Let $f:\mathbb{R}^n \to \mathbb{R}^m$ be a smooth open map ($n \ge m$). Then it cannot happen that $\text{rank}(df)<m$ everywhere. I would like to find an elementary argument for that, which only uses elementary calculus and topology.
This fact is an immediate corollary of Sard's theorem. Indeed, suppose that every point in $\mathbb{R}^n$ is critical. Then the measure of $f(\mathbb{R}^n)$ is zero, so it cannot be open.
Is there a more elementary argument which does not use Sard's theorem? (ideally does not involve measure theory at all).
Proposition. Let $f:M\to N$ be a smooth map between smooth manifolds and suppose that $\mathrm{rk}(df_p)<\dim N$ for all $p\in M$. Then, $f$ is not an open map.
Proof. Let $k=\sup\{\mathrm{rk}(df_p):p\in M\}$. Then, $k<\dim N$. Moreover, $$M_k:=\{p\in M:\mathrm{rk}(df_p)=k\}$$ is open in $M$. Thus, the restriction $f:M_k\to N$ is smooth of constant rank $k<\dim N$. By the rank theorem (see, e.g. Lee's book on smooth manifolds), $f$ is locally of the form $$\mathbb{R}^m\to\mathbb{R}^n,\quad(x_1,\ldots,x_k,x_{k+1},\ldots,x_{m})\mapsto(x_1,\ldots,x_k,0,\ldots,0)$$ near $0$. In particular, $f(M_k)$ is not open in $N$. $\square$