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.



Suspension of an excisive pair
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. 


