Let $f: U\subseteq \mathbb{R}^n \to \mathbb{R}^m$ be $C^1$ s.t. $n \leq m$, $U$ open, $\mathrm{rank}{D_pf}=n$. Prove $f$ is locally injective at $p$.

Question

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.

Answer 1

I don't think you need all that. The differential $Df_p$ is injective, and the question is how to measure this injectivity and transfer that to $f(x) -f(p)$ which is only approximated by $Df$ near $p$. Here is one way:

Take ball around $p$ such that for any $\vec{v}$ with $\|\vec{v}\|=1$ and any $q$ in this ball we have $Df_p(\vec{v})\cdot Df_q(\vec{v})>0$ (Such neighborhood exists since $Df_p(\vec{v})\cdot Df_p(\vec{v})>m$ for some positive $m$, the sphere $\|\vec{v}\|=1$ is compact, and the map $(\vec{v}, q) \to Df_p(\vec{v})\cdot Df_q(\vec{v})$ is continuous; this implies that the function $q\mapsto \min_{\vec{v}}\big\{\, Df_p(\vec{v})\cdot Df_q(\vec{v})\,\big| \,\,\|v\|=1\,\big\}$ is continuous in $q$, and hence positive near $p$.)

Now suppose $p_1$ and $p_2$ are in this ball. We will show that $f(p_1)\neq f(p_2)$ by showing that $f(p_2)\cdot \vec{w}> f(p_1)\cdot \vec{w}$ for a well-chosen $\vec{w}$.

In fact, let $\frac{p_2-p_1}{|p_2-p_1|}=\vec{v}$. We take $\vec{w}=D_p f (\vec{v})$.

Now, take a unit speed straight line segment $\gamma(t)$ from $p_1$ to $p_2$, so that $\gamma'(t)=\vec{v}$. It is enough to show $(f(\gamma(t))\cdot \vec{w})'= (f(\gamma(t))'\cdot \vec{w}>0$ for all $t$.

By chain rule $D(f\cdot\gamma)= D f_{ \gamma(t) } (\vec{v})$. On the other hand, the path lies inside the ball neighborhood (the ball is convex), so $D f_{ \gamma(t) } (\vec{v}) \cdot \vec{w}=D f_{ \gamma(t) } (\vec{v})\cdot D f_{p} (\vec{v}) >0$. This completes the proof.

Answer 2

You cannot do without some form of the inverse function theorem.

As ${\rm rank}\bigl(Df(p)\bigr)=n$ the matrix $\bigl[Df(p)\bigr]$ has an $(n\times n)$-submatrix with nonvanishing determinant; say $$\det \left[{\partial f_i\over\partial x_k}(p)\right]_{1\leq i\leq n, \>1\leq k\leq n}\ \ne0\ .$$ Let $f(p)=:q\in{\mathbb R}^m$, and let $\pi$ be the projection of the full $y$-space ${\mathbb R}^m$ onto its $(y_1,\ldots, y_n)$ coordinate plane $Y'$. Then the map $$f':=\pi\circ f:\quad{\mathbb R}^n\to Y',\qquad (x_1,\ldots, x_n)\mapsto\bigl(f_1(x),\ldots, f_n(x)\bigr)$$ has Jacobian matrix $$\left[{\partial f_i\over\partial x_k}(p)\right]_{1\leq i\leq n, \>1\leq k\leq n}$$ at $p$. The inverse function theorem then implies that $f'$ maps a neighborhood $V$ of $p$ injectively onto a neighborhood $V'$ of $q':=\bigl(f_1(p),\ldots, f_n(p)\bigr)$. This immediately implies that $f$ is injective on $V$.