Consider a (smooth) bundle E_→_B$E\to B$, and a (smooth) function f: E → R$f: E\to\mathbf{R}$ on the total space. Then it makes sense to talk about the derivatives of f$f$ along the fibers. Let C$C$ be the subspace of E$E$ consisting of all points for which all fiber-wise derivatives of f$f$ vanish, so that upon intersecting with any fiber C$C$ consists of the critical points of the restriction of f$f$ to the fiber. If the fiber is n$n$-dimensional, then C$C$ is carved out by n$n$ equations, and so generically has codimension n$n$ in E$E$.
Let's say that c$c$ is a point in C$C$ so that the second derivative of f$f$ in the fiber is nondenegerate (i.e. has non-degenerate Hessian; i.e. f$f$ restricts to a Morse function on the fiber through c$c$). Does it follow that the projection C_→_B$C\to B$ is a local diffeomorphism near c$c$?
The answer is yes when everything is finite-dimensional (and I believe the statement is iff). I am interested in the case when B$B$ is finite-dimensional but the fibers of E$E$ 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$M$ be a smooth manifold and f:M→R$f:M\to\mathbb{R}$ a smooth map. What type of object is the second derivative f(2)$f^{(2)}$? 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))$(f^{(1)},f^{(2)})$ is a vector in the 2$2$-jet bundle over M$M$. But if c$c$ is a critical point of f$f$, then f(2)(c)$f^{(2)}(c)$ is naturally a symmetric bilinear form TcM x TcM → R$\mathsf T\!_cM \times \mathsf T\!_cM \to\mathbf{R}$. Thus it is a map TcM→T*cM$\mathsf T\!_cM\to \mathsf T^\ast\!\!\!_c M$. 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.