Let $f\colon \mathbb{R}^{n} \to \mathbb{R}^{n} $ with $\left\| \mathrm{D}_{}f(\mathbf{x})\right\|_{2}\le c< 1$ and \begin{align*} g\colon \mathbb{R}^{n} \to \mathbb{R}^{n}, \quad g(\mathbf{x})=f(\mathbf{x})+\mathbf{x} .\end{align*} I want to show that $g$ is surjective. My first thought was to maybe show that the function \begin{align*} d _{\mathbf{y}}(\mathbf{x}) = \left\|g(\mathbf{x})-\mathbf{y}\right\|_{2} ^2 \end{align*} and show that its minimum is attained and equals zero for each $\mathbf{y} \in \mathbb{R}^{n} $ but found that this probably won't work since $\mathrm{D}_{}f(\mathbf{x})$ is not guaranteed to be invertible. A second idea would be to use the fixed point theorem, but I don't know how I would use it in this case. Could someone maybe help me with this?
I came up with this:
For every $\mathbf{y} \in \mathbb{R}$ consider the function \begin{align*} h_\mathbf{y}(\mathbf{x}) =\mathbf{x}+\mathbf{y} - g(\mathbf{x}) .\end{align*} We have \begin{align*} \mathrm{D}_{}h_{\mathbf{y}}(\mathbf{x}) = \frac{\mathrm{d}}{\mathrm{d}\mathbf{x}} (\mathbf{x}+\mathbf{y}-\mathbf{x}+f(\mathbf{x})) = \frac{\mathrm{d}}{\mathrm{d}\mathbf{x}} (y+f(\mathbf{x})) = \mathrm{D}_{}f(\mathbf{x}) .\end{align*} By the mean value inequality, for every $\mathbf{x}_{1}, \mathbf{x}_{2} \in \mathbb{R}^{n} $ we have \begin{align*} \left\|h_{\mathbf{y}}(\mathbf{x}_{1})-h_{\mathbf{y}}(\mathbf{x}_{2})\right\|_{2} \le \left(\max_{\mathbf{x} \in [\mathbf{x}_{1}, \mathbf{x}_{2}]} \left\|\mathrm{D}_{}f(\mathbf{x})\right\|_{2} \right)\left\|\mathbf{x}_{1}-\mathbf{x}_{2}\right\|_{2} \le c \left\|\mathbf{x}_{1}-\mathbf{x}_{2}\right\|_{2} \end{align*} with $c < 1$ and hence it's a contraction mapping. By the fixed point theorem, the function $h_{\mathbf{y}}(\mathbf{x})$ has a unique fixed point, call it $\mathbf{x}_{0}$. This yields \begin{align*} h_{\mathbf{y}}(\mathbf{x}_{0}) =\mathbf{x}_{0} &\iff \mathbf{x}_{0} + \mathbf{y}- g(\mathbf{x}_{0})=\mathbf{x}_{0} \\ &\iff g(\mathbf{x}_{0}) = \mathbf{y} .\end{align*}