Why $df_{0}\simeq \mathrm{identity}$

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

First, $f$ is orientation-preserving. Second, $GL(n)^+$ is path-connected (e.g., use the $QR$ decomposition).
• Thanks, f is o.p. $\Rightarrow sign(det(df_{0}))=1>0\Rightarrow df_{0}\in GL^{+}(n)$ – TKM Feb 18 '14 at 4:55
• 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. – user135041 May 4 '14 at 17:01
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.