How to show it is a manifold.

Question

Let $f \colon \mathbb{R}^4 \rightarrow \mathbb{R}^2$ be the function $$f(x,y,z,w) = (xy+z, yz+w).$$ Prove that $f^{-1}(0,0)$ is a $2$-manifold.

Here is what I did: When $(xy+z,yz+w)=(0,0)$, we have $z=-xy$ and $w=-yz=-xy^2$. Therefore, we have $f^{-1}(0,0)=\{(x,y,-xy,-xy^2)\}$

I know I should use implicit function theorem to finish this proof, but I do not know how to use it.

Answer 1

Thanks to your computation we can find a global $C^\infty$ parametrization for $f^{-1}(0,0)$ which shows, by definition, that $f^{-1}(0,0)$ is a $2$-manifold.

Consider the maps: \begin{gather} \varphi:\mathbb R^2\rightarrow f^{-1}(0,0)\subset \mathbb R^{4}, \qquad \varphi(x,y) = (x,y,-xy,-xy^2)\\ \psi: f^{-1}(0,0)\rightarrow \mathbb R^2, \qquad \psi(x,y,z,w) = (x,y)\\ \end{gather} They are $C^\infty$ maps and one is the inverse of the other, hence $\varphi$ is a global parametrization of $f^{-1}(0,0)$.

---

We have shown something more: $\varphi$ is infact a diffeomorphism, hence $f^{-1}(0,0)$ is diffeomorphic to $\mathbb R^2$.

Answer 2

Hint : Let $g : \Omega \subset \mathbb{R}^{n} \longmapsto \mathbb{R}^{n-k}$ of class $C^{1}$ such that $\forall x \in g^{-1}(p)$ it holds that $Dg(x)$ (The differential of $g$ in the point $x$) is surjective, then $g^{-1}(p)$ is a submanifold.