Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$, $m<n$ be a mapping that is linear in each variable separately (i.e., in each of the functions $f_i(x_1,\cdots,x_n)$, $1\leq i\leq m$, the degree of each variable $x_j$ is at most 1, but there can be a product of several variables, e.g., $7x_1x_3x_{n-1}$).

I define a local boundary point as a point $\boldsymbol{x}\in \mathbb{R}^n$ with $f(\boldsymbol{x})$ not on the global boundary of $f$, that for some $\delta>0$ there exist a ball $\mathcal{B}_\delta(\boldsymbol{x})$ of radius $\delta$ around $\boldsymbol{x}$ such that there is no $\epsilon>0$ for which $f(\mathcal{B}_\delta(\boldsymbol{x}))$ will contain a ball of radius $\epsilon$ around $f(\boldsymbol{x})$. Or in simple words: there is some direction in the $f$-plane I can't step to from $\boldsymbol{x}$.

Can this mapping have a local boundary point? The Jacobian can definitely be rank deficient, but it is counterintuitive for such mapping to have a local boundary point.