Suspension of an excisive pair - MathOverflow

Suspension of an excisive pair

2012-01-05T16:21:08Z

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.

Answer by Tom Goodwillie for Suspension of an excisive pair

2012-01-05T19:12:11Z

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.

Answer by Karol Szumiło for Suspension of an excisive pair

2012-01-05T19:32:44Z

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.

Answer by Ronnie Brown for Suspension of an excisive pair

2012-01-07T22:31:07Z

Since this question is about homotopical excision, I refer to my recent answer to

http://mathoverflow.net/questions/56435/what-is-the-intuition-behind-the-freudenthal-suspension-theorem/85158#85158