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?

  • $\begingroup$ 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 $(X \times Y, A \times Y \cup X \times B)$? $\endgroup$ Feb 20 at 17:58
  • $\begingroup$ @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/… $\endgroup$
    – Nico
    Feb 20 at 18:03
  • $\begingroup$ @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! :) $\endgroup$
    – Nico
    Feb 20 at 18:04
  • $\begingroup$ Ah I see, that's not notation I was familiar with :) $\endgroup$ Feb 20 at 18:08

1 Answer 1


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.$\blacksquare$


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