A diffeomorphism with $\det(df)>0$ and convexity of matrices Let $f: \mathbb{R}^{n}\rightarrow \mathbb{R}^n$ be a diffeomorphism such that $\det(df)>0$ at the origin (hence everywhere). Show that there is a continuous map $F: [0,1]\times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ such that for each $t\in [0, 1]$, $F(t, \cdot)$ is a diffeomorphism, $F(0, x)=Id, F(1, x)=f(x)$.
The hint is to use the fact that $\mathbb{R}^n$ is linearly contractible.
So this reduces to showing that $\det((1-t)df+tI)>0$. Here, I get stuck. At first I thought maybe there is some well known convexity relation that could be useful, but then I realized if I take the map $(x_1, x_2, x_3, \ldots, x_n)\to (-x_1, -x_2, x_3, \ldots x_n)$, it has $\det(df)>0$ but when $t=1/2$ $\det((1/2)df+(1/2)I)=0$, so the above idea fails.
Any input would be appreciated. 

Since $Gl_n^+$ is connected, up to a translation and a rotation I may suppose $f(0)=0$ and $df(0)=I$.
Now, for $t>0$ define
$$F_t(x)=f(tx)/t$$
and
$$F_0(x)=x$$
Since $df(0)=I=F_0$, $F$ is continuous, $dF_t(x)=df(tx)>0$ for every $t$, $F_1=f$ and $F_0=I$, we have finished.
