Let $f: \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ mapping such that $df_{\bf a}: \mathbb{R}^n \to \mathbb{R}^n$ is one-to-one, so that $f$ is one-to-one in a neighborhood of ${\bf a}$. How would I go about showing there exists $\epsilon > 0$ such that if $g: \mathbb{R}^n \to \mathbb{R}^n$ is a $C^1$ mapping with $g({\bf a}) = {\bf a}$ and $||dg_{\bf a}|| < \epsilon$, then the mapping $h: \mathbb{R}^n \to \mathbb{R}^n$, defined by $h({\bf x}) = f({\bf x}) + g({\bf x})$, is also one-to-one in a neighborhood of ${\bf a}$?
My work so far. I can show the following:
- Let $F: \mathbb{R}^n \to \mathbb{R}^m$ be $C^1$ at ${\bf b}$. If $dF_{\bf b}: \mathbb{R}^n \to \mathbb{R}^m$ is one-to-one, then $F$ itself is one-to-one on some neighborhood of ${\bf b}$.
- Let $T: \mathbb{R}^n \to \mathbb{R}^m$ be a one-to-one linear mapping with $|T({\bf{x}})|_0 \ge a|{\bf{x}}|_0$ for all ${\bf{x}} \in \mathbb{R}^n,$ where $a>0.$ If $||S-T|| \le \epsilon < a, |S({\bf{x}})|_0 \ge (a-\epsilon)|{\bf{x}}|_0$ for all ${\bf{x}} \in \mathbb{R}^n,$ so $S$ is also one-to-one. Hence the set of all one-to-one linear mappings $\mathbb{R}^n \to \mathbb{R}^m$ forms an open subset of $\mathscr{L}_{mn} \approx \mathbb{R}^{mn}.$ $($$\mathscr{L}_{mn}$ denotes the vector space of all linear mappings $\mathbb{R}^n \to \mathbb{R}^m.$$)$
We give two different solutions. They are the same on some fundamental level.
First Solution
1. Fix $n$ and $\epsilon > 0$ such that $n\epsilon <1$. Let $A$ be an $n \times n$ real matrix such that $$\|A\| = \max_{1 \le i, j \le n} |a_{ij}| < \epsilon.$$ Then $I + A$ is invertible. If not, there exists ${\bf x} = (x_1 \dots x_n)^\text{T} \neq {\bf 0}$ such that $$(I + A){\bf x} = {\bf 0} \implies A{\bf x} = -{\bf x} \implies -x_i = \sum_{j=1}^n a_{ij}x_j \text{ for all }j.$$Let $|x_k| \ge |x_j|$ for all $j$. Then $|x_k| > 0$, so$$|-x_k| = \left|\sum_{j=1}^n a_{k_j}x_j\right| \le \sum_{j=1}^n |a_{k_j}||x_j|<\epsilon(|x_1| + \dots + |x_n|) \le \epsilon n |x_k| < |x_k|,$$which is a contradiction.
2. Let $A$ be an invertible $n \times n$ matrix. Then there exists $\epsilon > 0$ if $B = (b_{ij})_{n \times n}$ and $|b_{ij}| < \epsilon$ for all $i, j$. Then $A + B$ is invertible. Let $\epsilon > 0$ such that if $B = (b_{ij})_{n \times n}$, $|b_{ij}| < \epsilon$ for all $i, j$ then $A^{-1}B = (c_{ij})$, $|c_{ij}| < \eta$ for all $i, j$ where $\eta > 0$ is fixed such that $n\eta < 1$. By $(1)$, $I + A^{-1}B$ is invertible, hence $A(I + A^{-1}B) = A + B$ is invertible.
3. By $(2)$, there exists $\epsilon > 0$ if $\|dg_{\bf a}\| < \epsilon$, so $d(f + g)_{\bf a}$ is invertible $($$df_{\bf a}$ is invertible$)$. Therefore, $h$ is $1$-$1$ in a neighborhood of ${\bf a}$, as we wanted to show.
Second Solution
Since $df_{\bf a}: \mathbb{R}^n \to \mathbb{R}^n$ is $1$-$1$, there exists $\epsilon > 0$ if $A: \mathbb{R}^n \to \mathbb{R}^n$ and $\|df_{\bf a} - A\| < \epsilon$, so $A$ is also $1$-$1$. Let $g: \mathbb{R}^n \to \mathbb{R}$ be $C^1$ such that $\|dg_{\bf a}\| < \epsilon$. So we have$$\|d(f + g)_{\bf a} - df_{\bf a}\| = \|dg_{\bf a}\| < \epsilon \implies d(f+g)_{\bf a}: \mathbb{R}^n \to \mathbb{R} \text{ is }1\text{-}1.$$Hence $h = f + g$ is one to one in a neighborhood of ${\bf a}$.