I'm starting to learn a little basics about smooth manifolds, starting from embedded manifolds.
In a linear space, say $\mathbb{R}^{n}$, I know we can define embedded submanifolds with a locally defined function, which is equivalent to the canonical definition via diffeomorphisms. Following is the definition found in An introduction to optimization on manifolds
Definition 3.10. Let $\mathcal{E}$ be a linear space of dimension d. A non-empty subset $\mathcal{M}$ of $\mathcal{E}$ is a (smooth) embedded submanifold of $\mathcal{E}$ of dimension $n$ if either
- $n=d$ and $\mathcal{M}$ is open in $\mathcal{E}$-we also call this an open submanifold; or
- $n=d-k$ for some $k \geq 1$ and, for each $x \in \mathcal{M}$, there exists a neighborhood $U$ of $x$ in $\mathcal{E}$ and a smooth function $h: U \rightarrow \mathbb{R}^k$ such that (a) If $y$ is in $U$, then $h(y)=0$ if and only if $y \in \mathcal{M}$; and (b) $\operatorname{rank} \mathrm{D} h(x)=k$. Such a function $h$ is called a local defining function for $\mathcal{M}$ at $x$.
So suppose I'm given a $C^{\infty}$ function $f: \mathbb{R}^{n} \to \mathbb{R}^{k}$, and consider a set $S := \{ x \in \mathbb{R}^{n}|f(x)=0 \}$, I know for different $x \in S$, $\operatorname{rank}Df(x)$ can be different, so in this case, $f$ is not a local defining function in the above definition for all $x$.
So I wonder based on the above definition, can we conclude $S = \sqcup_{i \in I}M_{i}$, where $M_{i}$ has dimension $d(i)$ for some $d:I \to \{d-k, \dots, d \}$?
I tried to constructing partition $S$ based on $\operatorname{rank}Df(x)$, and consider $p \circ f$, for some projection map $p$, but although I can satisfy part $b$ in the above definition, I cannot satisfy $(a)$. Maybe I need to consider a different function or use the diffeomorphism definition? Or perhaps it is not true, then what is a counterexample?
Edit: I think as the comment stated, this is not true, but what about the case we allow boundaries?