Talk:Serre–Swan theorem

I would be more comfortable with the statement if M were assumed to be connected. Is that necessary? AxelBoldt 14:13, 4 Dec 2003 (UTC)

Oh, we're in a compact situation, so we have at most finitely many connected components, and I don't think we need connectedness after all. AxelBoldt 14:54, 4 Dec 2003 (UTC)

Compactness is sufficient. Phys 14:58, 4 Dec 2003 (UTC)

I assume there is also a version of the theorem in algebraic geometry? AxelBoldt 23:04, 11 Dec 2003 (UTC)

I've always wondered, is compactness really the most natural hypothesis for this important result? The usual proof seems to work without a hitch on any smooth manifold that may be a covered by a finite collection of contractible sets. And that class of smooth manifolds is very large indeed: it contains every manifold with finitely many connected components. (You need to add paracompactness as an assumption, if that isn't already part of your definition of what a manifold is.) PerVognsen (talk)

Indeed in the smooth case compactness is not necessary (a proof without this condition is found in the book: Jet Nestruev, Smooth manifolds and observables, Springer graduate text) 1 Sep 2008 —Preceding unsigned comment added by Matterink (talk • contribs) 11:44, 1 September 2008 (UTC)[reply]

An interesting book. Thanks for suggesting it. siℓℓy rabbit (talk) 13:18, 1 September 2008 (UTC)[reply]

Merge to Serre-Swan theorem[edit]

I propose merging to Serre-Swan theorem. silly rabbit (talk) 21:34, 5 May 2008 (UTC)[reply]