Let $g:\mathbb{R}^n \to \mathbb{R}^n$, such that $g$ is differentiable.
We suppose that :
$$|||\mathrm{d}g(x)||| \leq k < 1, \forall x$$
Then I need to proove that : $f = Id + g$ is one-to-one.
First if $f$ is one-to-one then it must be bijective because $f:\mathbb{R}^n \to \mathbb{R}^n$.
Moreover $f$ is also differentable because the sum of two differentiable functions is differentiable. So we have :
$$\mathrm{d}f(x) = x + \mathrm{d}g(x)$$
Using the operator norm we hance have :
$$||| \mathrm{d}f(x) ||| \leq ||| x ||| + ||| \mathrm{d}g(x) ||| \leq 1 + k $$
Yet, I don't understand how it gives an information on $f$ at this point...
Injectivity:
$f(x)=f(y)$ is equivalent to $x+g(x)=y+g(y)$ and $g(x)-g(y)=y-x$. We deduce that $\|g(x)-g(y)\|=\|x-y\|$. But $\|dg(x)\|<k$ implies that $\|g(x)-g(y)\|\leq k\|x-y\|$. We deduce that $\|x-y\|\leq k\|x-y\|, k<1$ and $\|x-y\|=0$.
Surjectivity.
$df(x).u=0$ implies that $u+dg(x)(u)=0$ implies that $\|u\|=\|dg(x)(u)\|\leq k\|u\|$. We deduce that $\|u\|=0$. The local inversion theorem implies that $f$ is a local homeomorphism and $f(\mathbb{R}^n)$ is open.
$f(x)=\int_0^1df(tx).xdt=\int_0^1x+dg(tx).xdt$. We deduce that $\|f(x)\|=\|\int_0^1x+dg(tx).xdt\|\geq |\|x\|-\|\int_0^1\|dg(tx).xdt\||\geq |\|x\|-k\|x\||=(1-k)\|x\|$ Let $y$ in the adherence of $f(\mathbb{R}^n)$, there exists a sequence $x_n$, such that $lim_nf(x_n)=y$. We have $\|f(x_n)\|\geq (1-k)\|x_n\|$. Since $(f(x_n)$ is bounded, we deduce that $x_n$ is bounded, consider a subsequence $x_{n_k}$ which converges towards $x$, since $f$ is continuous, $limf(x_{n_k})=f(x)=y$, we deduce that $f(\mathbb{R}^n)$ is open and closed and $f(\mathbb{R}^n)=\mathbb{R}^n$.