Visualisation of the smash product

Solution 1:

The smash product of two $1$-spheres gives a nice start in understanding things. You start with $S^1 \times S^1$, a torus. Draw a latitude curve and a longitude curve (i.e., $\{a\} \times S^1$ and $S^1 \times \{b\}$). These intersect at a point $(a, b)$ on the torus, and their union is $S^1 \vee S^1$.

The smash product is the result of collapsing this figure-eight shape to a single point.

It's probably easiest to see this by visualizing the torus as a square with pairwise opposite edges identified. The wedge-product of the two circles then corresponds exactly to the boundary of the square, which each point being seen twice in the square (on opposite sides) except for the join-point, which corresponds to all four vertices.

The smash product then identifies all these boundary points to a single point, i.e., the quotient is a disk with its boundary collapsed to a point, i.e., a 2-sphere.

If I recall correctly, in general $S^k \wedge S^p$ is just $S^{k+p}$; one sort of intuitive argument for the smash-product being a fairly simple space, at least at the homotopy level, is that in the cross product, there's some $\pi_k$ and $\pi_p$, coming from the injection map on each factor, but collapsing the wedge product to a point kills off both of these. More useful, however, is to work the analogy of the 2D example: $S^k \times S^p$ looks like a rectangle in $\mathbb R^{k+p}$ with some identifications; the stuff that ends up identified is exactly the wedge product. If you work this out for $S^1 \times S^2$ in 3-space, I think it'll be fairly clear how it generalizes.