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Milnor lemma 2 pg 34 "Any orientation preserving diffeomorphism f on $R^m$ is smoothly homotopic to the identity"

So he proves that $f\simeq df_0$ ,which he says is clearly homotopic to the identity. Can you explain me why?

Here I found two explanations I don't understand: 1) $Gl^{+}(m,\mathbb{R})$ is path connected. Why is $df_0 \in Gl^{+}(m,\mathbb{R})$? What prevents $df_{0}\in Gl^{-}(m,\mathbb{R})$?

2) $df_0$ is isomorphic everywhere and thus(why?) isotopic to identity.

Thanks

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First, $f$ is orientation-preserving. Second, $GL(n)^+$ is path-connected (e.g., use the $QR$ decomposition).

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  • $\begingroup$ Thanks, f is o.p. $\Rightarrow sign(det(df_{0}))=1>0\Rightarrow df_{0}\in GL^{+}(n)$ $\endgroup$ – TKM Feb 18 '14 at 4:55
  • $\begingroup$ Nice answer Ted; so the isotopy in question is defined by $(x,t)\mapsto\gamma(t)x$, where $\gamma$ is a smooth path joining $df_0$ to $\mathrm{id}_{\mathbf{R}^n}$? @TKM: this answer fully answers your question, and is certainly more appropriate than the other one. Please consider accepting it. $\endgroup$ – user135041 May 4 '14 at 17:01
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This is not really an answer to your specific question, but another way of proving the lemma.

Since $f$ is an orientation preserving diffeomorphism on $\mathbb R^m$ it must have degree 1. By the Hopf degree theorem (p. 51 in Milnor), it is smoothly homotopic to the identity.

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    $\begingroup$ Oh come on! A cannon to kill a fly! (Not to mention almost surely circular reasoning.) $\endgroup$ – Ted Shifrin Feb 18 '14 at 2:14

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