I am interested in knowing whether the zero locus of a nonlinear function forms a manifold. In essence, I am looking for an extension of the preimage theorem.
The preimage theorem, in simple terms, requires a smooth function $f: \mathbb{R}^n \to \mathbb{R}^n$ with the property that the derivative $Df_x: \mathbb{R}^n \to \mathbb{R}^n$ at any point $$ x\in\mathcal{M} := \{ x\in\mathbb{R}^n ~|~ f(x) = 0 \} $$ is surjective. The theorem then states that the set $\mathcal{M}$ is a manifold, and that the tangent space $T_x \mathcal{M}$ for every $x\in\mathcal{M}$ is given by $$ T_x \mathcal{M} = \ker(Df_x). $$
I am curious if we can relax the assumptions on both $f$ and its derivative $Df_x$. My intuition is that it should be sufficient to demand that $f$ is continously differentiable and that $Df_x$ has constant rank for all $x\in G$, where $G$ is an open neighborhood of $\mathcal{M}$.
My thoughts are the following: Under these assumptions, in a neighborhood of $x\in\mathcal{M}$ the function $f$ "looks like" $Df_x$. Since $Df_x$ is a matrix, the kernel $\ker(Df_x)$ forms a manifold. And since $Df_x$ resembles $f$ locally, also $\ker(Df_x)$ should resemble $\mathcal{M}$ locally. The assumption on constant rank of $Df_x$ then should ensure that the manifold $\mathcal{M}$ has constant dimension.
To illustrate this further, take a linear map $f:\mathbb{R}^n \to \mathbb{R}^n, ~ x \mapsto Ax$ with a matrix $A$ that has a nontrivial kernel. As $\ker(A) = \mathcal{M}$ is a subspace of $\mathbb{R}^n$, it is clearly a manifold. But the preimage theorem can not be applied in this situation, since $Df_x = A$ is not surjective.