I'm interested in showing that $S^m * S^n \approx S^{m+n+1} $, as discussed in exercise 0.18 of Hatcher's Algebraic Topology. One way to show it would be to show that $X * Y \approx \Sigma(X \wedge Y)$. This fact shows up only in a later exercise in the chapter, so I have to think he's looking for something else.
This, this, and this question allude to answers. In particular, there is this answer, which I don't understand:
If $A$ and $B$ are subsets of $\mathbb R^n$ then you can map $A\times I\times B$ to $\mathbb R^n$ by
$(a,t,b)\mapsto (1−t)a+tb$. If the resulting continuous map $A∗B\to \mathbb R^n$ happens to
be one to one then you can ask whether it gives a homeomorphism to its
image. If $A$ and $B$ are compact, then the answer is necessarily yes,
since a continuous bijection from a compact space to a Hausdorff space
is always a homeomorphism.This suffices to see that the join of spheres is a sphere.
Best Answer
$S^m * S^n = (S^m \times S^n \times [0, 1])/\sim$ where $\sim$ identifies the top $S^m \times S^n \times \{0\}$ to $S^m$ and the bottom $S^m \times S^n \times \{1\}$ to $S^n$. Cutting this in half gives $$S^m * S^n = (S^m \times S^n \times [0, 1/2])/\!\!\sim \cup_{S^m \times S^n \times \{1/2\}}\; (S^m \times S^n \times [1/2, 1])/\!\!\sim$$
In the first piece, $\sim$ does nothing except pinching the copy of $S^m \times \{0\}$ to a point. Thus, the first piece is homeomorphic to $C(S^m) \times S^n \cong D^{m+1} \times S^n$. Similarly, $\sim$ just pinches the copy of $S^n \times \{0\}$ in the second piece, so that one is homeomorphic to $S^m \times C(S^n) \cong S^m \times D^{n+1}$. So the space is homeomorphic to
$$D^{m+1} \times S^n \cup_{S^m \times S^n} S^m \times D^{n+1} \cong D^{m+1} \times \partial(D^{n+1}) \cup_\partial \partial(D^{m+1}) \times D^{n+1} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; \cong \partial(D^{m+1} \times D^{n+1}) \\ \;\;\;\;\;\;\ \cong \partial(D^{m+n+2})$$
Which is just $S^{m+n+1}$ $\blacksquare$