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.