Assume $f:\mathbb R^n\times\mathbb R\to\mathbb R^n$ is given by $f(x,u)=Ax+bu$ for some choice of coordinates $(x,u):x\in\mathbb R^n,u\in\mathbb R$, some constant matrix $A\in M_n(\mathbb R)$, and a constant vector $b\in\mathbb R^n$. Assume that the matrix $S=(b,Ab,\dots,A^{n-1}b)$ has full rank. That is $r(S)=n$.
Question: Is it true that whenever $D\subset\mathbb R^n$ is a set with a non-empty interior in $\mathbb R^{n+1}$, then so is $f[D]$ in $\mathbb R^{n}$? Why is that?
p.s. I am not sure about the appropriate tags for this question. I added the control-theory tag, since the system $\dot x=f(x,u)$ is a controllable system by Kalman's criterion.
Let us write
$$f(x,u) = \begin{bmatrix}A & b\end{bmatrix} \begin{bmatrix}x \\ u\end{bmatrix}$$
By the Open Mapping Theorem if $f$ is surjective, then it is an open mapping (it is continuous and linear). This means $f$ is an open mapping if $\begin{bmatrix}A & b\end{bmatrix}$ has full rank.
But if you mean the mapping
$$f(x_0,u) = e^{A(t-t_0)}x_0 + \int_{t_0}^t e^{A(t-\tau)}Bu(\tau)d\tau $$
which is the solution of $\dot{x} = Ax + bu$ for $x(t_0) = x_0$, then it is well-known that $\text{Im} f = \text{Im} \begin{bmatrix}b & Ab & \dots A^{n-1}b\end{bmatrix}$. So, we can conclude that $f$ is an open mapping if $(A, b)$ is controllable.