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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.

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Do you know about the inverse function theorem? – Tim kinsella Dec 22 '12 at 20:05
Since you are new here, here are some tips. 1) for better results show some work or where you get stuck in the proof 2) when youve found a satisfactory answer please accept it by clicking the check-mark next to the answer. – MSEoris Dec 22 '12 at 20:18
@ Timkinsella I've only seen IFT for mappings from a space to itself. I'll be looking into it. – Babamots Dec 22 '12 at 21:39
@ Timkinsella Yeah, I guess that's all I need. Sorry for making noise. – Babamots Dec 22 '12 at 21:44
up vote 2 down vote accepted

If $J_f(z_0)$ has maximal rank for $z_0\in U$, then we can find coordinates $(x_1,\ldots, x_m, y_1,\ldots, y_{n-m})=(x,y)$ such that the minor of $J_f(x,y)$ formed by the first $m$ columns is non-vanishing for $(x,y)=z$ in a neighborhood of $z_0$.

Now, consider the map $v(x,y)=(f(x,y), y)$, $v:U\to\mathbb{R}^n$. The Jacobian matrix is $$\begin{pmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\\0 & I_{n-m}\end{pmatrix}$$ and note that its determinant is non-vanishing for $(x,y)$ in the same neighborhood of $z_0$. Hence, it is invertible with smooth inverse; let us write $v^{-1}(x,y)=(a(x,y), b(x,y))$ and compute $$(x,y)=v(v^{-1}(x,y))=(f(a(x,y), b(x,y)), b(x,y))$$ which means $b(x,y)=y$ and $f(a(x,y),y)=x$. So $f\circ v^{-1} (x,y)=x$, for $(x,y)$ in the given neighborhood of $z_0$.

Now, the map $f\circ v^{-1}$ is clearly open, as it is a projection; the map $v$ is continuosly invertible, hence open. Therefore the map $f=f\circ v^{-1}\circ v$ is open in a neighborhood of $z_0$.

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