Let $M$ be a smooth n-manifold($n\geq 1$).Prove that $M$ admits a diffeomorphism $f : M \to M$ which is not the identity I don't know how to come up with a proof for this. I think I have to come up with a bump function and do it locally on each chart and then extend it to the whole manifold??
May be I am wrong..
Thanks for any help
 A: I assume $n = \dim M \ge 1$.
The easiest way I know to see this is to pick any smooth, not-identically-zero section $X$ of $TM$, and consider its flow $\phi_X(x, t)$:
$\dot \phi_X(x, t) = X(\phi_X(x, t)); \tag 1$
we have 
$\phi_X(x, 0) = x, \; \forall x \in M, \tag 2$
but it is easy to see from (1) (by if necessary using local coordinates near $x$) that if $X(x) \ne 0$ for sufficiently small $\epsilon$
$\phi_X(x, \epsilon) \ne x. \tag 3$
Thus $\phi_X(x, \epsilon)$ is a non-identity diffeomorphism of $M$.
It appears that it is also possible, as the comment stream suggests, to pick a coordinate ball about some point and directly construct an appropriate diffeomorphism in that ball, a process which I think is also relatively straightforward.
A: You can do this without vector fields.
Let us see first how the case $n=2$ is dealt. For any $0\leq \theta\leq 2\pi$, denote by $R_\theta$ the rotation matrix of $\mathbb R^2$ of angle $\theta$. Let $D^2$ be the unit closed disk of $\mathbb R^2$. 
Consider the function $D^2\xrightarrow \psi D^2$ given by
$$x\mapsto R_{2\pi||x||^2}(x).$$
As the function $x\mapsto ||x||^2$ is $C^\infty$ on $D^2,$ so is $\psi$. 
The function $x\mapsto R_{-2\pi||x||^2}(x)$ is $C^\infty$ and it is the inverse function of $\psi$, hence $\psi$ is a diffeomorphism.
Notice that $\psi\upharpoonright\partial D^2$ is the identity. Thus if $M$ is a surface, we can get a subset $D$ of $M$ diffeomorphic to $D^2$, and using $\psi$ we can get a diffeomorphism $D\rightarrow D$ which is not the identity, but it is the identity on the boundary of $D$. This function can be extended to a diffeomorphism of $M$ which is not the identity.
For the case $n\geq 3$  consider the function $D^n\rightarrow D^n$ given by 
$$x=(x_1,x_2,x_3,\ldots,x_n)\mapsto (R_{2\pi||x||^2}(x_1,x_2),x_3,\ldots,x_n),$$
and argue as before.
