Propagation of connectivity for level lines of a submersion Let $1\leq m\leq n$ integers and $\mathcal{O}$ an open connected set of $\mathbb{R}^n$. Let $f\in\mathcal{C}^1(\mathcal{O},\mathbb{R}^m)$, such that the differential of $f$ at any $x\in\mathcal{O}$ is surjective. Assume that for a certain $x_0\in\mathcal{O}$, the level line $f^{-1}\big(f\big(\{x_0\}\big)\big)$ is connected.
Question:  For any $x\in\mathcal{O}$, is the level line $f^{-1}\big(f\big(\{x\}\big)\big)$ connected?
 A: No. Let $\mathcal{O} = \{ (x,y) \in \mathbb{R}^2, y > 0 \}$ and $f \colon \mathcal{O} \rightarrow \mathbb{R}$ be $f(x,y) = x^2 - y^2$. For $(0,1) \in \mathcal{O} $, the level set
$$ f^{-1} \left( f((0,1)) \right) = f^{-1} ( \{ -1 \}) = \{ (x,y) \in \mathbb{R}^2 \, | \, x^2 - y^2 = -1, y > 0 \} = \{ (x, \sqrt{x^2 + 1}) \, | \, x \in \mathbb{R} \} $$
is connected while for $(1,1) \in \mathcal{O}$, the level set
$$ f^{-1} \left( f((1,1)) \right) = f^{-1} ( \{ 0 \}) = \{ (x,y) \in \mathbb{R}^2 \, | \, x^2 - y^2 = 0, y > 0 \} = \{ (x,x) \, | \, x > 0 
\} \cup \{ (x,-x) \, | \, x < 0 \} $$
is not connected.

For an example where the connectivity of the fibers isn't even preserved locally, take $\mathcal{O} = \mathbb{R}^3 \setminus \{ (x,y,0) \, | \, xy = 1 \}$ and $f \colon \mathcal{O} \rightarrow \mathbb{R}^2$ to be $f(x,y,z) = (y,z)$. For $(0,0,0) \in \mathcal{O}$, the level set
$$ f^{-1} \left( f \left( (0,0,0) \right) \right) = f^{-1} \left( \{ (0,0) \} \right) = \{ (x, 0, 0) \, | \, x \in \mathbb{R} \}$$
is connected but near $(0,0,0)$ you have points of the form $(\varepsilon, \varepsilon, 0)$ with $\varepsilon > 0$ and for them the level sets
$$ f^{-1} \left( f \left( (\varepsilon,\varepsilon,0) \right) \right) = f^{-1} \left( \{ (\varepsilon, 0) \} \right) = \left \{ (x,\varepsilon,0) \, | \, x \neq \frac{1}{\varepsilon} \right \}$$
are disconnected.
