How to show that a continuously differentiable function $f:\mathbb{R}^{n}\to \mathbb{R}^m$ can't be a 1-1 when $n>m$? This is an exercise in Spivak's "Calculus on manifolds".
I can solve the problem in the case $m=1$. To see this, note that the result is obvious if the first partial derivative $D_1 f(x)=0$ for all $x$ as then $f$ will be independent of the first variable. Otherwise there exists $a\in \mathbb{R}^n$ s.t. $D_1 f(a)\not=0$. Put $g:A\to \mathbb{R}^n, g(x)=(f(x),x_2, \ldots, x_n)$ (with $x=(x_1,\ldots, x_n)$). Now the Jacobian
$$
g'(x) = \left[
\begin{array}{cc}
D_1 f(x) & 0 \\
0 & I_{n-1}
\end{array}
\right]
$$
so that $\text{det}\, g'(a)=D_1 f(a)\not=0$. By the Inverse Function Theorem we have an open set $B\subseteq A$ s.t. $g:B\to g(B)$ is bijective with a differentiable inverse. In particular, $g(B)$ is open.
Pick any $g(b)=(f(b),b_2,\ldots, b_n)\in g(B)$. Since $g(B)$ is open there exists $\varepsilon > 0$ such that $(f(b), b_2, \ldots, b_n+\varepsilon) \in g(B)$. Thus we can find $b'\in B$ s.t. $g(b')=(f(b), b_2, \ldots, b_n+\varepsilon)$. By the definition of $g$, $f(b')=f(b)$ with $b'$ and $b$ differing in the last coordinate. Thus $f$ isn't injective.
However, I cannot generalize this argument to higher dimenssions.
Best Answer
Let $k\leq m$ be the maximum rank of $f'(x)$ when $x$ runs through $\mathbb R^n$ and $a\in \mathbb R^n$ be a point where the maximum is attained: $rank(f'(a))=k$.
We will have $rank(f'(u))=k$ for all $u$ in some open neighbourhood $U$ of $a$ [obtained by choosing a non-zero $k\times k$ minor of $f'(a)$].
The constant rank theorem will then assure us that locally near $a$ the map $f$ looks like $ (x_1,...,x_n) \mapsto (x_1,...,x_k,0,...,0)$ and is thus not injective since $n\gt k$ (because $n\gt m\geq k$)
Edit
Here is a 13-line proof of the constant rank theorem, written by a high-quality contributor to this site: @Pierre-Yves Gaillard