Talk:Bredon cohomology

This article was nominated for deletion. Please review the prior discussions if you are considering re-nomination:

Merge, 16 February 2018, see discussion.
Promoted to article, 1 March 2019, see discussion.

notability, current shape[edit]

I came here from a notification at WT:WPM -- there is no doubt the topic is notable (an easy mathscinet search will show lots of publications on the topic), but the current shape of the article is crying for improvement! Jakob.scholbach (talk) 20:41, 1 March 2019 (UTC)[reply]

Well, this page was a draft page but was somehow prematurely moved to mainspace and that explains a somehow poor quality. —- Taku (talk) 18:56, 2 March 2019 (UTC)[reply]

"Somehow" moved by Taku [1] Legacypac (talk) 19:02, 2 March 2019 (UTC)[reply]

Well, you wanted to force the immediate action (by threatening the deletion) and that’s what you got; I was merely explaining that the quality got suffered as a result. The quality works take time because it requires looking for good refs, etc. —- Taku (talk) 19:08, 2 March 2019 (UTC)[reply]

Material moved from the article text[edit]

The following was moved from the article text. Perhaps an expert can make sense of it (I am no expert and can’t make sense of it)

If one does the contour integral, carefully following a line that goes up one leg of the problem climbing towards the geometree, or bush. An edging solution to the anomaly, using functions and pullbacks [[2]] can be found.

For a topological space, XXX, the definition of singular cohomology should be expanded to a set of at least two.

functors h_xxx (for integers i) from the category of U&I-pairs (XXX, U'→ I) (so XXX is a U&I complex and U is a subcomplex) to the category of non abelian groups, U&I paired is symmetric when ensconced and has a commutator, together with a natural transformation $\partial$ xxx: hxxx(U, I) → hu&i'(U) called the boundary homomoerotirphism the solution to this problem is the union of U&I.

—- Taku (talk) 08:00, 14 December 2022 (UTC)[reply]