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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 $V$?

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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.

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    If $f$ is orientation-preserving and there is a \epsilon>0 such that |f-id|<\epsilon, then the conclusion holds.2012-04-28