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Where can one find the proof of the following fact:

If there are two orientation-preserving diffeomorphisms $\phi_0$ and $\phi_1$ of $R^n$, then there exists a homotopy $\phi(t)$, such that $\phi(0)$ = $\phi_0$, $\phi(1)$ = $\phi_1$, and $\phi(t)$ is a diffeomorphism for any $t$ (and $\phi$ is smooth for all variables).

This is easy to prove for $n=1$, and relatively easy for $n=2$, but for an arbitrary $n$ seems difficult.

This fact is used in the knot theory to justify the coincidence of two definitions of the knot equivalence.

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An explicit deformation retraction of $\mathrm{Diff}(\mathbb R^n)$ onto $\mathrm{Aff}(\mathbb R^n)$ is given by $$f_t(x)=f(0)+\frac{f(tx)-f(0)}{t}$$ for $t\in (0,1]$ and $f_0(x)=f(0)+f^\prime(0)x$.

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  • $\begingroup$ Actually, both answers give the same approach. For me, the second one has a more compact form. $\endgroup$
    – user50311
    Commented May 19, 2018 at 0:48
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By a one parameter family of translations, get the origin fixed. By a one parameter family of linear transformations, get the derivative at the origin to be the identity matrix. Conjugate by a dilation. Send the dilation parameter to infinity. Look at the Taylor series: it is going to the identity. I think the rest is easy from there.

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