For completeness, I should post a follow-up to Chris's answer above.  Chris's argument shows that the map _dp_: T<sub>_c_</sub>_C_ &rarr; T<sub>_p(c)_</sub>_B_ is an injection, where _p_ is (the restriction of) the projection _E_ &rarr; _B_.  But the conditions I originally proposed do not force this to be a surjection except in the finite-dimensional case (when a (co)dimension count shows that dim _C_ is at least dim _B_), as the following example shows.

Let _N_ be a compact connected manifold, _B_ = _N_ x _N_, _E_ = Smooth Maps ( [0,1] &rarr; _N_ ), and the projection is &pi;(_e_) = (_e_(0),_e_(1)).  Pick a function _L_ : T<i>N</i> &rarr; **R**, and define _f_ : _E_ &rarr; **R** to by _f_(_e_) = &int;<sub>[0,1]</sub> _L_(_e_'(t),_e_(t)) dt.  Then the set _C_ consists of solutions to the Euler-Lagrange equations d<sub>t</sub>[\partial<sub>v</sub> _L_] = \partial<sub>q</sub> _L_.  Generically, this is a non-degenerate second-order differential equation, and so dim _C_ = 2 dim _N_ = dim _B_.  This is in fact the application I am interested in.

On the other hand, let's pick a one-form _b_ and a function _c_ on _N_, and let's suppose that the exterior derivative d_b_ is nondegenerate, so that d_b_ is a symplectic structure on _N_.  Let _L_(v,q) = _b_v + _c_.  Then one can check (it's straightforward) that the Euler-Lagrange equations are a non-degenerate _first_ order differential equation on _N_, equivalent to the Hamilton equation for the sympectic manifold (_N_,d_b_) with Hamiltonian _c_ (or perhaps -_c_ depending on your conventions).  Thus a solution is determined by its initial location, and so dim _C_ = dim _N_.

On the other hand, the Hessian _H_ defined in Chris's answer is now a nondegenerate first-order linear differential equation, and the condition says that the only solution &phi; to this equation with &phi;(0) = 0 = &phi;(1) is the trivial solution &phi; = 0.  For a general Lagrangian, the Hessian is a second-order operator, and this condition is nontrivial, but when the Lagrangian is first-order, the Hessian necessarily has no kernel &mdash; a solution is determined by a single value.

Thus the condition that Chris thought I was imposing &mdash; that _C_ be a manifold with dimension the same as _B_ &mdash; cannot be dropped.

A final remark is that in finite dimensions, _C_ is cut out by dim _F_ = dim _E_ - dim _B_ equations, and so dim _C_ is at least dim _B_.  The point is that this dimension count fails when dim _F_ = &infin;.