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Let $M$ be a smooth manifold, $x,y\in M$. Must there exist a diffeomorphism $f : M \rightarrow M$ with $f(x) = y$?

I tried proving this via vector fields, i.e. trying to find a vector field whose flow through $x$ passes through $y$, without much success. Besides, this only has a chance of working on complete manifolds. Anyone know the answer to this?

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Do you assume that $M$ is connected? – mixedmath Oct 2 '12 at 0:37
Yes, lets assume $M$ is connected. Otherwise this is rather trivially false. – user15464 Oct 2 '12 at 0:41
up vote 9 down vote accepted

No; take $M$ to be the disjoint union of two smooth manifolds which are not diffeomorphic.

However, the statement is true if $M$ is connected. You do not need completeness. It suffices to show that the set of all points that can be reached from $x$ via some diffeomorphism is both open and closed.

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Yeah, I was thinking an open disk disjoint union and an open interval. – ncmathsadist Oct 2 '12 at 0:51
Maybe a better way to phrase the hint is as follows: define $x \sim y$ if $x$ can be reached from $y$ via some diffeomorphism. Show that this is an equivalence relation and that the equivalence classes are open. – Qiaochu Yuan Oct 2 '12 at 1:47

You can find a demonstration of this fact (if M is connected) in the book of Milnor - Topology from the differentiable viewpoint. It is the lemma of homogeneity. In fact you have more :

Homogeneity Lemma: Let $y$ and $z$ be arbitray interior points of the smooth, connected manifold M. Then there exists a diffeomorphism $f:M\rightarrow M$ that is smoothly isotopic to the identity and carries $y$ into $z$.

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Way cool, Kaye. +1 for you! – ncmathsadist Oct 2 '12 at 0:52
@ncmathsadist or Tomas could you please make me understand the lemma? I am not able to grasp after some stage for example from the step where he introduced some differential equation. – Un Chien Andalou Mar 9 '13 at 9:47

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