Show that $f^{-1}$ of a critical value isn't a submanifold $$f: \Bbb R^2 \to \Bbb R, \:\: f(x_1, x_2)=x_1x_2$$
I've determined the critical and regular points and critical and regular values of f (see: Regular value: intuition about surjectivity condition): 


*

*regular points are all $(x, y)$ with $x \not =  0$ OR $y \not = 0$, in other words $\Bbb R^2 \backslash (0,0)$

*critical point: $(0,0)$

*regular values: $\Bbb R \backslash \{0\}$

*critical values: $\{0\}$


Now I need to show that that $f^{-1}(0)$ isn't a submanifold of $\Bbb R^2$. I'm fairly new to the subject and don't quite know how to do this. Any help is greatly appreciated!
 A: Notice that $f^{-1}(\{0\})\setminus\{(0,0)\}$ has four connected-components.


*

*If $f^{-1}(\{0\})$ is a $0$-submanifold, then it is discrete, which is a contradiction since for all $t\in\mathbb{R}$, one has:
$$(0,t)\in f^{-1}(\{0\}).$$

*If $f^{-1}(\{0\})$ is a $1$-submanifold, then $f^{-1}(\{0\})\setminus\{(0,0)\}$ is diffeomorphic to $\mathbb{R}\setminus\{t\}$, $t\in\mathbb{R}$.

*If $f^{-1}(\{0\})$ is a $2$-submanifold, then $f^{-1}(\{0\})\setminus\{(0,0)\}$ is diffeomorphic to $\mathbb{R^2}\setminus\{p\}$, $p\in\mathbb{R}^2$.
Alternatively, one can proceed using the intuition of tangent spaces. If $f^{-1}(\{0\})$ is a $d$-submanifold of $\mathbb{R}^2$, then there exists $V$ an open neighborhoods of $(0,0)$ in $\mathbb{R}^2$, $W$ an open neighborhoods of $(0,0)$ in $\mathbb{R}^d$ and $\varphi\colon V\rightarrow W$ a diffeomorphism such that:
$$\varphi(f^{-1}(\{0\})\cap V)=\mathbb{R}^d\cap W.$$
Hence, $\mathrm{d}_{(0,0)}\varphi\cdot (1,0)$ and $\mathrm{d}_{(0,0)}\varphi\cdot (0,1)$ are linearly independent vectors in  $\mathbb{R}^2$ so that one has $d=2$. Therefore, $f^{-1}(\{0\})$ is an open subset of $\mathbb{R}^2$, which is a contradiction.
