# f differentiable map of finite dimensional vector spaces, with derivative injective. Why is f injective?

Suppose A and B are finite dimensional vector spaces, $U\subseteq{A}$ is an open subset, $a\in U$ and $f:U\rightarrow B$ is $C^\infty$ with $(Df)_a$ injective. I need helping showing that there exists an open neighbourhood $V$ of $a$ in $U$ (i.e. $a \in V \subseteq U$) such that f is injective on V.

I know this is similar to the inverse function theorem, but in order to do a similar proof I need $(Df)_a$ invertible.

I tried doing this by contradiction, showing that there is a sequence $(v_n)_n$ in A such that $v_n \not= 0$ and $lim_{n\to\infty} v_n=0$ and $f(a+v_n)=f(a)$. But here I don't know exactly how to keep going.

If $m$ is the rank of $DF_a$ and $n\ge m$ the dimension of $B$: if $m=n$ use the inverse function theorem. Otherwise there is a complementary subspace of dimension $k=n-m$ of the image of $Df_a$ in $B$. Add a factor $\mathbb{R}^k$ to $A$ and look at $A\times \mathbb{R}^k :(x, y) \mapsto (f(x),y)$ with $y\in \mathbb{R}^k$ Verify this satisfies the assumptions of the inverse function theorem an apply it. Then conclude back to properties of $f$.