The diffeomorphism of $\mathbb R^n$ If $f$ is a diffeomorphism of $\mathbb R^n$ and $K$ is a compact set in $\mathbb R^n$, can we find another diffeomorphism $\tilde f$ of $\mathbb R^n$ such that:
(1)$f=\tilde f$ on a neighborhood of $K$.
(2)There is a bounded set $V$ and $\tilde f=id$ outside $V$?
 A: The function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=-x$ and $K=[-1,1]$ provides a counterexample - any continuous map $\tilde{f}$ satisfying properties 1) and 2) would necessarily fail to be injective by the intermediate value theorem. 
As Jason DeVito points out below, we can use a similar setup to create a counterexample for any $n$. Letting $f:\mathbb{R}^n\to\mathbb{R}^n$ be defined by $f(x_1,x_2,\ldots,x_n)=(-x_1,x_2,\ldots,x_n)$, we have for all $p\in\mathbb{R}^n$
$$\det(df_p)=\begin{vmatrix}
-1 & 0 & \cdots & 0\\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1
\end{vmatrix}=-1.$$
Thus, any diffeomorphism $\tilde{f}$ satisfying 1) would also have to have $\det(d\tilde{f}_p)=-1$ for some $p\in\mathbb{R}^n$. Because $\tilde{f}$ is a diffeomorphism, it can't have $\det(d\tilde{f}_p)=0$ for any $p\in\mathbb{R}^n$. Because $\det(d\tilde{f}_p)$ varies continuously with $p$, we must have $\det(d\tilde{f}_p)<0$ for all $p\in\mathbb{R}^n$. But $\det(dI_p)=1>0$ for all $p\in\mathbb{R}^n$, so we can't have $\tilde{f}=I$ anywhere.
