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It is well-known (e.g. Lang, Fundamentals of differential geometry, Prop. 2.3 in Chapter II) that the following extension of the regular value theorem holds for Banach manifolds:

Let $\phi : M\rightarrow N$ be a $C^k$-map between two Banach manifolds and let $y\in N$ be so that at every $x\in \phi^{-1}(y)$, the kernel of the differential $d\phi_x$ splits (i.e. can be complemented by a closed subspace) and $d\phi_x$ is surjective. Then $\phi^{-1}(y)$ is a (Banach) $C^k$-submanifold of $X$.

But what if $d\phi_x: T_xM\rightarrow T_y N$ is at every $x\in \phi^{-1}(y)$ not necessarily surjective, but something "a bit worse". For example, what if there exists a fixed Banach space $X$ so that $\operatorname{im}(d\phi_x)\cong X$ at every $x\in \phi^{-1}(y)$?

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    $\begingroup$ @PietroMajer Thank you for reminding me, I totally forgot to add that part ^^ $\endgroup$
    – Martin
    Commented Mar 24 at 16:52
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    $\begingroup$ Note that $\Phi\subset M$ being a submanifold also implies by definition that $T_p\Phi$ is a complemented subspace of $T_p M$. $\endgroup$
    – Pietro Majer
    Commented Mar 24 at 17:53
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    $\begingroup$ This will fail even in the finite dimensional case, say for functions whose derivative vanishes identically on the zero-level set. $\endgroup$
    – Moishe Kohan
    Commented Mar 24 at 18:00
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    $\begingroup$ What you really want is an analogue of the constant rank theorem for smooth maps of Banach spaces, see the references given in the answers here: mathoverflow.net/questions/204378/… $\endgroup$
    – Moishe Kohan
    Commented Mar 24 at 19:23

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There are some issues in infinite dimensions such as not all linear subspaces are closed etc. So I will state the result in finite dimensions first.

Suppose $F:M\to N$ is a smooth map between smooth manifolds (the regularity can be relaxed) and $S\subset N$ is a proper submanifold (it is a closed subset of $N$) We say that $F$ is transversal to $S$ if for any $x\in M$ such that $F(x)\in S$, we have

$$ D_x F(T_xM)+T_{F(x)} S= T_{F(x)} N. $$ More about transversality here. Then $F^{-1}(S)$ is a submanifold of $M$. This more general result reduces to the implicit function theorem. The implicit function theorem corresponds to the simplest case when $S$ consists of a single point.

Comment. Without surjectivity of the differential things can get tricky. The best one can hope for is so called Kuranishi description requiring for example the map be Fredholm. Here is a finite dimensional example. $\newcommand{\bR}{\mathbb{R}}$.

Consider $\phi:\bR^2\oplus \bR^2\to\bR^2\oplus\bR$ given by $$ \bR^2\oplus \bR^2\ni (x,y)\mapsto \phi(x,y)= (x,P(y))\in\bR^2\oplus \bR $$ where $P:\bR^2\to\bR$ is a polynomial map in the variables $y=(y_1,y_2)$.

If $P=0$, then $\phi^{-1}(0)$ is the $2$-plane $0\oplus \bR^2$.

If $P(y)=(y_1-y_2)(y_1+y_2)$, then $\phi^{-1}(0)$ is the union of two lines, $y_2=\pm y_1$. This is a $1$-dimensional algebraic set with a singularity at the origin.

In both cases the image of $d\phi(0)$ is $\mathbb{R}^2\oplus 0$. The map $P:\bR^2\to\bR$ is an example of Kuranishi map.

This example is in a sense universal. If $\phi$ is Fredholm then the zero set of $\phi$ is locally the zero set of a nonlinear map between finite dimensional spaces, the so called Kuranishi map.

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    $\begingroup$ Does it really address the question? OP is asking about preimage of a singleton, in which case transversality is the same as regularity, same as surjectivity of the derivative. $\endgroup$
    – Moishe Kohan
    Commented Mar 24 at 19:13
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    $\begingroup$ I'll add a comment to my answer. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 25 at 10:16
  • $\begingroup$ As a minor comment, one doesn't necessary need that the derivative $D_x F$ is Fredholm. It suffices that it is an relatively open operator with closed complemented kernel and image. In this case, the singularities of the level set are described by a non-linear map from the kernel to the complement of the image of $D_x F$. (Reference: Section 3 of arxiv.org/abs/2010.10165) $\endgroup$
    – Tobias Diez
    Commented May 2 at 9:50

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