Consider a (smooth) bundle $E\to B$, and a (smooth) function $f: E\to\mathbf{R}$ on the total space. Then it makes sense to talk about the derivatives of $f$ along the fibers. Let $C$ be the subspace of $E$ consisting of all points for which all fiber-wise derivatives of $f$ vanish, so that upon intersecting with any fiber $C$ consists of the critical points of the restriction of $f$ to the fiber. If the fiber is $n$-dimensional, then $C$ is carved out by $n$ equations, and so generically has codimension $n$ in $E$.


Let's say that $c$ is a point in $C$ so that the second derivative of $f$ in the fiber is nondenegerate (i.e. has non-degenerate Hessian; i.e. $f$ restricts to a Morse function on the fiber through $c$). Does it follow that the projection $C\to B$ is a local diffeomorphism near $c$?


The answer is yes when everything is finite-dimensional (and I believe the statement is iff). I am interested in the case when $B$ is finite-dimensional but the fibers of $E$ are infinite-dimensional.


**Edit:** This is a response to Andrew's question below (since answering in the comments proves difficult).


I'm using the word "Morse" loosely, largely because I don't actually know Morse theory. I would suggest that the definition I give is better than what's traditionally used. What I actually mean is this:


Let $M$ be a smooth manifold and $f:M\to\mathbb{R}$ a smooth map. What type of object is the second derivative $f^{(2)}$? In general, you should not talk about it by itself, because it does not transform as a tensor, although the pair $(f^{(1)},f^{(2)})$ is a vector in the $2$-jet bundle over $M$. But if $c$ is a critical point of $f$, then $f^{(2)}(c)$ is naturally a symmetric bilinear form $\mathsf T\!\_cM \times \mathsf T\!\_cM \to\mathbf{R}$. Thus it is a map $\mathsf T\!\_cM\to \mathsf T^\ast\!\!\!\_c M$. All I ask is that this map have zero kernel.


But if this condition is too weak, whereas a reasonable stronger condition works, I'd love to hear it.



Critical points on a fiber bundle
The part I'm still hesitant about is that the manifold you call $C$ is $n$-dimensional. Codimension arguments in the infinite dimensional setting are always a little sticky. So, I'm just going to treat it as an assumption.


Then the answer to your question is yes, and doesn't depend on what type of infinite dimensional vector spaces you use as your local model. Let's see why.


First you have a space $E$ which is a fiber bundle over $B$, via the projection $p$. Then at any point $x$ in $E$ we have the sequence of vector spaces,
$$
\mathsf T\!\_xF =\operatorname{ker}\mathrm{d}p\to \mathsf T\!\_xE \to\mathsf T\!\_{p(x)}B
$$
You also have a map $f: E \to R$, and hence a map $\mathrm{d}f: \mathsf T\!\_xE \to R =\mathsf T\!\_{f(x)}R$ for all points $x$ in $E$.


$C$ is defined to be the points in $E$ such that the restriction of $\mathrm{d}f$ to $\mathsf T\!\_{x}F$ vanishes. By assumption, $C$ is an $n$-dimensional manifold.


Now let's talk about second derivatives. There are two problems with second derivatives. The first arrises because we are working in the infinite dimensional setting. This means that we might not be able to think of the second derivative as a bilinear form due to possible convergence problems. However, as you pointed out, we can think of it as a map,
$$
\mathsf T\!\_x E\to \mathsf T^\ast\!\!\!\_x E.
$$
Now depending on the type of infinite dimensional manifold you are considering, this dual space may take on different meanings (banach dual, Frechet dual, etc). You at least get something in the algebraic dual (of possible non-continuous functionals).


The second problem with second derivatives is that they are usually not defined independent of a choice of coordinates. Again, as you pointed out, the pair ($\mathrm{d}f$, second derivative of $f$) transforms as a section of the second jet bundle, i.e. the second derivative doesn't change like a tensor but in an affine fashion depending on $\mathrm{d}f$.


This means that in general there is no intrinsic way to say the second derivative is "non-degenerate" at a random point of $E$. However, along $C$, a portion of the second derivative is still well defined, independently of coordinate choices. It is not the whole second derivative, but just the composite,
$$
H:\mathsf T\!\_cE\to \mathsf T^\ast\!\!\!\_c E \to \mathsf T^\ast\!\!\!\_c F
$$
(Here $H$ is for Hessian or some such thing). Under coordinate changes, $H$ transforms like a tensor because, along $C$, the restriction of $\mathrm{d}f$ to $\mathsf T\!\_c F$ vanishes.


Here is the key fact: The tangent space of $C$ is the kernel of $H$. This is an easy calculation, which can be done in local coordinates.


Finally, we are supposed look at the points of $C$ and we are supposed to consider the restriction of $H$ to $\mathsf T\!\_c F$. We want to consider a point $c$ in $C$, where the kernel of this restriction is zero.


But this condition is equivalent to saying that the intersection of $\mathsf T\!\_c F$ with the kernel of $H$ is zero, i.e.


* At the point $c$, $\mathsf T\!\_c F$ has zero intersection with $\mathsf T\!\_c C$.


Now since $\mathsf T\!\_c F$ is the kernel of $\mathrm{d}p$, this means the the restriction of $\mathrm{d}p$ to $\mathsf T\!\_c C$ also has zero kernel, and hence for dimension reasons is an isomorphism. Hence at the point $c$ in $C$, under your assumptions, the map $f: C \to B$ is a local diffeomorphism.


