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Let $U\subset \mathbb R^m$ be an open and $f:U\to \mathbb R^m$ be a function of class $C^1$.

If the points where the determinant of the Jacobian Matrix of $f$ is zero are isolated ones and $m>1$, how do I prove this function must be open?

I don't know even how to begin, I need some hints.

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    Have you tried considering an open subset $W\subseteq U$ and an arbitrary point $y\in f(W)$ and showing that there exists an open subset $V$ of $f(W)$ for which $y\in V$ using the Inverse Function Theorem? I think the fact that the zeros of $f'$ are isolated will permit you to consider an $x\in W$ such that $f(x)=y$ and for which $f'(x)\in M_m^{-1}$ so that you can indeed apply the Inverse Function Theorem. – 2017-01-11
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    @Guest Thank you for your comment. I'm trying to understand what do you mean by your comment. What is $M_m^{-1}$? – 2017-01-11
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    The set of invertible $m\times m$ real matrices. – 2017-01-11
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    @Guest Is this $W$ arbitrary? I didn't understand where you applied the Inverse function theorem. – 2017-01-11

2 Answers 2

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The thing that you are missing is the notion of the degree of a map. You can start by reading this wikipedia article (you only need the section about degree of maps from a closed region), then take a look at the references they list. But the best reference I know is "Differential Topology" by Guillemin and Pollack (not on the wikipedia list).

The setup is that you have two oriented smooth manifolds $M, N$ of dimension $m$, where $M$ possibly has boundary while $N$ has empty boundary. Suppose that $f: M\to N$ is a $C^1$-smooth proper map (proper means that preimages of compacts are compact). The degree $deg_f(y)$ is defined for $y\in N - f(\partial M)$. In the case when $y$ is a regular value of $f$, $$ deg_f(y)=\sum_{x\in f^{-1}(y)} (\pm 1), $$ where $1$ is used when $df: T_xM\to T_yN$ is orientation-preserving and $-1$ otherwise. In particular, if $f$ preserves orientation (at its regular points) and $y$, as above is a regular value, then $deg_f(y)$ is the cardinality of $f^{-1}(y)$. An important property of degree is that it is locally constant, in particular, it is constant on each component of $N -f(\partial M)$.

Given all this, here is a proof of the claim that you are after. It suffices to consider the case when $U$ is connected. Note that the set of critical points of your map $f$ is 0-dimensional, $m>1$, hence, the critical set of $f$ does not separate $U$. Therefore, without loss of generality, we may assume that $f$ has positive jacobian determinant $J_f(x)$ at its regular points. Openness is a local property. It is clear (by the inverse function theorem) at the points where $J_f(x)\ne 0$, hence, consider an (isolated) critical point $x_0\in U$ of $f$. Let $B$ be a sufficiently small closed ball centered at $x_0$, $B\subset U$, I am assuming that $x_0$ is the only critical points of $f$ in $B$. Without loss of generality (for generic choice of the radius of $B$), $y_0:=f(x_0)\notin f(\partial B)$. Now, I will take $M:=B$ and consider $deg_g$ for the restriction $g$ of $f$ to $M=B$. There is a sequence of points $x_i\in B- \{x_0\}$ converging to $x_0$. Then, since $J_g$ is positive away from $x_0$ (in $B$), $deg_g(y_i)>0$, $y_i=f(x_i)$. Hence, by continuity of degree, $deg_g(y_0)>0$. Now, if $y_0$ is not an interior point of $f(B)$, there is a sequence $z_j\notin f(B)$ converging to $y_0$. Hence, $deg_g(z_i)=0$ for all large $j$. Hence, $deg_g$ is discontinuous at $y_0$. A contradiction. Thus, $f$ is open at $x_0$. qed

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    So could you say a bit more about what you mean by $deg_g(y_0)$ (i.e. the degree of a critical value)? I'm worried, for instance, that the pre-image of $y_0$ could consist of sequence of points converging to the critical point, $x_0$. – 2017-01-14
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    I mean, obviously one thing you could do is look at the degree of the map $f|_{\partial B}$ when you post compose it with a projection onto a sphere around $y_0$. But with infinitely many elements of $f^{-1}(y_0)$ running around inside every neighborhood of $x_0$, I'm less certain that notion of degree is well behaved. But maybe it is and I'm just being stupid. – 2017-01-14
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    @Timkinsella: Not me, this is a standard definition: Make a small perturbation of $f$ making $y_0$ a regular value, then compute as above. Use transversality to show that the definition is independent of perturbation. – 2017-01-15
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    I see. Thanks. . – 2017-01-15
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    @Timkinsella: The best reference for such arguments that I know is Guillemin and Pollack, "Differential Topology". – 2017-01-15
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    So, one thing that might help me is to see where your argument fails for $m=1$... – 2017-01-16
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    @Timkinsella: A point can separate $R^1$ but not $R^m$, $m>1$. Therefore, $J_f$ has constant sign whenever nonzero. – 2017-01-16
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    @Timkinsella: The sign change of $f'(x)$ is exactly what happens in your example $f(x)=x^2$. If $f'(x)\ge 0$ or $f'(x)\le 0$ for all $x\in {\mathbb R}$ then the function $f: {\mathbb R}\to {\mathbb R}$ is, of course, monotonic and, hence, open (provided that its set of critical points has empty interior). You can also see that for arbitrary $m$ the "right" condition is not that the set of critical points is isolated but that it does not separate $R^m$. With this condition, the case $m=1$ is no longer exceptional. – 2017-01-16
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Here is an elementary proof, using only the Inverse Function Theorem.

It suffices to prove the question in the case of just one singular point, i.e., one point where the Jacobian of $f$ (denoted as $\det f'(\boldsymbol{x}))$, vanishes. We shall then show the following:

Claim. Let $m>1$ and $B_R$ be the open ball of radius $R$ in $\mathbb R^m$, centered at the origin, and $f:B_R\to\mathbb R^m$, continuously differentiable. If the determinant of the Jacobian of $f$ vanishes only at $x=0$, then $f$ is open.

In order to show that the above claim, it suffices to show that $f[B_r]$ is open, for every $r

We assume for simplicity that $f(0)=0$. If there exists a sequence $\{\boldsymbol{x}_n\}\subset B_R\smallsetminus\{0\}$, with $\boldsymbol{x}_n\to 0$ and $f(\boldsymbol{x}_n)=0$, then for every $r1$.) The set $V$ is open as a consequence of the Inverse Function Theorem. It remains to show that $W$ is also open.

Let $\,\boldsymbol{w}\in W$. Since $\,\boldsymbol{w}\not\in f[B_\varrho]$, then the function $\,g:\overline{B}_\varrho\to\mathbb R^m$, defined as $$ g(\boldsymbol{x})=|f(\boldsymbol{x})-\boldsymbol{w}|^2, $$ does not vanish and hence it attains a minimum in some $\boldsymbol{x}_0\in\overline{B}_\varrho$, where $\overline{B}_\varrho$ is the closed ball of radius $\varrho$ centered at the origin. In particular, $|\boldsymbol{x}_0|<\varrho$, since for $\boldsymbol{x}\in C_\varrho$ we have $$ g(\boldsymbol{x})=|f(\boldsymbol{x})-\boldsymbol{w}|^2>\big(|f(\boldsymbol{x})|-|\boldsymbol{w}|\big)^2> \big(2d-d\big)^2=d^2>|\boldsymbol{w}|^2=g(0). $$ Therefore, since the continuously differentiable function $g$ attains a local minimum at $\boldsymbol{x}_0$, then its gradient
vanishes at $\boldsymbol{x}_0$ $$ 0=\nabla g(\boldsymbol{x}_0)=2f'(\boldsymbol{x}_0)\big( f(\boldsymbol{x}_0)-\boldsymbol{w}\big), $$ and since $f(\boldsymbol{x}_0)-\boldsymbol{w}\ne 0$, the matrix $f'(\boldsymbol{x}_0)$ is not invertible and therefore $\det f'(\boldsymbol{x}_0)=0$. Now, since $\det f'(\boldsymbol{x})$ vanishes only for $\boldsymbol{x}=0$, then $\boldsymbol{x}_0=0$, which implies that for every $\boldsymbol{x}\in \overline{B}_\varrho$ $$ |f(\boldsymbol{x})-\boldsymbol{w}|^2=g(\boldsymbol{x})\ge g(0)=|\boldsymbol{w}|^2. $$ This means that the whole open ball of radius $|\boldsymbol{w}|$ centered at $\boldsymbol{w}$ does not intersect the set $f\big[\overline{B}_\varrho\big]$, and hence $W$ is open, which concludes the proof of the Claim.