Consider a (smooth) bundle *E*→_B_, and a (smooth) function *f*: *E* → **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*→_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*→**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 T_{c}*M* x T_{c}*M* → **R**. Thus it is a map T_{c}*M*→T^{*}_{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.