Talk:Relative homology

Could someone add relative cohomology? Charles Matthews 21:32, 19 October 2005 (UTC)[reply]

What is the definition of a morphism of pairs of topological spaces (as implied by the reference to a "category" of such pairs)? - 74.112.133.181 01:29, 12 May 2007 (UTC)[reply]

If (X,A) and (X',A') are both pairs, then a morphism between them is a continuous map which maps A to (a subset of) A'. - Saibod 18:33, 14 May 2007 (UTC)[reply]

Is it always true that or only when ? 69.234.22.18 21:09, 19 October 2007 (UTC)[reply]

only when n ≠ 0. the equality is true for all n if the right hand side is replaced by reduced homology. Mct mht (talk) 10:56, 14 December 2007 (UTC)[reply]
This should be made clear in the article. I'll edit it to reflect that. Hurkyl (talk) 04:08, 21 February 2008 (UTC)[reply]

The literature often requires (X,A) to be a good pair. I.e., A is closed and is a deformation retract of a neighborhood in X. Shouldn't this be mentioned, I wonder?69.202.65.222 (talk) 23:10, 14 September 2008 (UTC)[reply]

IIRC, the relative homology is a functor between Top2 and Ab. I think that this is more important than the functoriality of and that it should be added in the proper section. --SMP - talk (en) - talk (ca) 11:50, 21 June 2009 (UTC)[reply]

Relative homology of a triple

The article says that "Relative homology readily extends to the triple (X, Y, Z) for ZYX", but I have never heard of relative homology of a triple and no references are given. What is this? There is a long exact sequence of a triple, but this is worded strangely if that's what is being referred to. (Over on the [Lefschetz duality] page, there appears to be a use of this.) --Dylan Thurston (talk) 10:13, 16 October 2015 (UTC)[reply]

Local Homology

The definition for local homology should be the one in page 8 of https://www3.nd.edu/~lnicolae/MS.pdf . In addition, there should be inclusions of computations for

  • a manifold minus a point — Preceding unsigned comment added by 161.98.8.4 (talk) 19:45, 9 August 2017 (UTC)[reply]
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