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This question is motivated by

Does excision imply that these inclusions are isomorphisms in homology?

There are two variants of the excision axiom for a homology theory:

(E) Let $U, A \subset X$. If $\overline{U} \subset int(A)$, then the inclusion map $i : (X \backslash U, A \backslash U) \to (X,A)$ induces isomorphisms $i_\ast : H_n(X \backslash U, A \backslash U) \to H_n(X,A)$ for each $n$.

(E-open) As above, but restricted to open $U$.

When axiomatic homology theory was introduced by Eilenberg and Steenrod, they used axiom (E-open). However, they also showed that singular homology theory satisfies (E).

In the literature (E) prevails, but it is certainly not a universally accepted standard.

It seems that (E) is strictly stronger than (E-open). To verify this we would need an example of a homology theory satisfying (E-open) but not (E), but I could not find any.

Does anybody know such an example?

Edit:

Eilenberg and Steenrod define their axioms for homology theories living on an admissible category $\mathfrak{A}$ of pairs of spaces and maps of such pairs. This means that $\mathfrak{A}$ contains a single point space and is closed with respect to a few elementary operations. Some well-known theories live on the category $\mathfrak{T}^2$ of all pairs of topological spaces (for example singular homology), other theories are only defined on smaller categories.

The answer to my question depends on $\mathfrak{A}$. For example, if $\mathfrak{A} = \mathfrak{C}^2$ = category of all compact pairs, then (E) and (E-open) are equivalent simply because only an open $U$ gives us a compact pair $(X \backslash U, A \backslash U)$.

This means that the desired example must live on an admissible $\mathfrak{A}$ having the following property ($\ast$):

Call a triple $(X,A,U)$ of spaces excision enabled if both $(X,A)$ and $(X \backslash U, A \backslash U)$ belong to $\mathfrak{A}$ and $\overline{U} \subset int(A)$.

($\ast$) The class of excision enabled triples is strictly larger than class of excision enabled triples with an open $U$.

This is certainly true for $\mathfrak{T}^2$.

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  • $\begingroup$ Interesting. What was their proof that the open version implies the stronger? Does it work for any theory? $\endgroup$ – Randall Aug 24 '18 at 11:53
  • $\begingroup$ @Randall Eilenberg and Steenrod proved directly from the definition of the singular chain complex that the stronger variant is satisfied. They did not show that the open version implies the stronger, that is, their proof does not generalize to other theories. $\endgroup$ – Paul Frost Aug 24 '18 at 12:24

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