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Show that if $X$ is a connected manifold then for any $x,y\in X$, there exists a diffeomorphism $h$ sending $x\xrightarrow{h}y$. This is what I have so far:

Given a connected manifold $X$, for any $x,y\in\mathbb{R}$ there is an embedding $\gamma_x:[0,1]=I\rightarrow X$ with $\gamma_x(0) = x, \gamma_x(1) = y$. Let $K = \gamma(I)$, and note that this is compact. Then we can define a vector field $v$ on $X$ to be $v(r) = \gamma'_x(t_r)$ if $\gamma_x(t_r) = r$ i.e. $r\in K$ and $0$ otherwise. Then the results follows from the following theorem:

If $v$ is a vector field on $X$ such that $v$ is zero outside of a compact $K\subset int X$. Then $\exists !$ flow $H$ on $X$ whose velocity field is $v$.

From this we get a diffeomorphism $h_t$ such that $h_t(x) = y$ so $X$ is homogeneous.

A couple issues I have with this so far: 1) is $v$ a smooth vector field as defined or do I need to use partitions of unity to smooth it out? 2) Since $K$ is required to be in the interior of $X$, does this not work if either $x$ or $y$ are in $\partial X$?

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  • $\begingroup$ In response to question (1): Unfortunately, your vector field typically isn't even continuous. Partitions of unity are not going to help. $\endgroup$ – Jack Lee Apr 20 '17 at 0:03
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    $\begingroup$ You're right. Though, I think there should be a way to salvage this. What if we replace K as follows: Assuming $dimX = k$ and that $X$ is embedded in euclidean space, let $\gamma_{i,t}$ be a homotopy from our original $\gamma$ to some constant map, such that the derivative with respect to t at any point $x$ points in the direction of the i'th basis vector of $T_x(X)$. Then taking $K$ to be the image of these homotopies $(1\leq i \leq k)$, this is compact and the velocity fields would continuously go to zero outside $K$. @JackLee $\endgroup$ – Juan Chi Apr 20 '17 at 2:04
  • $\begingroup$ You can use a bump function to extend your vector field defined along the curve to the whole manifold. $\endgroup$ – Thomas Rot Apr 20 '17 at 13:05

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