# The extension of diffeomorphism

Let ${\Omega _1}$,${\Omega _2}$ be two open sets in $\mathbb R^n$ and $f$ is a diffeomorphism between them. For every $x$ in ${\Omega _1}$, is there an open set $\Omega_{x} \subset \Omega_1$ and a diffeomorphism $g$ of $\mathbb R^n$ such that $g=f$ when restricted to $\Omega_{x}$?

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Assume $x\in \Omega_1$ such that $Df(x) = A$. Wolg $x=0$ and $f(0) =0$, otherwise the following reasoning needs to be 'translated'. Now let $\phi (x)= \psi(||x||)$ be a smooth radially symmetric cutoff function which is identically $0$ outside of $B_t(0)$ and $=1$ inside, say, $B_{t/2}(0)$, s.t. $\psi^'(t) \le O(1/t)$ . Assume $B_t(0)\subset \Omega_1.$

Now look at

$$F(x):= (1-\phi(x)) Ax + \phi(x) f(x)$$

which is defined for every $x\in\mathbb{R}^n$ (it's easy to see that this is possible).

For $||x|| \ge t$ this is just $Ax$.

For $||x||< t$ denote $r=||x||$ and calculate for $v$ such that $||v||= 1$ $$DF(x) v = -\psi^'(r)<\frac{x}{r},v> Ax +(1-\phi(x))Av +\psi^'(r)<\frac{x}{r}, v> f(x) + \phi(x) Df(x)v\\ = Av + \phi(x)(Df(x)v-Av) +\psi^'(r) <\frac{x}{r}, v>(f(x)-Ax)$$

The first term on the rhs is nonzero for nonzero $v$, since $A$ is invertible. Note that it does not depend on $t$, and since the sphere is compact it's norm is bounded from below by a constant.

The second term is arbitrarily small, since $f$ is continuously differentiable (I do assume continuity of $Df$) and $\phi$ is bounded, if only $t$ is small enough.

Since $f(x)-Ax = o(x)$ the third term also becomes arbitrarily small given the growth rate of $\psi$.

In response to a comment from Leonid Kovalev: from $$F(x)-F(y) =Ax-Ay - \phi(x)(Ax-f(x)) +\phi(y)(Ay - f(y))$$ it is easy to see that $F$ is 1-1, using the $f(x)-Ax=o(||x||)$ property and the fact that $A$ invertible. The fact that $F$ is onto follows, e.g., by elementary degree theory from the fact that $F=A$ on the boundary of $B_t(0)$. It is probably possible to see this with a more elementary reasoning, for which I don't have the time right now.

An alternative reasoning which you may find more satisfying: according to Thm. 1.7 in Hirsch's Differential Topology the set of diffeomorphisms is open in $C^1(\mathbb{R}^n, \mathbb{R}^n)$. The $F$ I have constructed is arbitrarily close to $x\mapsto Ax$ in $C^1$.

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Something should be said about the global invertibility of $F$. –  user31373 May 23 '12 at 18:42
@LeonidKovalev ok, valid point. I did say something about global invertibility of $F$. –  user20266 May 23 '12 at 19:24
Thanks! I did not doubt that your proof works, just wanted to make this point for the sake of other readers. For the same reason I'll add that the computations would be a bit tidier with the normalization $Df(x)=I$ (achieved by composition with a linear invertible map). –  user31373 May 23 '12 at 19:27
@LeonidKovalev: this time I disagree. If, by accident or transformation, $A=I$, it is (in my opinion, your mileage may vary) less transparent, which term is a derivative and which is evaluation of a function. –  user20266 May 23 '12 at 19:34
Nice proof. BTW, that $F$ is a diffeomorphism is easy if we notice that $F$ is proper. –  Hezudao May 24 '12 at 1:45