# excisive couples, sufficient conditions

Question: What are sufficient conditions on pairs $$(X,A)$$ and $$(Y,B)$$ so that $$(A\times Y,X\times B)$$ is an excisive couple?

In more detail: Given pairs of spaces $$(X,A)$$ and $$(Y,B)$$, consider the inclusion $$\Delta(A\times Y) + \Delta(X \times B) \to \Delta(A\times Y \cup X\times B)$$ Here $$\Delta(A\times Y) + \Delta(X \times B)$$ denotes the subcomplex of the singular chain complex $$\Delta(A\times Y \cup X\times B)$$ generated by those singular simplices whose image is completely contained in one of $$A\times Y$$ or $$X\times B$$.

Question: Under which circumstances is this map a weak equivalence (i.e. induces isomorphisms in homology)?

It is clear to me that $$A=B$$, one of $$A,B$$ is empty and $$A,B$$ are both open are sufficient. I am learning from lecture notes which claim that it also follows "feom excision results" in case the inclusions are closed cofibrations, but I have no idea how to prove this. I'd love to get a reference. Also what happens when $$(X,A)$$ and $$(Y,B)$$ are NDR pairs?

• I don't think $(A \times Y, X \times B)$ is what you want to consider (for one, it is not generally a couple on the nose), do you perhaps mean $(X \times Y, A \times Y \cup X \times B)$? Feb 20 at 17:58
• @BenSteffan I am working with the definition of an excisive couple from Spanier, Algebraic Topology, page 188. It is the same definition as the one used in this question here: math.stackexchange.com/questions/2681767/…
– Nico
Feb 20 at 18:03
• @BenSteffan I believe the chain map above is the one that needs to be a weak equivalence if one wants to define relative products, so I believe it is what I want to consider! :)
– Nico
Feb 20 at 18:04
• Ah I see, that's not notation I was familiar with :) Feb 20 at 18:08

Here are some observations (no claims of exhaustiveness):

• The map is an isomorphism if and only if $$A$$ is a union of path-components of $$X$$ or $$B$$ is a union of path-components of $$Y$$. (E.g. if $$A=\emptyset$$ or $$B=\emptyset$$.)

• The map is a chain-homotopy equivalence if and only if it is a quasi-isomorphism (you call it weak equivalence) since both complexes are free.

• The excision theorem tells us that a sufficient condition is that $$X\times B\cup A\times Y$$ is the union of the interiors of $$X\times B$$ and $$A\times Y$$ respectively in $$X\times B\cup A\times Y$$. This is the case if $$A\subseteq X$$ and $$B\subseteq Y$$ are open (I don't see any other practical sufficient condition).

• The map is a quasi-isomorphism if the inclusions $$A\subseteq X$$ and $$B\subseteq Y$$ are closed and one of them is a cofibration. Firstly, this implies that $$X\times B$$ and $$A\times Y$$ are closed in $$X\times Y$$, hence closed in $$X\times B\cup A\times Y$$. Thus, we have a pushout diagram $$\require{AMScd} \begin{CD} A\times B @>>> X\times B\\ @VVV @VVV\\ A\times Y @>>> X\times B\cup A\times Y \end{CD}.$$ Now, closed cofibrations are compatible with products so the claim is implied by the following

Lemma: Let $$X$$ be a top. space and $$A,B\subseteq X$$ closed subspaces s.t. $$X=A\cup B$$. If $$(A,A\cap B)$$ is cofibered, the inclusion $$\Delta(A)+\Delta(B)\le\Delta(X)$$ is a quasi-isomorphism, i.e. $$(X;A,B)$$ is excisive.

Proof: The closedness implies that $$X$$ is the pushout of $$A$$ and $$B$$ along $$A\cap B$$. Let $$Z$$ be the corresponding homotopy pushout, i.e. $$Z=A\cup_{A\cap B}(A\cap B)\times I\cup_{A\cap B}B$$, and $$p\colon Z\rightarrow X$$ the natural projection map. Now, $$Z$$ is the union of the open subsets $$U=Z\setminus B$$ and $$V=Z\setminus A$$ and $$p$$ restricts to homotopy equivalences $$p_A\colon U\rightarrow A$$, $$p_B\colon V\rightarrow B$$ and $$p_{A\cap B}\colon U\cap V\rightarrow A\cap B$$. Furthermore, the condition that $$(A,A\cap B)$$ is cofibered implies that $$p$$ is a homotopy equivalence. Consider the diagram of short exact sequences $$\begin{CD} 0 @>>> \Delta(U\cap V) @>>> \Delta(U)\oplus\Delta(V) @>>> \Delta(U)+\Delta(V) @>>> 0\\ @. @VV{p_{A\cap B,\ast}}V @VV{p_{A,\ast}\oplus p_{B,\ast}}V @VV{p_{A,\ast}+p_{B,\ast}}V @.\\ 0 @>>> \Delta(A\cap B) @>>> \Delta(A)\oplus\Delta(B) @>>> \Delta(A)+\Delta(B) @>>> 0 \end{CD}.$$ The natural long exact sequence in homology implies, since $$p_{A,\ast}$$, $$p_{B,\ast}$$ and $$p_{A\cap B,\ast}$$ are quasi-isomorphisms, that $$p_{A,\ast}+p_{B,\ast}\colon\Delta(U)+\Delta(V)\rightarrow\Delta(A)+\Delta(B)$$ is a quasi-isomorphism. Now, consider the diagram $$\begin{CD} \Delta(U)+\Delta(V) @>>> \Delta(Z)\\ @VV{p_{A,\ast}+p_{B,\ast}}V @V{p_{\ast}}VV\\ \Delta(A)+\Delta(B) @>>> \Delta(X) \end{CD}.$$ The left vertical map is a quasi-isomorphism as demonstrated. The right vertical map is a quasi-isomorphism since $$p$$ is a homotopy-equivalence. The top horizontal map is a quasi-isomorphism by the excision theorem. Thus, the bottom horizontal map is a quasi-isomorphism.$$\blacksquare$$

• The map is a quasi-isomorphism if $$A\subseteq X$$ and $$B\subseteq Y$$ are closed and one of the pairs $$(X,A)$$ and $$(Y,B)$$ is good. In the same way as the previous bullet point, this reduces to the lemma (which is precisely the problem statement of the linked question)

Lemma: Let $$X$$ be a top. space and $$A,B\subseteq X$$ closed subspaces s.t. $$X=A\cup B$$. If $$(B,A\cap B)$$ is a good pair, the inclusion $$\Delta(A)+\Delta(B)\le\Delta(X)$$ is a quasi-isomorphism, i.e. $$(X;A,B)$$ is excisive.

Proof: The assumption implies that there is an open $$V\supseteq A\cap B$$ in $$B$$ s.t. $$V$$ deformation retracts to $$A\cap B$$. There is an open $$U$$ in $$X$$ s.t. $$V=B\cap U$$. Note that $$A\cup V=A\cup U$$ and $$B\setminus A\cap B$$ are open in $$X$$. The excision theorem thus implies that $$\Delta(A\cup V)+\Delta(B)\rightarrow\Delta(X)$$ is a quasi-isomorphism. A deformation retraction of $$V$$ to $$A\cap B$$ induces a deformation retraction of $$A\cup V$$ to $$A$$ that is constant on $$A$$. This implies (comparing long exact sequences as in the previous lemma) that $$\Delta(A)+\Delta(B)\rightarrow\Delta(A\cup V)+\Delta(B)$$ is a quasi-isomorphism. Thus, the composite $$\Delta(A)+\Delta(B)\rightarrow\Delta(X)$$ is a quasi-isomorphism.$$\blacksquare$$