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 <i>M</i> be a smooth manifold and <i>f</i>:<i>M</i>→<b>R</b> a smooth map.  What type of object is the second derivative <i>f</i><sup>(2)</sup>?  In general, you should not talk about it by itself, because it does not transform as a tensor, although the pair (<i>f</i><sup>(1)</sup>,<i>f</i><sup>(2)</sup>) is a vector in the 2-jet bundle over <i>M</i>.  But if <i>c</i> is a critical point of <i>f</i>, then <i>f</i><sup>(2)</sup>(<i>c</i>) is naturally a symmetric bilinear form T<sub>c</sub><i>M</i> x T<sub>c</sub><i>M</i> → <b>R</b>.  Thus it is a map T<sub>c</sub><i>M</i>→T<sup>*</sup><sub>c</sub><i>M</i>.  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.