? If it is the later, $[0,1]$ as an open subset of the space $[0,1]$ is an easy counterexample. – Martin Sleziak Jan 17 '15 at 10:34
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1Yes I am working with $\mathbb{R}^n$. With Euclidean topology. – R_D Jan 17 '15 at 10:39
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2Thanks for clarification @Raisa. It is useful to add such information also to post, where it is easier to spot than in comments. I'll add that I am not sure whether ([tag:fixed-point-theorems]) is really a good tag for this question. You might see some information about what this tag is for in the tag-info. (And the tag-excerpt was shown to you when adding the tag and you can view it simply by hovering above the tag with your mouse.) – Martin Sleziak Jan 17 '15 at 10:43
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1Do you assume that $U$ is connected? I am not an expert in this area at all, but if $a, b$ are in different connected components of $U$ (e.g. a disk and a punctured disk) then it might be impossible. – Martin R Jan 17 '15 at 10:48
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@Martin. No I don't assume connectedness of $U$. But I do want to know if I need to assume it or not. Hence I ask if one needs any further condition on $U$ or the points. – R_D Jan 17 '15 at 10:52
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2Related: Are all connected manifolds homogeneous – Martin Sleziak Jan 17 '15 at 10:55
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I think @MartinR's comment basically answers your question. You only need to find two disjoint connected subsets $A$, $B$ which are not homeomorphic. Then $U=A\cup B$ is not homogeneous. (Since any homeomorphism must map connected component to another connected component.) – Martin Sleziak Jan 17 '15 at 11:09
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@MartinSleziak: So if $a, b$ are in the same connected component the answer is "yes", and if $a, b$ lie in different components which are not homeomorphic the answer is "no". I am not sure about the case where $a, b$ lie in different components $A$, $B$ of $U$ which are homeomorphic. From the answer that you linked to I conclude that there is a homeomorphism $f: A \to B$ such that $f(a) = b$. But can this $f$ always be extended to a homeomorphism $f: U \to U$? (Topology was never my main area of expertise, so this may be nonsense.) – Martin R Jan 17 '15 at 12:23
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Since connected components are clopen subset which are disjoint and their union is the whole space, any topological space is [topological sum](http://en.wikipedia.org/wiki/Topological sum) of the connected components. A map which "exchanges" two homeomorphic summands/components (and leave all other points unchanges) is a homeomorphism. – Martin Sleziak Jan 17 '15 at 12:40
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@Martin: I see, so simple, thanks! Are you going to post an answer? – Martin R Jan 17 '15 at 13:03
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@MartinR If you wish to do so, feel free to summarize (some of the) comments in an answer. – Martin Sleziak Jan 17 '15 at 13:09
1 Answers
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I seems that all puzzle pieces to answer your question have been found in the above discussion. I'll try to summarize it:
Let $U$ be an open subset of $\mathbb R^n$ and $a, b \in U$ be distinct. A homeomorphism $f: U \to U$ with $f(a) = b$ exists if and only if the connected components $A, B$ of $U$ containing $a$ and $b$ respectively, are homeomorphic.
In particular, such a homeomorphism exists if $a, b$ lie in the same connected component of $U$.
The "only if" part is the easier one: If a homeomorphism $f: U \to U$ with $f(a) = b$ exists then $f(A) = B$ because homeomorphisms map connected components onto connected components, so $A$ and $B$ are homeomorphic.
For the "if" part assume that there is a homeomorphism $g: A \to B$. The crucial tool is to use the result of
which states that there is a homeomorphism $h: B \to B$ such that $h(g(a)) = b$. Then $F := h \circ g : A \to B$ is a homeomorphism from $A$ onto $B$ satisfying $F(a) = b$. $F$ can be extended to a homeomorphism $f:U \to U$: In the case $A \ne B$ define $$ \begin{align} f(x) &= F(x) &\text{ if } x \in A, \, \\ f(x) &= F^{-1}(x) &\text{ if } x \in B, \,\\ f(x) &= x \quad &\text{ if } x \in U \setminus (A \cup B) \, . \end{align} $$ $f$ "swaps" the components $A$ and $B$ and leaves all other components unchanged. And if $A = B$ then define $f$ via \begin{align} f(x) &= F(x) &\text{ if } x \in A, \, \\ f(x) &= x \quad &\text{ if } x \in U \setminus A \, . \end{align}
