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Let us consider two pair of spaces $(X, A)$ and $(Y, B)$. We set $(X, A) \wedge (Y, B) := (X \times Y, (X \times B) \cup (A \times Y))$. Given a cohomology theory $h^{\bullet}$, we can define a product $h^{\bullet}(X, A) \times h^{\bullet}(Y, B) \rightarrow h^{\bullet}((X, A) \wedge (Y, B))$. In the case of singular cohomology, this can be done via a suitable construction using cochains. In general, if $(X, A)$ and $(Y, B)$ are CW-pairs, we have that $h^{\bullet}(X, A) \simeq \tilde{h}^{\bullet}(X/A)$, $h^{\bullet}(Y, B) \simeq \tilde{h}^{\bullet}(Y/B)$ and $h^{\bullet}((X, A) \wedge (Y, B)) \simeq \tilde{h}^{\bullet}(X \times Y /(X \times B) \cup (A \times Y)) \simeq \tilde{h}^{\bullet}(X/A \wedge Y/B)$, hence we easily get the product.

Let us consider the relative cones $C(X, A)$ and $C(Y, B)$. We have that $h^{\bullet}(X, A) \simeq \tilde{h}^{\bullet}(C(X, A))$ and $h^{\bullet}(Y, B) \simeq \tilde{h}^{\bullet}(C(Y, B))$. We get a natural product with value in $C(X, A) \wedge C(Y, B)$, but the exterior product, previously defined, takes value in $C((X, A) \wedge (Y, B)) = (C(X \times Y, (X \times B) \cup (A \times Y))$.

We do not manage to find a natural map from $C((X, A) \wedge (Y, B))$ to $C(X, A) \wedge C(Y, B)$, in order to define the product via the pull-back. If $(X, A)$ and $(Y, B)$ are CW-pairs, then $C(X, A)$ has the same homotopy type of $C(X, A)/CA \simeq X/A$, and the same holds for the other pairs, hence $C((X, A) \wedge (Y, B))$ and $C(X, A) \wedge C(Y, B)$ have the same homotopy type. Does this hold for generic pairs? If the answer is negative, is there a natural way to define the product via the cones?

The motivation of the question is the following. We want to define the relative product for two generic maps $\rho: A \rightarrow X$ and $\eta: B \rightarrow Y$, not necessarily embeddings. We have that $h^{\bullet}(\rho) := \tilde{h}^{\bullet}(C(\rho))$ and $h^{\bullet}(\eta) := \tilde{h}^{\bullet}(C(\eta))$, hence we get a product with value in $\tilde{h}^{\bullet}(C(\rho) \wedge C(\eta))$. We define the map $\rho \wedge \eta: (X \times B) \cup_{A \times B} (A \times Y) \rightarrow X \times Y$ by $(x, b) \mapsto (x, \eta(b))$ and $(a, y) \mapsto (\rho(a),y)$, and we need to define a product with values in $C(\rho \wedge \eta)$. The question is: do $C(\rho) \wedge C(\eta)$ and $C(\rho \wedge \eta)$ have the same homotopy type? If not, how can we define the product?

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  • $\begingroup$ What is the domain of $\rho\wedge\eta$? Is it simply the pushout of $X\times B\leftarrow X\times Y\rightarrow A\times Y$? If so, then it is not what you want in general. $\endgroup$
    – Tom Goodwillie
    Commented Dec 3, 2015 at 0:34
  • $\begingroup$ Tom: there are no maps $X\times Y \to X \times B$, $X\times Y \to A\times Y$. I think the $X\times Y$ should be $A\times B$. $\endgroup$
    – John Klein
    Commented Dec 3, 2015 at 0:42
  • $\begingroup$ The domain is $(X \times B) \sqcup_{X \times Y} (A \times Y)$, I corrected the text of the question. This means that $(x, b) \sim (a, y)$ if and only if $(x, \eta(b)) = (\rho(a), y)$. When $\rho$ and $\eta$ are embeddings, it coincides with $(X \times B) \cup (A \times Y)$. $\endgroup$
    – Fabio
    Commented Dec 3, 2015 at 4:01
  • $\begingroup$ @Fabio : I was confused by your notation, which doesn't seem to be standard. I guess the standard way to write it is this: let $P$ be the pullback of the maps $X\times B \to X \times Y$ and $A\times Y \to X\times Y$. Then your space is the pushout of the diagram $X\times B \leftarrow P \to A\times Y$. Right? $\endgroup$
    – John Klein
    Commented Dec 4, 2015 at 2:35
  • $\begingroup$ I think the domain of $\rho \wedge \eta$ is the push-out of $A \times Y \leftarrow A \times B \rightarrow X \times B$. We consider the disjoint union $(A \times Y) \sqcup (X \times B)$ and identify $(\rho(a), b) \sim (a, \eta(b))$. The map $\rho \wedge \eta$, with codomain $X \times Y$, is $(a, y) \mapsto (\rho(a), y)$ and $(x, b) \mapsto (x, \eta(b))$. I would like to show that $C(\rho \wedge \eta)$ has the same homotopy type of $C(\rho) \wedge C(\eta)$. $\endgroup$
    – Fabio
    Commented Dec 4, 2015 at 15:28

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