Why is this pre-image a disconnected manifold?

Question

I am working on the following problem: consider $f:\mathbb{R}^3\rightarrow\mathbb{R}^2$ given by $f(x,y,z)=(y^2-x^2, x^2+z^2)$. Prove that $f^{-1}(1,4)$ is a disconnected manifold of dimension $1$.

I have already computed the set of critical points and critical values of $f$ and concluded that $(1,4)$ is a regular value of $f$. By the Regular Value Theorem, I know that $f^{-1}(-1,4)$ is a manifold of dimension $3-2=1$. Now I can't prove that it is disconnected. I know that connected $1-$ dimensional manifolds are either diffeomorphic to one interval or to $\mathbb{S}^1$. Does this help?

Thank you very much!

Answer 1

Well you can directly parametrize. Let $X$ be your set. Then if $(x,y,z) \in X$ then $x^2+z^2 =4$ so that $x = 2\cos{\varphi}, z = 2\sin{\varphi}, \varphi \in [0, 2\pi)$. Also we have $y^2 - 4\cos^2{\varphi} = 1$ and thus

$$ y = \pm \sqrt{1+4\cos^2{\varphi}} $$ which gives you the two components, i.e. $$ X = \{\, (2\cos{\varphi}, \sqrt{1+4\cos^2{\varphi}}, 2\sin{\varphi})\, |\, \varphi \in [0, 2\pi) \, \} \cup \{\, (2\cos{\varphi}, -\sqrt{1+4\cos^2{\varphi}}, 2\sin{\varphi})\, |\, \varphi \in [0, 2\pi) \, \}. $$ Here is a 3D plot of your set, generated by Wolfram Mathematica

Answer 2

Your manifold, let's call it $M$, contains, in particular, the points $(0, 1, 2)$ and $(0, -1, 2)$. Right? But it does not contain any points with $y=0$.

This means that the projection $p:M\to\Bbb{R},\; p(x,y,z)=y$ - a continuous function - has a disconnected image (it contains 1 and -1 but it does not contain 0). Which means that $M$ itself cannot be connected.