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Let $f: U \rightarrow \mathbb{R}^m$, $C^1$, such that all singularities are isolated and $m > 1$. Show that $f$ is an open application, i.e., $f$ maps open subsets to open subsets.

Well, I am trying so solve this exercise but any of my ideas seem to work. I am trying to think first in the case when $f$ has one isolated singularity (it must be at maximum a countable set of such points).

Since the point, let's call $p$, is isolated, it means that there is an $\epsilon > 0$ such that $\forall x \in B(p,\epsilon)-\{p\}$ we have det$J(x) \neq 0$. I am trying to show that for some $y$ sufficient close to $x$, by the inverse function theorem and since det$J(y) \neq 0$ there is an open subset $W$ such that $f(W)$ is open, and somehow we must have $x \in W$.

I'm facing two big problems: first, I don't know if this is true, i.e, as we get closer to $x$ the open set $W$ containing $y$ might get so small such that we never have $x \in W$, for every $y$ we choose.

The second problem is the fact that I am not using the fact that $m >1$.

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    $\begingroup$ Where does $U$ live, and what is your definition of singularity? Are you assuming $U\subset\Bbb R^m$ (same $m$?) so that the Jacobian is a square matrix? (You'd better use $m>1$, or else the function $f(x)=x^2$ on $\Bbb R$ gives you a counterexample.) $\endgroup$ Commented Aug 19, 2016 at 22:18
  • $\begingroup$ Well $U \subset \mathbb{R}^m$ and a singularity of f is a point $p$ such dat det$f'(p) = 0$. I am aware of counter examples but I cannot find how to use the fact that $m > 1$. $\endgroup$ Commented Aug 19, 2016 at 23:27
  • $\begingroup$ Have you tried to prove that if $f(x) = p$ then $f$ maps a small punctured neighborhood $U\setminus \{x\}$ of $x$ on a punctured neighborhood of $p$? $\endgroup$ Commented Jun 18, 2018 at 19:07
  • $\begingroup$ It's not a real function by the way. $\endgroup$ Commented Jul 5, 2018 at 22:25

1 Answer 1

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I'm assuming you mean that $U$ is open. We will show that the image of any ball of small enough radius around singular point $p$ contains an open ball round $f(p)$. Since the same is true for non-singular points by inverse function theorem, this implies that map $f$ is open.

Let $r>0$ be small enough that the closed ball $\bar{B}(p, 2r)$ is inside $U$ and does not contain any other singular points of $f$. Let $A=\bar{B}(p,2r)\setminus B(p,r)$. By compactness of $A$, the operator norm of $(Df)^{-1}$ attains a positive maximum, say $M$ on $A$. Consider an open ball $B(q=f(p), r/2M)$ and the punctured open ball $O=B(q,r/2M)\setminus q$. Our goal is to show that $O$ is contained in the $f$-image of $B(p, 2r)$.

Since $f$ is continuous and sends $p$ to $q=f(p)$, there exists $p_0\neq p$ in $B(p, r)$ such that $q_0=f(p_0)\in O$. Take any $q_1 \in O$. Since $m>1$, the punctured ball $O$ is path connected, and in fact any two points in it are connected by a smooth path of length less than $r/M$. Let $\gamma(t)$ be such a path from $q_0$ to $q_1$ in $O$ (i.e. $\gamma:[0,1] \mapsto O$, $\gamma(0)=q_0$, $\gamma(1)=q_1)$.

I now claim that there exists a path $\tilde{\gamma}:[0,1]\to B(p, 2r)$ lifting $\gamma$ (i.e. with $f(\tilde{\gamma}(t))=\gamma(t)$). In particular, $f$ maps the "other end" $ \tilde{\gamma}(1)$ to $q_1$; since $q_1$ was arbitrary, $O$ is in the image of $B(p, 2r)$.

Why does $\tilde{\gamma}$ exist? By inverse function theorem, there is a (unique) lift of $\gamma_{t\in[0, \tau]}$ for some $\tau >0$. We need to show that maximal such $\tau$ is $1$. But that is ok: $\tilde{\gamma}(\tau)$ is in $B(p, 2r)$ (this follows from the fact that $\tilde{\gamma}$ lifts $\gamma$, so whenever it is in $A$ its velocity is at most $M$ times larger than that of $\gamma$, and hence the the total length of the part of $\tilde{\gamma}$ in $A$ is bounded by $r/M \times M=r$, so $\tilde{\gamma}$ can not escape $B(p,2r)$). Now if $\tau \neq 1$, we can apply inverse function theorem at $\tilde{\gamma}(\tau)$ and extend the lift to a larger segment.

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