Let $g:\mathbb{R}^m \longrightarrow \mathbb{R}^m,g\in C^1(\mathbb{R}^m)$ such that:
$\|g'(x)(v)\|\geq\|v\|,\forall v\in \mathbb{R}^m,\forall x \in \mathbb{R}^m$
show that any rectilinear path starting from $y_0\in g(\mathbb{R}^m)$ can be lifted from any point $x_0 \in g^{-1}(y_0)$.
Any hints would be appreciated.
If $ g $ be surjective get the result, but how do I prove this.
The map is surjective:
From the assumption on the differential of $g$ it follows that $g$ is a local $\mathcal C^1$ diffeomorphism since the differentials kernel is everywhere zero. Hence $g(\mathbb R^m)$ is open in $\mathbb R^m$. Without loss of generality i assume $g(0) = 0$. Assume $g$ is not surjective. Let $$r := \sup \{s \geq 0 \vert B_s(0) \subseteq g(\mathbb R^m)\}.$$ Then $0 < r < \infty$ since the image of $g$ is open and $g$ is not surjective. Again since $g$ has open image and $B_s(0)$ is compact for all $s < \infty$ it follows that there exists some $y \in \partial B_r(0)$ with $y \notin g(\mathbb R^m)$. Let $\sigma : [0,1] \to \mathbb R^m$ be a straight line from $0$ to $y$ (parametrized by arc length). By construction $\sigma([0,1[) \subset g(\mathbb R^m)$ and $\sigma(1) = y \notin g(\mathbb R^m)$. Let $0 < t_n < 1$ be an increasing sequence with $t_n \to 1$. Since $g$ is a local diffeomorphism it is easy to see that for all $t_n$ there exists a unique lift $\tilde \sigma_n$ with respect to $g$ of $\sigma_{\vert [0,t_n]}$ to $\mathbb R^m$ starting at $0$ (by compactness of $\sigma_{\vert [0,t_n]}$, which is contained in $g(\mathbb R^m)$, we can cover $\sigma_{\vert [0,t_n]}$ by a finite number of balls where we have a lift on every ball and paste these lifts together). We claim that the sequence $\tilde \sigma_n(t_n)$ is bounded. In fact we can estimate the length of $\tilde \sigma_n$ as follows: $$1 > \mathcal L (\sigma_{\vert [0,t_n]}) = \int_0^{t_n} \vert \vert \sigma'(t) \vert \vert dt = \int_0^{t_n} ||(g \circ \tilde \sigma_n)'(t)|| dt = \int_0^{t_n} ||g'_{\tilde \sigma_n(t)}(\tilde \sigma_n'(t))|| dt \\ \geq \int_0^{t_n} ||\tilde \sigma_n'(t)|| dt = \mathcal L(\tilde \sigma_n).$$
Thus we may extract a convergent subsequence $\tilde \sigma_n(t_n) \to p$. Consequently by continouity $g(p) = \lim g(\tilde \sigma_n(t_n)) = \lim \sigma(t_n) = \sigma(1) = y$, contradicting the fact that $y \notin g(\mathbb R^m)$. So $g$ is surjective.
Using the methods of this proof one can directly show that any path of finte length admits such a desired lift, or more generally every path that has finite length restricted to any compact subset of the domain.