I'd like to show that if $U\subset R^n$ is open, $f:U\to R^m$ is smooth, and $J_f(x)$ is surjective (full rank) for every $x\in U$, then $f(U)$ is open.
My thoughts so far:
For any $f(x)\in f(U)$, Taylor's theorem provides $\varepsilon>0$ so that whenever $\|z\|<\varepsilon$, $f(x+z)=f(x)+J_f(x)z+O(\|z\|^2)$. Choose $\delta>0$ so that whenever $r\in R^m$ and $\|r\|<\delta$ there is a $z\in R^n$ so that $r=J_f(x)z$, $\|z\|<\varepsilon$ and $x+z\in U$.
Choose any $r\in R^m$ so that $\|r\|<\delta$. If we can find $z$ so that $f(x+z)-f(x)=r$, we will have shown that $f(x)$ is an interior point of $f(U)$.
From the choice of $\delta$, we can choose $z\in R^n$ so that $\|z\|<\varepsilon$, and $x+z\in U$ and therefore $f(x+z)=f(x)+J_f(x)z+O(\|z\|^2)$ in other words $f(x+z)-f(x)=r+O(\|z\|^2)$.
I don't know what to do about the $O(\|z\|^2)$ terms and I'd appreciate some more eyes checking my work so far. I suspect there's something much easier anyway. Internet searches have led me to discussions of manifolds, but I haven't studied them yet. This is my first question, so I apologize for all the expectations I'm breaking.