$S^m * S^n \approx S^{m+n+1}$ 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.

 A: Let $S^n \subset \mathbb R^{n+1}$ and $S^m\subset \mathbb R^{m+1}$, then we can consider that they are both in $\mathbb R^{n+m+2} \cong \mathbb R^{n+1} \oplus \mathbb R^{m+1}$. Then define the map 
$$ \phi : \mathbb S^n \times [0,\pi/2] \times \mathbb S^m \to S^{n+m+1} \subset \mathbb R^{n+m+2}$$ 
by $\phi(a, t, b) = (\cos t) a + (\sin t) b$. This map descend to the quotient (still call) $\phi :S^n *S^m \to S^{n+m+1}$, which is bijective. Then as the domain is compact and the image is Hausdorff, $\phi$ is a homeomorphism. (This is essentially the idea given in your question)
A: Consider $\phi : S^n \times S^m \times [0,1] \to S^{n+m+1}$ defined by,
$$\phi((x_0,\dots,x_n),(y_0,\dots,y_m),t)=(\sqrt{1-t})x_0,\dots(\sqrt{1-t})x_n,\sqrt{t}y_0,\dots,\sqrt{t}y_m)$$
Then it is easy to check that $\phi$ is Continuous , onto and fails to be one-one precisely at $S^n \times S^m \times \{0\}$ and $S^n \times S^m \times \{1\}$ .
For $t=0$, $\phi((x_0,\dots,x_n),(y_0,\dots,y_m),0)=(x_0,\dots ,x_n,0,\dots,0)=\phi((x_0,\dots,x_n),(\tilde{y_0},\dots,\tilde{y_m}),0)$ 
Thus, $S^n \times S^m \times \{0\}$ is 'collapsed' to $S^n$ and similarly, $S^n \times S^m \times \{1\}$ is 'collapsed' to $S^m$  .
Now, invoking Universal Property of Quotient, we get an onto,continuous map $ \bar{\phi} : S^n * S^m \to S^{n+m+1}$ which is clearly one-one and thus $\bar{\phi}$ is a continuous bijection from a Compact space to a Hausdorff space and thus it turns out to be a homeomorphism!
A: There are other ways of defining the join $X=X_1*X_2$ with a topology. In Topology and Groupoids, Chapter 5, the analogy is taken with the join of two subsets of a high dimensional $\mathbb R^n$  by saying a point of the join is a formal sum $r_1x_1+r_2x_2$ where $x_i  \in X_i$ and $r_1+r_2=1, r_1,r_2 \in [0,1]$, except that if $r_1$ or $r_2$ is $0$ then we ignore that term. There are partial functions $\xi_i:X \to [0,1], \eta_i:X \to X_i$ defined by $r_1x_1+r_2x_2 \mapsto r_i, r_1x_1 +r_2x_2 \mapsto x_i$, respectively. Here $\eta_i$ has domain $\xi_i^{-1}(0,1]$ . We give $X$ the initial topology with respect to these functions.  
One advantage of using initial topologies here is that the join becomes associative.  
A homeomorphism $S^p * S^q \to S^{p+q+1}$ is defined by 
$$rx+sy \mapsto (x\sin r\pi /2, y \sin s\pi /2).   $$   
This is related to this stackexchange answer and picture. 
A: $S^m * S^n = (S^m \times S^n \times [0, 1])/\sim$ 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 $S^{m+n+1}$ $\blacksquare$
A: Consider the homeomorphism $\Delta^{n-1}\ast\Delta^{m-1}\cong\Delta^{n+m-1}$ given by $$((t_0,\dots,t_n),(s_0,\dots,s_m),t)\mapsto(tt_0,\dots,tt_n,(1-t)s_0,\dots,(1-t)s_m).$$
A: On page 9, Hatcher tells us that $S^{n-1}\approx X_1*\cdots *X_n$, where $X_i$ is the two unit vectors along the $i^{th}$ coordinate axis in $\mathbb R^n$. Hence $$S^{m+n+1}\approx X_1*\cdots*X_{m+n+2}\approx (X_1*\cdots *X_{m+1})*X_{m+2}*\cdots*X_{m+n+2}\approx S^m*(X_{m+2}*\cdots*X_{m+n+2})\approx S^m*S^n.$$
Note that the iterated join $X_1*X_2*X_3=(X_1*X_2)*X_3$.
