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.

  • 2
    $\begingroup$ Do you know about the inverse function theorem? $\endgroup$ – Tim kinsella Dec 22 '12 at 20:05
  • 1
    $\begingroup$ 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. $\endgroup$ – still_learning Dec 22 '12 at 20:18
  • $\begingroup$ @ Timkinsella I've only seen IFT for mappings from a space to itself. I'll be looking into it. $\endgroup$ – Babamots Dec 22 '12 at 21:39
  • 1
    $\begingroup$ @SeleneAuckland It's a special case of "Are submersions open?" $\endgroup$ – Babamots Jul 23 at 21:42
  • 1
    $\begingroup$ @SeleneAuckland Yes, that's what I mean. It's specifically about a submersion between Euclidean spaces. In retrospect, it was a very simple question, but I don't use functional analysis regularly and I wasn't thinking about it right. $\endgroup$ – Babamots Jul 25 at 16:59

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 $z_0$.

  • $\begingroup$ Is this question the same as "Are submersions open?" please? $\endgroup$ – user636532 Jul 23 at 5:53
  • 1
    $\begingroup$ If you refer to submersions between smooth manifolds, then this is the local version of that result (as you can read here: math.stackexchange.com/questions/1607049/… ) $\endgroup$ – wisefool Jul 24 at 9:00
  • $\begingroup$ Thanks wisefool! $\endgroup$ – user636532 Jul 24 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.