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Recently I've been reading "Topology" by Klaus Janich. I find this book very entertaining as it contains lots of graphical illustrations that appeal to my "geometrical" imagination. In paragraph 3.6 Janich gives nice illustrations of concepts such as "cone over a set" or "suspension". In the same paragraph he defines a smash product of two topological spaces. Yet, in the version of a book that I posses, there is no image that would present this concept. My question is therefore as in the title:

Is there a way to graphically imagine smash product of two topological spaces?

I'm not sure whether this question is suitable for MO. Perhaps I should put it into a comunity-wiki mode?

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Is there a pure geometric proof of $S^n \wedge S^m = S^{n+m}$? – Martin Brandenburg Jan 29 '11 at 9:59
What do you mean by a pure geometric proof? The obvious geometric intuition seems to work -- imagine $S^n$ as the one-point compactification of $\mathbb{R}^n$ (with the point at infinity as the base-point), so that when you smash two of these gadgets together, you get your larger Euclidean space $\mathbb{R}^{n+m}$ with a lot of stuff at $\infty$ that all gets contracted together. Of course it is hard to visualise except for $n + m \le 3$, but the idea is clear from those cases. – Arnav Tripathy Jan 29 '11 at 15:34
up vote 13 down vote accepted

Here's a picture of a smash product that I drew for this talk, as far as I can tell it's what "Qfwfq" is describing in the middle paragraph.

smash product of two intervals

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Great! That's exactly what I imagined when I was reading unknowngoogle's post. If you now glue appropriate points you'll eventually get $\mathbb{S}^2$ as latter comments argue. – Michal Oszmaniec Jan 29 '11 at 21:34
Well, yes, i meant a picture like that. – Qfwfq Jan 29 '11 at 23:29
I'd like to see the talk, but I'm getting a "document not found." :/ – jdc Apr 1 at 4:27
@jdc Sorry! My old website has gone so I'll need to rehost these files. I'll update them after the weekend. – Loop Space Apr 1 at 7:09

To eloborate on what Arnav Tripathy said in a comment: if $X$ and $Y$ are compact, then $X \wedge Y$ is the one-point compactification of $(X \setminus \{x_0\}) \times (Y \setminus \{y_0\})$.

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As far as I remember, the smash product $X\wedge Y$ of two (pointed) spaces $(X,x_0)$ and $(Y,y_0)$ is obtained by taking the product of the two spaces $X\times Y$ and collapsing both the vertical "line" {$x_0$}$\times Y$ and the horizontal "line" $X\times${y_0} to a point.

So, if you (very loosely) imagine $X$ and $Y$ as segments, you can imagine $X\wedge Y$ as a square handkerchief which is shrinked along the central vertical and horizontal line, and the result is four "overhangs" coming out from the base point. [Of course the 4 overhangs are just an "artefact" of your simplified mental picture of the spaces $X$ and $Y$ as segments, it's not something general!]

Was I "graphical" enough? :)

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Yeah, that's graphical enough ;) I'm not sure whether I understand "Of course the 4 overhangs are just an "artefact" of your simplified mental picture of the spaces and as segments, it's not something general!". Clearly, your description fits for the simple case $[0,1]\wedge[0,1]$. That's good enough form me :) – Michal Oszmaniec Jan 29 '11 at 10:15
If you think of $\mathbb{S}^1\wedge \mathbb{S}^1$ you'll 'see' what I meant. – Qfwfq Jan 29 '11 at 11:40
(there you don't obtain 4 copies of a space attached at a point) – Qfwfq Jan 29 '11 at 11:41

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