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Let $M$ be a connected topological manifold of dimension $n \geq 1$. It is well known that if $p,q \in M$ then there exists a homeomorphism $\phi : M \to M$ such that $\phi(p) = q$. This can be summed up by saying that every connected topological manifold is topologically homogeneous.

I have wondered a few times whether:

given points $m \in M$ and $p,q \in M \backslash \{m\}$ there exists a homeomorphism $f : M \to M$ which fixes $m$ and sends $p$ to $q$.

Now this is true when $M$ is a compact connected topological manifold of dimension $n \geq 2$. Here is why. Since $n \geq 2$, $M \backslash \{m\}$ is a connected manifold whose one point compactification is $M$. By topological homogeneity, there exists a homeomorphism $f : M \backslash \{m\} \to M \backslash \{m\}$ which maps $p$ to $q$. $f$ is a proper map, therefore it lifts to an automorphism $M \to M$ of the 1 point compactification which maps infinity to infinity, i.e it maps $m$ to $m$.

It is also true for $\mathbb{R}$. A dilation centered at $m$ does the trick.

is the above boxed statement true for other non compact manifolds?

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2 Answers 2

up vote 3 down vote accepted

Define an equivalence relation on $N = M \smallsetminus \{m\}$ by declaring $p \sim q$ if there exists a homeomorphism of $M$ fixing a neighborhood of $m$ and carrying $p$ into $q$. This equivalence relation has open equivalence classes because every point is equivalent to all points in a coordinate patch around it(1). Since $N$ is connected (here I'm using that the dimension is $\geq 2$), there's only one equivalence class.

Added:

(1) To see this, it suffices to prove that a Euclidean ball is homogeneous via a diffeomorphism fixing a neighborhood of its boundary. To this end, let $a,b \in B_{r}(0)$ and choose a smooth bump function $f: B_{r}(0) \to [0,1]$ supported inside $B_{r}(0)$ and constant equal to one in a ball containing both $a$ and $b$. The vector field $X(p) = f(p)(b-a)$ generates a flow $\varphi^t$ whose time-one map $\varphi^1$ is a diffeomorphism satisfying $\varphi^1(a) = b$ and fixing everything outside the support of $f$. See also Georges Elencwajg's “fishy answer” here and the discussion in the comments.

Incidentally, a similar argument also proves the statement in Mariano's answer on the $k$-transitivity of the homeomorphism group of a connected $n$-manifold of dimension at least two: $N = M^{k} \smallsetminus \{\text{large diagonal}\}$ is connected and the equivalence relation on $N$ given by $(p_1,\ldots,p_k) \sim (q_1,\ldots,q_k)$ if and only if there is a homeomorphism $\varphi$ of $M$ such that $(q_1,\ldots,q_k) = (\varphi(q_1),\ldots,\varphi(q_k))$ has open equivalence classes, hence it must be all of $N$.

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In fact, the group of homeomorphisms (when the dimension is at least 2) acts $k$-transitively for all $k\geq0$, so the answer to your question is: it is true for all manifolds.

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Since you know how to prove that the action is transitive, try to extend that proof to one of $k$-transitivity. If that does not work, ask and I or someone else can provide details. –  Mariano Suárez-Alvarez Apr 23 '12 at 8:14
    
ok i will give it a go! thanks Mariano –  DBr Apr 23 '12 at 8:17

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