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?