Prove there exists a diffeomorphism map $f:M\rightarrow M$ such that $f(x)=y$

Question

Let M be a connected smooth manifold, $x,y\in M$,then there exists a diffeomorphism map $f:M\rightarrow M$ such that $f(x)=y$.

Answer 1

Let $x$ be an element of $M$, we may find a chart $x\in U$. Let $f:U\rightarrow f(U)\subset \mathbb{R}^n$ be diffeomorphism. There exists $r>0$, such that $B(f(x),r)\subset f(U)$. For every $u\in\mathbb{R}^n$ with $\|u\|<r/2$, consider the vector field $X$ defined on $f(U)$ such that the restriction of $X$ to $B(f(x),r/2)$ is $u$ and the restriction of $X$ to $f(U)-B(f(x),r)$ is zero. This can be done with bump functions. We can pullback $X$ on $U$ with $f$ and it defines a vector fields $Y'$ on $U$ such that the restriction of $Y'$ on $U-f^{-1}(B(f(x),r))$ is zero. $Y'$ can be extended to $M$ by zero in $M-f^{-1}(B(f(x),r))$. Let $Y$ be that extension. The support of $Y$ is compact, so we can define the flow $\phi_t^u$ of $Y$ remark that the family of flow $\phi_t^u$ act transitively in a neighborhood of $x$. Since $M$ is connected, we deduce that $Diff(M)$ acts transitively on $M$.