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 $d_0f(x)=lim_{t\to 0} \frac{f(tx)}{t}$.
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 $f(x)=\int_0^1\frac{d}{ds}f(sx)ds=\sum_i\int_0^1x_i\frac{\partial}{\partial x_i}f(sx)ds$, so you just set $g_i(x)=\int_0^1\frac{\partial}{\partial x_i}f(sx)ds$. 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 $\det (d_0f)>0$.