Smash and join products of spheres How to prove rigorously that $\mathbb S^n \wedge \mathbb S^m =\mathbb S^{n+m}$ and $\mathbb S^n \ast \mathbb S^m = \mathbb S^{n+m+1}$?
And what intuition should i have for compute $\wedge,\ast$ for difficult spaces?

I would like to give another proof for the smash product using the definition. By the definition of the smash product, we have $$S^n\wedge S^m\cong S^n\times S^m/(S^n\vee S^m).$$
Suppose $I^n=[0,1]^n\subset \mathbb R^n$ and consider $S^n$ as $I^n/\partial I^n\cong (I^n)^\mathrm o\cup O$, where $O$ is the origin of $\mathbb R^n$. Then
$$S^n\times S^m \cong (I^{n+m})^\mathrm o\cup(\partial I^n\times (I^m)^\mathrm o)\cup((I^n)^\mathrm o\times \partial I^m),$$
$$S^n\vee S^m\cong I^n\cup I^m.$$
Since $(I^{n+m})^\mathrm o/(I^n\cup I^m)\cong(I^{n+m})^\mathrm o$, $(\partial I^n\times (I^m)^\mathrm o)\cup((I^n)^\mathrm o\times \partial I^m)/(I^n\cup I^m)\cong O$, we have $S^n\wedge S^m\cong (I^{n+m})^\mathrm o\cup O\cong S^{n+m}$.
