I was trying to solve a problem in differential geometry that I realized the following statement is the core of my argument
Let $f: U\subseteq \mathbb{R}^n \to \mathbb{R}^m$ be a $C^1$ function on an open set $U$ where $n \leqslant m$ such that $\mathrm{rank}{Df}=n$ at some $p\in U$. Show that $f$ is injective in a neighborhood of $p$.
After thinking about it, I think that it can be proven using the constant rank theorem. Firstly, since $f$ is $C^1$, we have $\mathrm{rank}Df\geq n$ in a neighborhood of $p$. Since $n$ is the maximum possible rank, we have $\mathrm{rank}Df = n$ near $p$. So, the constant rank theorem applies.
Now the constant rank theorem says that I can find two open sets $V \subseteq U$ and $W\subseteq \mathbb{R}^m$ such that $f(V) \subseteq W$ and two diffeomorphisms $\psi:\mathbb{R}^n \to V$ and $\varphi:\mathbb{R}^m \to W$ such that $\varphi^{-1}\circ f\circ \psi: \mathbb{R}^n \to \mathbb{R}^m$ has the canonical form $(x_1,\cdots,x_n) \mapsto (x_1,\cdots,x_n,0,\cdots,0)$.
Since $\varphi^{-1}\circ f\circ \psi$ is clearly injective, and $\varphi$ and $\psi$ are diffeomorphisms, $f = \varphi \circ \big(\varphi^{-1}\circ f\circ \psi \big) \circ \psi^{-1}$ is injective on $V$.
Assuming my proof is correct (well, is it?) I still think it's overkill. Is there a proof that is more elementary? Ideally, a proof without using the Inverse Function Theorem. Or if it uses the Inverse Function Theorem, it should not be longer than this one since the constant rank theorem can be proven using the Inverse Function Theorem and hence, it's obvious that a longer proof exists.