This function must be open if these points are isolated

Question

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.

Answer 1

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

Answer 2

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 $r<R$.

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 $r<R$, the ball $B_r$ contains a term $\boldsymbol{x}_n$ and as $\det f'(\boldsymbol{x}_n)\ne 0$, then $f[B_r]$ contains an open ball centered at $f(\boldsymbol{x}_n)=0$. Let's now assume that no such sequence exists, which implies that there exists an $\varrho_1\in (0,R)$, such that $$ f(x)\ne 0\quad\text{whenever $0<|\boldsymbol{x}|\le \varrho_1$}. $$ We pick an arbitrary $\varrho\in(0,\varrho_1)$ and let $$ \min_{\boldsymbol{x}\in C_\varrho} |f(\boldsymbol{x})|=2d, \quad \text{where $C_\varrho=\{\boldsymbol{x}: |\boldsymbol{x}|=\varrho \}$}. $$ We shall show that $B_d\subset f[B_\varrho]$. Clearly, it suffices to show that $B_d\!\smallsetminus\!\{0\}\subset f[B_\varrho\!\smallsetminus\!\{0\}]$. Let $$ V=(B_d\!\smallsetminus\!\{0\})\cap f[B_\varrho\!\smallsetminus\!\{0\}] \quad\text{and}\quad W=(B_d\!\smallsetminus\!\{0\})\!\smallsetminus\! f[B_\varrho\!\smallsetminus\!\{0\}]. $$ It suffices to show that both $V$ and $W$ are open, since $B_d\!\smallsetminus\!\{0\}$ is connected. (In fact, here is the only place where we use the requirement that $m>1$.) 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.