Let \(f:\mathbb R^3\to\mathbb R^2\), and suppose \(0\) is a regular value of \(f\); that is, \(Df\) has rank \(2\) at every point of \(f^{-1}(0)\). Prove that \[ \mathbb R^3\setminus f^{-1}(0) \] is arcwise connected.
Solution
Put \[ Z=f^{-1}(0). \]
1. The set \(Z\) is a closed smooth embedded submanifold of \(\RR^3\)of dimension \(1\).
The set \(Z\) is closed because \(f\) is continuous and \(\{0\}\) is closed. Since \(0\) is a regular value and \[ \operatorname{rank}Df_z=2 \] for every \(z\in Z\), the regular-level-set theorem gives \[ \dim Z=3-2=1. \] If \(Z\) is empty, its complement is \(\RR^3\), which is arcwise connected.
2. Let \(a,b\in\RR^3\setminus Z\) be distinct, and define thestraight arc\[ \gamma_0(t)=(1-t)a+tb, \qquad 0\leq t\leq1. \]There are neighborhoods of \(0\) and \(1\) in \([0,1]\) whose images under\(\gamma_0\) are disjoint from \(Z\).
Since \(Z\) is closed and \(a,b\notin Z\), there are open balls about \(a\) and \(b\) disjoint from \(Z\). Continuity of \(\gamma_0\) gives intervals near \(0\) and \(1\) mapped into those balls.
3. There exists a smooth embedded arc\[ \gamma:[0,1]\longrightarrow\RR^3 \]with\[ \gamma(0)=a, \qquad \gamma(1)=b, \]which agrees with \(\gamma_0\) near the endpoints and is transverse to\(Z\).
2. It gives an arbitrarily small smooth perturbation\(\gamma\) that is transverse to \(Z\) and agrees with \(\gamma_0\) there.The original map \(\gamma_0\) is an embedding because \(a\neq b\).Embeddings of the compact interval into \(\RR^3\) are open in the\(C^1\) topology, so the perturbation may be chosen small enough that\(\gamma\) is still an embedding.
4. The arc \(\gamma\) does not meet \(Z\).
Suppose instead that \[ z=\gamma(t)\in Z \] for some \(t\in[0,1]\). Transversality would require \[ D\gamma_t(T_t[0,1])+T_zZ=T_z\RR^3. \] The first summand has dimension at most \(1\), and step 1 gives \[ \dim T_zZ=1. \] Hence the left-hand side has dimension at most \(2\), whereas \[ \dim T_z\RR^3=3. \] This is impossible. Therefore \[ \gamma([0,1])\cap Z=\varnothing. \]
5. Any two distinct points of \(\RR^3\setminus Z\) are joined by anarc contained in \(\RR^3\setminus Z\).
For arbitrary distinct \(a,b\in\RR^3\setminus Z\), step 3 gives an embedded arc from \(a\) to \(b\), and step 4 shows that its image lies entirely in the complement.
6. Therefore\[ \boxed{\RR^3\setminus f^{-1}(0)\text{ is arcwise connected}}. \]
Step 5 is exactly the defining property of arcwise connectedness.
7. Q.E.D.
Step 6 is the required conclusion.