# Smash product of pointed spaces is not associative

I've read many times that the smash product of pointed topological spaces is not associative, for example $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q})$ is not homeomorphic to $(\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$. How to prove this?

In general $\mathbb{N} \wedge X = \bigvee_{n>0} X$, so that $\mathbb{N} \wedge (X \wedge Y) = \bigvee_{n>0} (X \wedge Y)$ and $(\mathbb{N} \wedge X) \wedge Y = (\bigvee_{n>0} X) \wedge Y$. By the universal property of the wedge sum, we get a canonical continuous map $$\mathbb{N} \wedge (X \wedge Y) \to (\mathbb{N} \wedge X) \wedge Y,~(n,(x,y)) \mapsto ((n,x),y).$$ It is clearly bijective. For $X=Y=\mathbb{Q}$ it should not be open - why? Probably one has to take some open subset of $\mathbb{Q} \times \mathbb{Q}$ which approaches the base point $(0,0)$ in a weird way?

Can someone explain/supplement Joriki's proof?

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If $X,Y,\mbox{ and }Z$ are all locally compact Hausdorff then the triple smash products agree. Note that the rationals based at $0$ are not locally compact. – Sanath K. Devalapurkar Feb 28 '14 at 23:44
Have you seen the proof in section 1.5 of math.uiuc.edu/K-theory/0716/MaySig.pdf – Niels Diepeveen Mar 2 '14 at 18:50
@Niels: Wow, that's exactly what I was looking for! When you post this as an answer, I will accept it. – Martin Brandenburg Mar 2 '14 at 19:09

## 3 Answers

There appears to be a detailed proof due to Kathleen Lewis in section 1.5 of Parametrized Homotopy Theory by J.P. May and J. Sigurdsson.

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Notice that they don't prove what they claim: They only prove that the canonical map $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q}) \to (\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ is not a homeomorphism. But they claim that there is no homeomorphism at all. – Martin Brandenburg Mar 2 '14 at 19:38
Good point. I wonder if there is an easy argument that there can be no other homeomorphism. – Niels Diepeveen Mar 2 '14 at 19:42

Look at the following diagram (created using http://Presheaf.com. I didn't know how to do the
\Bbb-letters there)

Since both paths simply identify $(\Bbb N×\Bbb Q×\{0\})\cup(\Bbb N×\{0\}×\Bbb Q)∪(\{0\}×\Bbb Q×\Bbb Q)$ to a point, we have induced set maps $s$ and $\tilde s$, where the later is a bijection. Since the lower left arrow is a quotient map, $\tilde s$ is continuous if and only if $$s:(\Bbb N\wedge\Bbb Q)×\Bbb Q\to \Bbb N∧(\Bbb Q∧\Bbb Q)$$ is continuous. In order to show that it is not, we have to find an open set in $\Bbb N∧(\Bbb Q∧\Bbb Q)$ whose preimage is not open. Since the preimage of $V$ under $s$ is the same as the image of $q^{-1}(V)$ under $l×1$, this amounts to finding an open $q$-saturated set $U$ in $\Bbb N×\Bbb Q×\Bbb Q$ whose image is not open.
The product $\Bbb N×\Bbb Q×\Bbb Q$ is a disjoint union of $\Bbb Q×\Bbb Q$'s, one for each natural number, starting at $0$. A set in $\Bbb N×\Bbb Q×\Bbb Q$ is $q$-saturated if it contains the entire $(\Bbb Q×\Bbb Q)$-copy at $n=0$, and for each $n>0$ it contains the coordinate axes of $\Bbb Q×\Bbb Q$.
Consider the set $U:=(\{0\}×\Bbb Q×\Bbb Q)\cup\bigcup_{n>0} U_n$, where $$U_n=\{(n,x,y)\mid y<\pi/n\text{ or }y>x+\pi/n\}$$ Each $U_n$ is open and contains the axes, so $U$ is saturated and open. Its image $V=(l×1)(U)$ contains $(b,0)$, where $b$ denotes the base point of $\Bbb N∧\Bbb Q$. However, we can show that $(b,0)$ is not interior point of $V$:
For $(b,0)$ to be in $\text{int }V$, there had to exist an open neighborhood $B$ of $b$ and an $ϵ>0$ such that $B×[-ϵ,ϵ]⊆V$. One can think of $\Bbb N∧\Bbb Q$ as the wedge sum $\bigvee_{n>0}(\Bbb Q,0)$, countably many copies of $\Bbb Q,$ with all $0$'s identified. The set $B$ thus contains as a subspace the wedge $\bigvee_{n>0}((-\delta_n,δ_n),0)$, where all $δ_n>0$. Now for $k\in\Bbb N$ such that $\pi/k<ϵ$, the product $(-δ_k,δ_k)×[-ϵ,ϵ]$ is not contained in $U_k$.

This also shows that for a quotient map $q$ the product $q\times 1_Y$ need not be a quotient map if $Y$ is not locally compact. The reason why it fails here is that the neighborhood $K:=[-ϵ,ϵ]$ of $0$ in $\Bbb Q$ is not compact. Otherwise we could consider the set $W$ which is the largest subset of $\Bbb N×\Bbb Q$ such that $W×K⊆U$. This set turns out to be open and saturated. But here the openness would require that we can apply the Tube Lemma to $\{(k,0)\}×[-ϵ,ϵ]⊆U$, which is not possible as the interval isn't compact in $\Bbb Q$.

By the way, a smash product $X∧Y∧Z$ is associative if $X$ and $Z$ are locally compact, but also if two out of the three spaces are compact Hausdorff. The reason is that if $Y,Z$ are compact Hausdorff, then so is $Y∧Z$, which means that $p:Y×Z\to Y∧Z$ is perfect. For perfect $p$ a map $1_X×p$ is closed, hence a quotient map.

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Thank you! This is very similar to the proof by Kathleen Lewis mentioned in the answer by Niels Diepeveen. But your construction of $U_n$ (there $V(\alpha/n)$) is simpler. – Martin Brandenburg Mar 2 '14 at 20:36
I guess it is similar, from what I've seen in that document, though I haven't gone into the details, there. I was a bit disappointed to see that the bounty had already been awarded to the Niels' answer, which he posted when I was making up the proof. But that's the risk when you wait until the last hours of the grace period. – Stefan Hamcke Mar 2 '14 at 20:47
@MartinBrandenburg: I suggest that you accept this answer instead of mine. – Niels Diepeveen Mar 2 '14 at 21:58
@NielsDiepeveen: Thank you for passing on the bounty to me :-) – Stefan Hamcke Mar 2 '14 at 23:16

The preimage in $(\mathbb N\wedge\mathbb Q)\times\mathbb Q$ of an open neighbourhood of the origin of $(\mathbb N\wedge\mathbb Q)\wedge\mathbb Q$ must contain a product of open neighbourhoods of the origins of $\mathbb N\wedge\mathbb Q$ and $\mathbb Q$, so its preimage in $\mathbb N\times\mathbb Q\times\mathbb Q$ must contain a set of the form $\{0\}\times\mathbb Q\times U$, where $U\subseteq\mathbb Q$ is a neighbourhood of $0$. However, $\{(n,x,y)\mid|xy|\lt1\}$ doesn't contain such a set, and its image in $\mathbb N\wedge(\mathbb Q\wedge\mathbb Q)$ is open.

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Thank you. But I don't understand why the set must contain $\{0\} \times \mathbb{Q} \times U$; why not $\{0\} \times V \times U$ for some $V$? Also, I argued that $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q}) \to (\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ is continuous, but your proof seems to produce an open subset of whose preimage is not open. Therefore I am confused. – Martin Brandenburg Jan 9 '13 at 23:32
@Martin: The preimage in $(\mathbb N\wedge\mathbb Q)\times\mathbb Q$ must contain $V\times U$ with $V\subseteq\mathbb N\wedge\mathbb Q$ containing the origin of $\mathbb N\wedge\mathbb Q$, and the preimage of that origin in $\mathbb N\times\mathbb Q$ contains $\{0\}\times\mathbb Q$, so the preimage in $\mathbb N\times\mathbb Q\times\mathbb Q$ of the preimage in $(\mathbb N\wedge\mathbb Q)\times\mathbb Q$ must contain $\{0\}\times\mathbb Q\times U$. Regarding the continuity, you not so much argued as claimed this :-) Why do you believe that map is continuous? – joriki Jan 10 '13 at 0:17
@Martin: Actually I think I didn't really understand what you're doing with that map. Does $\bigvee$ denote the disjoint union? Then I don't think that's equivalent, since in $\mathbb N\wedge\mathbb Q$ all of $\{0\}\times\mathbb Q$ is collapsed in the origin whereas in the disjoint union all copies look the same and are uncollapsed. – joriki Jan 10 '13 at 0:27
If $X$ is locally compact, then $- \times X$ preserves quotient maps, right? And this would then imply that it also preserves wedges. – Martin Brandenburg Jan 16 '13 at 13:25
There is also a result by Puppe (in Math. Zeitschrift, Vol. 69) which says that the smash product is associative for well-pointed spaces (where the inclusion of the base point is a cofibration). – Martin Brandenburg Jan 16 '13 at 13:54