Suspension of an excisive pair - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T16:11:51Z http://mathoverflow.net/feeds/question/84971 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84971/suspension-of-an-excisive-pair Suspension of an excisive pair Fabian Hebestreit 2012-01-05T16:21:08Z 2012-01-07T22:31:07Z <p>Hi, </p> <p>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?</p> <p>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. </p> <p>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:</p> <p>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.</p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/84971/suspension-of-an-excisive-pair/84991#84991 Answer by Tom Goodwillie for Suspension of an excisive pair Tom Goodwillie 2012-01-05T19:12:11Z 2012-01-05T19:28:31Z <p>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 $n+1$.</p> <p>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. </p> <p>To the question in the last paragraph: yes. </p> http://mathoverflow.net/questions/84971/suspension-of-an-excisive-pair/84996#84996 Answer by Karol Szumiło for Suspension of an excisive pair Karol Szumiło 2012-01-05T19:32:44Z 2012-01-06T08:01:38Z <p>The answer to your first question is negative. Before I give a counterexample, let me rephrase the problem in terms I consider more natural.</p> <p>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 $B = X \setminus U$.</p> <p>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.)</p> <p>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), <code>$U = \{-1, 1\}$</code> and <code>$A = X \setminus \{-i, i\}$</code>. 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 $S X$.</p> <p>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.</p> <p>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 <em>Algebraic Topology</em>. 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.</p> http://mathoverflow.net/questions/84971/suspension-of-an-excisive-pair/85159#85159 Answer by Ronnie Brown for Suspension of an excisive pair Ronnie Brown 2012-01-07T22:31:07Z 2012-01-07T22:31:07Z <p>Since this question is about homotopical excision, I refer to my recent answer to</p> <p><a href="http://mathoverflow.net/questions/56435/what-is-the-intuition-behind-the-freudenthal-suspension-theorem/85158#85158" rel="nofollow">http://mathoverflow.net/questions/56435/what-is-the-intuition-behind-the-freudenthal-suspension-theorem/85158#85158</a></p>