15
$\begingroup$

In Section 1.7 of Parametrized Homotopy Theory by May and Sigurdsson it is stated that the smash product of pointed topological spaces is not associative (which is just another hint that $\mathrm{Top}$ is the "wrong category"). Specifically, they claim that $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q})$ is not isomorphic to $(\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ in $\mathrm{Top}_*$. But actually they just prove that the canonical bijection $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q}) \to (\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ is not an isomorphism. Equivalently, there is no isomorphism in the slice category $(\mathbb{N} \times \mathbb{Q} \times \mathbb{Q}) / \mathrm{Top}_*$. Therefore, my question is as follows:

How to prove that there is no isomorphism between $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q})$ and $(\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ in $\mathrm{Top}_*$?

I had asked the same question on math.SE.

$\endgroup$
5
  • 3
    $\begingroup$ It seems to me that however the result was stated literally, what they probably meant (and probably what they ought to have said) is precisely what they proved, that the canonical map is not a homeomorphism. $\endgroup$
    – Todd Trimble
    Feb 9, 2015 at 15:13
  • $\begingroup$ Nevertheless it looks like an interesting question, to ask whether they are abstractly homeomorphic. $\endgroup$ Feb 9, 2015 at 15:34
  • $\begingroup$ Puppe (p.336 in Math. Zeitschrift Vol. 69) claims without proof that they are not homeomorphic and refers to a similar example in Bourbaki Top. Gen. Ch.1 §9 Exer.11 (1951 edition). I could not track down this reference though. $\endgroup$
    – johndoe
    Feb 9, 2015 at 15:43
  • 3
    $\begingroup$ @johndoe in the reference mentioned in the question, the authors say that they intend to provide a published account of that claim by Puppe, since they don't know of any other one. $\endgroup$ Feb 9, 2015 at 15:59
  • $\begingroup$ @FernandoMuro Apologies, I didn't read the May-Sigurdsson reference. I found Puppe's ref through tom Dieck's monograph. Did anybody look into that Bourbaki's reference anyway? $\endgroup$
    – johndoe
    Feb 9, 2015 at 16:06

2 Answers 2

9
$\begingroup$

Building on Fernando's answer, here is a proof that they are not homeomorphic. By Fernando's answer, it suffices to show that if a sequence $(x_k,y_k,n_k)$ converges to the basepoint in $\mathbb{Q}\wedge(\mathbb{Q}\wedge\mathbb{N})$, it also converges to the basepoint in $(\mathbb{Q}\wedge\mathbb{Q})\wedge\mathbb{N}$. Let $(x_k,y_k,n_k)$ be such a sequence. We may assume that the points $(x_k,y_k,n_k)$ are all distinct from the basepoint. I claim $\{n_k\}\subseteq\mathbb{N}$ must be a finite set; the result then follows easily.

Suppose for a contradiction that $\{n_k\}$ is infinite. Passing to a subsequence, we may assume that for each $n\in\mathbb{N}$, there are only finitely many $k$ such that $n_k=n$. For definiteness of notation, let us say that $0$ is our chosen basepoint in both $\mathbb{Q}$ and $\mathbb{N}$. For each $n>0$, let $\epsilon_n>0$ be such that $\epsilon_n<|x_k|$ and $\epsilon_n<|y_k|$ for all $k$ such that $n_k=n$. Now let $$U=\{(x,y,n):|x|<\epsilon_n\text{ or }|y|<\epsilon_n\}\subset \mathbb{Q}\wedge(\mathbb{Q}\wedge\mathbb{N}).$$

Then $U$ contains the basepoint and is disjoint from our sequence. Furthermore, the pullback of $U$ to $\mathbb{Q}\times(\mathbb{Q}\wedge\mathbb{N})$ is open: it is the union of the open sets $$V=\mathbb{Q}\times\{(y,n):|y|<\epsilon_n\}$$ and $$W_n=\{x:|x|<\epsilon_n\}\times \{(y,n):y\neq0\},$$ the latter being a separate open set for each fixed $n>0$. Thus $U$ is open, contradicting the assumption that our sequence converges to the basepoint.

$\endgroup$
2
  • $\begingroup$ Thank you. Why does it suffice to look at sequences converging to the base point? What about $(0,y,n)$ for instance? $\endgroup$ Feb 12, 2015 at 10:10
  • $\begingroup$ The complement of the basepoint has the same topology in both spaces, which is just the subspace topology from $\mathbb{Q}\times\mathbb{Q}\times\mathbb{N}$. I'm not sure what you mean by your second question. $\endgroup$ Feb 12, 2015 at 14:39
5
$\begingroup$

This is not a real answer since there is a gap, a claim I'm not currently able to prove, but it is too long for a comment and I hope somebody can fix the gap. Consider the topological property of being sequential. The idea is that $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ is sequential but $\mathbb Q\wedge (\mathbb Q\wedge\mathbb N)$ should not.

First countable spaces, such as $\mathbb Q\times \mathbb Q$, are sequential. Sequential spaces are closed under quotients and disjoint unions, hence $\mathbb Q\wedge \mathbb Q$, $(\mathbb Q\wedge \mathbb Q)\times\mathbb N$ and $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ are sequential.

We can identify the underlying sets of $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ and $\mathbb Q\wedge (\mathbb Q\wedge\mathbb N)$ in the obvious way, so we're really speaking about different topologies on the same set. May and Sigurdsson show that the former has strictly more open sets than the latter. I think both of them have the same convergent sequences, but I haven't been able to prove it (the only problem is when the limit is the base point). If this is true, any open set in $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ would be sequentially open in $\mathbb Q\wedge (\mathbb Q\wedge\mathbb N)$, so the latter cannot be sequential, otherwise the two topologies would be the same.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.