Let M,N be real smooth manifolds and p:M-->N a smooth map. Then smooth functions on M form a module over the ring of smooth functions on N (via pullback). Is it know whether this module is flat when p is a submersion?

Recall that flatness is equivalent to the following: whenever h₁ ... h_k are smooth functions on N and g₁ ... g_k are smooth functions on M such that:

h₁g₁ + ... + h_kg_k = 0 (as function on M)

then there are functions G₁ ... G_r on M and a_i,j on N such that:

g_i= Σ_j a_i,jG_j for all i

and Σ_i h_i a_i,j= 0 for all j.

Some remarks:

• This condition of flatness is equivalent to the usual one (see the comments below).

• It is known that the inclusion of an open subset into N is flat since smooth functions on an open U are obtained from the smooth functions on N by localizing w.r.t. functions vanishing nowhere on U.

• It is also known that a smooth flat map has to be open. Proofs of both of these facts can be found for example in the book: Gonzales, Salas, C^\infty differentiable spaces, Lecture notes in Mathematics, Springer 2000.

• The equational condition of flatness I gave above seems to be the most reasonable thing to use trying to come up with a proof. But considering already the simplest situation here's what gets me stuck: suppose you want to check flatness of the standard projection of R^2 to R (by R I mean the reals), and consider the case of just one h(x) and g(x,y) such that hg=0. If you take h(x) to be a smooth function strictly positive for x<0 and 0 for x\geq 0, then the flatness condition translates into:

Any smooth function g(x,y) vanishing the half plane x\leq 0 admits a "factorization":

g(x,y)=Σ_j a_j(x)G_j(x,y)

where the a_j(x) are smooth functions vanishing on x\leq 0. (No restrictions on the G_j except smoothness).

Anyone has an idea how to prove this, or knows how to come up with a counter example?