Jtdylktuy

I am reading Milnor's ''Topology from the differentiable viewpoint'' and in the chapter about vector fields there is a Lemma that states that given $f:mathbb{R}^n to mathbb{R}^n$ an orientation preserving diffeomorphism, then $f$ is isotopic to the identity.

He starts the proof (somewhat) as follows:

Without loss of generality, you can suppose $f(0)=0$, Then you can define the derivative at $0$ by $d_0f(x)=lim_{tto 0} frac{f(tx)}{t}$.

Then define $F:mathbb{R}^n times [0,1] to mathbb{R}^n$ as $F(x,t)=frac{f(tx)}{t}$ for noncero $t$. And $F(x,0)=d_0f(x)$.

Now from here on out, my doubts start, so I am going to try to be loyal to the text on the proof:

To prove that $F$ is smooth, even as $t$ tends to $0$, you can write $f$ of the form: $f(x)= sum_{i=1}^{n} x_ig_i(x)$

(So I am guessing this is looking at $f$ as in $f=(g_1,...,g_n)$. If this is not it please correct me)

Then it justs says: ''Not that $F(x,t)=sum_{i=1}^{n} x_ig_i(tx)$ for all values of t.

Thus $f$ is isotopic to the linear mapping $d_0f$, which is clearly isotopic to the identity.''

So, why does this proves $F$ is smooth? And when did you use the hypothesis of $f$ preserving orientation?

Any help would be appreciated, thanks in advanced.

To get $f(x)=sum x_i g_i(x)$ you can use $f(x)=int_0^1frac{d}{ds}f(sx)ds=sum_iint_0^1x_ifrac{partial}{partial x_i}f(sx)ds$, so you just set $g_i(x)=int_0^1frac{partial}{partial x_i}f(sx)ds$. The function $F(x,t)=sum_{i=1}^{n} x_ig_i(tx)$ is evidently smooth as $g_i$'s are smooth. $F$ is an isotopy between $f$ (for $t=1$) and the linear map $d_0f$. So you still need to verify that $d_0f$ is isotopic to the identity, and that's because we suppose that $det (d_0f)>0$.