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A morphism of set-valued functors $\eta: F \to G$ on $\mathcal{C}$ is called smooth if for all epimorphisms $B \to A$, the natural morphism $F(B) \to F(A) \times_{G(A)} G(B)$ is surjective.

Obviusly "smooth => formally" smooth for $\mathcal{C} = \mathrm{Sch}$.

Now my question: Does the converse hold?

My thoughts: 1. Assume the morphism is of finite presentation.

  1. Assume it is in the local standard form: an étale morphism followed by an affine projection

  2. It is clear that an affine projection is smooth in the above sense, so we have reduced the problem to étale morphisms.

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Possible duplicate:… – Francesco Polizzi Sep 5 '11 at 10:42
I think my definition of "smooth" is a priori different from this. – user12832 Sep 5 '11 at 11:08
I don't have the book in front of me, but there is an appendix to Loday's "Cyclic Homology" where various definitions of 'smooth' are compared quite carefully. – Peter Samuelson Sep 5 '11 at 13:08
Your definition of "smooth" is for me the definition of "formally smooth", and "smooth" would be "formally smooth"+"locally of finite presentation". What in fact is your definition of formally smooth? – Keerthi Madapusi Pera Sep 5 '11 at 17:40

1 Answer 1

up vote 4 down vote accepted

In the case where $F\to G$ is representable by an epimorphism $C\to D$ of rings, then your smoothness condition implies that $C\to D$ has a section. (Take $A\to B$ to be the given map $C\to D$.) But it is not hard to find a formally smooth epimorphism that does not admit a section. Any localization will do. For instance, you can take $D$ to be the zero ring and $C$ any ring but the zero ring.

If you visualize the geometry here, it's pretty clear that your smoothness condition is much stronger than anything normally called smoothness.

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Ah, I missed the part where the OP allowed any epimorphism. – Keerthi Madapusi Pera Sep 5 '11 at 23:03

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