# Associativity of join for CW-complexes

I am currently self-studying a course in algebraic topology and one of the problems I encountered is to prove that the join operation defined as $$X \ast Y=X\times Y\times I/(x_1,y,1)\sim(x_2,y,1), (x,y_1,0)\sim(x,y_2,0)$$ is associative for CW-complexes.

I know that there is a natural bijection between $X\ast Y\ast Z$ as a quotient of $X\times Y\times Z \times \Delta$ and $(X\ast Y)\ast Z$, but I don't understand how to prove that it's a homeomorphism in the case of CW-complexes.

I found one related proposition in Fomenko's "Homotopical Topology" saying that the join operation is associative for compact spaces. However, it was left as an exercise for the reader. I think I understand how to prove something of the sort for compact Hausdorff spaces, but I am at a loss how compactness could be sufficient. And even then I have no idea how this could be generalized to CW complexes because the weak topology on a join of CW-complexes does not always coincide with the topology of the join.

Update: In the exercise mentioned in the comments it is recommended to consider a new join operation $$X\ \hat{\ast}\ Y=\{(x,\xi,y,\eta)\in CX \times CY\ |\ \xi+\eta=1\ \}$$ and show that for "good" spaces the two operations are equivalent. The problem is that my reasoning leads to the two operations being equivalent in all cases and I can't find the error. I suppose I don't understand quotient maps that well.

Consider $f: X\times Y\times I\times I\to X \times Y\times I, f(x,y,\xi,\eta)=(x,y,\xi(1-\eta))$. $f$ is continuous, as it is basically a map from $I\times I \to I$. Moreover, its restriction to the subset $A$ where $\xi+\eta=1$ is a homeomorphism. Now note that it behaves well with respect to quotients: if weconsider two equivalence relations in the join $(x_1,y,1)\sim(x_2,y,1)$ and in $CX\times CY$ $(x_1,y,1,\eta)\sim(x_2,y,1,\eta),$ then the preimages of equivalent points are equivalent and vice-versa. Hence, $f$ induces a homeomorphism between $A/(x_1,y,1,\eta)\sim(x_2,y,1,\eta)$ and $X\times Y\times I/(x_1,y,1)\sim(x_2,y,1)$. Proceeding similarly for the other pair of equivalence relations we obtain a homeomorphism betweem $X\ast Y$ and $X\ \hat{\ast}\ Y$.

This argument is clearly wrong. Right now I see two possible issues:

1) Is it true that if $f$ is a homeomorphism between $X$ and $Y$, and $A\subset X$, then $X/A \simeq Y/f(A)$? I think it is.

2) Something breaks down due to the order in which I go from $f: X\times Y\times I\times I$ to its subset, take the quotients and products. But again, I can't pinpoint the point at which these operations do not commute.

Any help would be greatly appreciated.

One other thing I thought of: is the original statement about the join operation being associative for CW-complexes even true? Maybe I misunderstood the problem and the aim is not to show that $(X\ast Y)\ast Z$ and $X\ast(Y\ast Z)$ are homeomorphic, but that they are homeomorphic as CW-complexes with the join structure?

Update 2: Thanks to another helpful comment by Steve D. I realised that the above argument (if written down more rigorously) works in the case when $X\times Y$ is Hausdorff and locally compact because of Whitehead's thorem.

Now it is true that if $X$ and $Y$ are CW-complexes and at least one of them is locally compact, then their join $X\times Y$ is also a CW-complex. Overall, for locally compact (equivalently, locally finite) CW-complexes there is no ambiguity in the question and it is indeed true that the join operation is associative. However, the case of general CW-complexes still remains out of reach for me.

• In my copy of Homotopical Topology, there is an exercise that lays out good detail about how to prove this when the spaces are locally compact and Hausdorff. It's exercise 11 in lecture 2. – Steve D Aug 8 '18 at 13:26
• @ Steve D. Thank you! This exercise is absent from the Russian version of the book. I will try solving it today. – Serg Aug 8 '18 at 15:06
• Regarding your update: it is not in general true that $id\times q$ is a quotient map when $q$ is. It is true when everything is locally compact Hausdorff. – Steve D Aug 8 '18 at 16:55