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Hi,

given a triple of spaces $(X,A,U)$, that is excisive with respect to some homology theory $H$, is the triple $(SX,SA,SU)$ again excisive?

Here SY means unreduced suspension of Y, and there's an obvious identfication in making $(SX,SA,SU)$ a triple. By being excisive I mean that the inclusion gives an isomorphism $H(X-U,A-U) \rightarrow H(X,A)$. The axioms guarantee this if the closure of $U$ is contained in the interior of $A$, but this property is certainly lost at the poles upon passage to suspensions.

The case of interest to me is homotopy groups (with appropriate connectivity assumptions and basepoints) under this assumption on $U$ and $A$. Here I have a second, somewhat related, question:

In tom Dieck-Kamps-Puppe's Homotopietheorie the statement of the homotopy excision theorem assumes $U$ to be closed and $A$ to be open, and I'm wondering whether the theorem also holds under the above slightly weaker condition.

Thanks in advance.

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3 Answers

up vote 4 down vote accepted

No. Suppose that $X$ is an $n$-sphere, $A$ is a closed hemisphere, and $U$ is a point in the interior of $A$. Then $SX$ is an $(n+1)$-sphere, $SA$ is a closed hemisphere, and $SU$ is a closed arc in $SA$ with endpoints in the boundary. $SX-SU$ is contractible and $SA-SU$ is homotopy-equivalent to an $(n-1)$-sphere, so $H(SX-SU,SA-SU)$ is in dimension $n$ whereas $H(SX,SA)$ is in dimension $n+1$.

EDIT: But maybe the following is what you wanted, or should have wanted. If $X=A\cup B$ and $C=A\cap B$ then we sometimes call the triad $(X;A,B)$ excisive if $H(B,C)\to H(X,A)$ is an isomorphism. As long as $S(A\cup B)=SA\cup SB$ and $S(A\cap B)=SA\cap SB$, the triad $(SX;SA,SB)$ will inherit the excision property from the original triad.

To the question in the last paragraph: yes.

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My answer is essentially the same as yours. (I didn't try to "improve" it, I'm just a slow typer.) –  Karol Szumiło Jan 5 '12 at 19:41
    
Off to embarrass myself some more: Of course you are absolutely correct in that I should have asked the second question (I used the forbidden distributive law S(X−U)=SX−SU when formulating the question here.) I accepted your answer; it nicely answers the question I posed. However I'm still at loss with the correctly formulated question... The 'interiors cover' property is again lost at the poles. I can clearly see the statement for example in the case of a CW-pair, but in general? –  old account Jan 10 '12 at 12:41
    
Does that warrant an entirely new thread? Or should I simply edit my question above? –  old account Jan 10 '12 at 12:41
    
As Karol Szumilo suggested in his answer, the "right" question is about square diagrams rather than triads. As further evidence for this, I point out that the suspension of the intersection is not actually the same as the intersection of the suspensions. In fact, it is not clear what one would mean by intersection of the suspensions of two subspaces, since the suspension of a subspace is not always a subspace of the suspension. The suspension of an open interval is not metrizable. –  Tom Goodwillie Jan 10 '12 at 14:02
    
And if you switch over to considering square diagrams then it's easy to see that when one such square is excisive (i.e. induces isomorphisms between homology of opposite sides) then the square of suspensions is excisive as well. –  Tom Goodwillie Jan 10 '12 at 14:05
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The answer to your first question is negative. Before I give a counterexample, let me rephrase the problem in terms I consider more natural.

First, I believe it is more convenient to consider excisive triads instead of excisive triples, i.e. I will replace a triple $(X, A, U)$ by a triad $(X; A, B)$ where $B = X \setminus U$.

Second, the excision (either in homotopy or homology) is not really a statement about excisive triples or triads, but about homotopy pushouts. Excisive triad is just a model for homotopy pushout with some specific point-set properties, which make topological arguments possible. By this I mean that $X$ is a homotopy pushout of $A$ and $B$ along $A \cap B$. (I don't think it is literally true that every excisive triad is a homotopy pushout, but those that aren't should be considered pathological anyway. However, every homotopy pushout is homotopy equivalent to an excisive triad.)

Thus your question could be rephrased as follows: given an excisive triple $(X, A, U)$ is the triad $(S X; S A, S X \setminus S U)$ excisive or at least a homotopy pushout? As you observed this triad is not excisive, which doesn't really tell us much since it still could be a homotopy pushout. However, this also doesn't have to be true. Let $X = S^1$ (as a subspace of $\mathbb{C}$ to fix the notation), $U = \{-1, 1\}$ and $A = X \setminus \{-i, i\}$. You can write down the suspended triad and observe that the homotopy pushout of $S A$ and $S X \setminus S U$ along $S A \setminus S U$ has the homotopy type of the wedge of three circles, so it cannot be $S X$.

On the other hand, it is easy to see that given an excisive triad $(X; A, B)$, the triad $(S X; S A, S B)$ is again excisive, which seems like a more natural thing to expect.

To answer your second question, I don't know the book you mention, but I assume that the proof of the Homotopy Excision Theorem is more or less the same as in tom Dieck's Algebraic Topology. In this proof the only moment when the point-set properties of $A$ and $B$ are used is when we map a cube into $X$ and use the Lebesgue Lemma to subdivide it into cubes mapping into $A$ or $B$. To do this we only need to assume that interiors of $A$ and $B$ cover $X$. This is equivalent to saying that closure of $U$ is contained in the interior of $A$ in the corresponding triple.

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You said it better than I did! –  Tom Goodwillie Jan 5 '12 at 20:38
    
I made a correction. My counterexample is not a homotopy pushout, but not for the reason I originally stated. –  Karol Szumiło Jan 6 '12 at 8:02
    
@Karol Szumiło, you said: "I don't think it is literally true that every excisive triad is a homotopy pushout, but those that aren't should be considered pathological anyway." You were talking about homotopy pushout in the Quillen Model Structure, or in the Strom Structure? (Or both?) –  Fernando Apr 7 '13 at 14:13
    
@Fernando: I don't remember what exactly I meant at the time of writing that answer and I was somewhat confused then. By now I have clarified it. All excisive triads are homotopy pushouts with respect to weak homotopy equivalences (see tom Dieck Algebraic Topology Theorem 6.7.9) and all numerable excisive triads are homotopy pushouts with respect to genuine homotopy equivalences (loc. cit. Proposition 4.2.3) –  Karol Szumiło Apr 20 '13 at 12:42
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Since this question is about homotopical excision, I refer to my recent answer to

What is the intuition behind the Freudenthal suspension theorem

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