Given pointed topological spaces $X$ and $Y$, their smash product is the space
$$ X \land Y = \frac{X \times Y}{X \times \{ y_0\} \cup \{x_0\} \times Y}, $$
where $x_0$ and $y_0$ are the distinguished points of $X$ and $Y$ respectively.
If $X$ and $Y$ are general topological spaces this product isn't associative; but if they are compactly generated (i.e a set is closed iff its intersection with every compact set is closed) then the smash product is associative. I've tried searching for a reference of this but I haven't found proof anywhere, only a counterexample for the general case. Can someone help me with a proof or at least give me a reference where the proof is done?
Thanks in advace.
Best Answer
The question is not entirely clear to me as to whether you want proof of associativity in the appropriate circumstances, or a counterexample.
To deal with the former, the problem is that in general the product of identification maps is not an identification map. An example for this failure is in section 4.3 of Topology and Groupoids (T&G) (as it was in the 1968, differently titled, edition of that book). It involves $$f \times 1: \mathbb Q \times \mathbb Q \to (\mathbb Q / \mathbb Z) \times \mathbb Q $$ where $f$ shrinks the closed subspace of integers in the rationals $ \mathbb Q$ to a single point. The problem is that $ \mathbb Q$ is not locally compact.
A solution proposed in the 1964 paper Function spaces and product topologies (pdf) was to change the category one is working in to a more convenient one, in which the product of identification maps is an identification map. This will be satisfied if the category is cartesian closed, i.e. has an exponential law giving a homeomorphism $$X^{Z \times Y} \cong (X^Y)^Z $$ for all spaces in the category, where $X^Y$ denotes the set of maps in this category with an appropriate topology. This situation is discussed in Section 5.9 of T&G. To get this law for compactly generated spaces you have also to make sure you have the right product and the right function space topology.
The theme is thus that you look for a category of spaces and maps which is adequate and convenient for your purposes, rather than insist that a particular definition of "space" is sacrosanct.
This gives some background to your question but I have not tried to work out a counterexample for the associativity of the smash product of based spaces in the usual topological category. It seems possible that you could get a counterexample using $\mathbb Q$ with base point $0$.