I was working in some exercises about multivariable calculus and I'm stuck. The exercise is
Lef $f:U\subseteq\mathbb{R}^{n}\to\mathbb{R}^{m}$ be a $C^1$ function where $U$ is an open set and $n<m$. If $x_0\in U$ and $\operatorname{rank}(Df(x_0))=n$ (rank of the matrix derivative) then there exist $A\subseteq U$ an open set such that $x_0\in A$ and $\operatorname{int}(f[A])=\emptyset$ (topological interior).
I really don't know how to procced but I have an idea: as $n<m$ and $\operatorname{rank}(Df(x_0))=n$ then $Df(x_0)$ has a submatrix $B$ of size $n\times n$ such that has non zero determinant. Suposse that $f=(f_1,\dots,f_m)$ where $f_{i}:U\subseteq\mathbb{R}^{n}\to\mathbb{R}$ and define $h:U\subseteq\mathbb{R}^{n}\to\mathbb{R}^{n}$ by $h(x)=(f_1(x),\dots,f_n(x))$ (w.l.g. the first $n$ rows of $Df(x_0)$ are l.i.). Then, by the inverse function theorem, there exist a neighborhood $V$ of $x_0$ where $h$ is invertible. But, from here, how can I go? Thanks for your help.
Indeed, $f(V)$ has empty interior under your assumptions. To see this, take $x \in V$ and write $y=(y_1,\ldots,y_m)=f(x)$. Observe that $y_{n+1}, \ldots, y_m$ are determined by $y_1,\ldots,y_n$ since $y=f(h^{-1}(y_1,\ldots,y_n))$. Thus for any $\epsilon \ne 0$ the point $(y_1,\ldots, y_{m-1},y_m+\epsilon)$ is not in $f(V)$.