Hatcher Algebraic Topology 0.24 This is my second question from Hatcher chapter 0 (and final I think). For $X$, $Y$ CW complexes, it asks one to show that
$$X \ast Y = S(X \wedge Y)$$
by showing
$$X \ast Y/(X \ast y_0 \cup x_0 \ast Y) = S(X \wedge Y) / S(x_0 \wedge y_0),$$
where $\ast$ is topological join, $\wedge$ is smash product, $S$ is suspension and $=$ is homeomorphism.
Unfortunately I am quite at loss on this question besides from knowing some definitions of these spaces. $S(x_0\wedge y_0)$ is suspension of a point which is an interval I. $X \wedge Y$ is the space obtained by the quotient $X \times Y / X \vee Y$ where $X \vee Y$ is wedge sum of $X$ and $Y$ at the point $x_0$ and $y_0$. Suspension of this is its product with interval $I$ and collapsing the end points. Overall we have
$$X \times Y \rightarrow X \times Y / X \vee Y \rightarrow  (X \times Y / X \vee Y) \times I \rightarrow Z = (X \times Y / X \vee Y) \times I/(\text{contract $t=0$ and $t=1$ to a point}) \rightarrow Z / S(x_0\wedge y_0).$$
$(X \ast {y_0} \cup {x_0} \ast Y)$ is the union of two cones over $X$ and $Y$ and finally $X \ast Y$ could be seen as union of cones $C(X,y)$ for all $y$ or the other way around. Then this space is
$$X \times Y \times I \rightarrow A= X \times Y \times I/(\text{contract $Y$ at $t=0$ and $X$ at $t=1$}) \rightarrow A/(X \ast {y_0} \cup {x_0} \ast Y).$$
However I cant see how to relate these operations. I know some theorems on these space making homotopy relations but this questions requires this to be a homeomorphism.
Thanks a lot.
 A: Let's show the reduced versions are homeomorphic, which will show the originals are homotopic (they are not equal in general).
The join can be thought of as "lines" from $X$ to $Y$, with some collapsing.  The relations are: 
$$ (x_1,y,0)\sim (x_2,y,0);$$
$$ (x,y_1,1)\sim (x,y_2,1).$$
The reduced version also collapses $x_0\ast Y$ and $X\ast y_0$.
So the additional relations are:
$$ (x_0,y,t)\sim (x_0,y_0,0);$$
$$ (x,y_0,t)\sim (x_0,y_0,0).$$
We can derive further relations too:
 $$(x,y,0)\sim(x_0,y,0)\sim(x_0,y_0,0);$$
$$(x,y,1)\sim (x,y_0,1)\sim(x_0,y_0,0).$$
The smash product is gotten from $X\times Y$ by collapsing $X\times y_0$ and $x_0\times Y$. The suspension of that can be thought of as $X\times Y\times I$, with the relations:
$$ (x,y_0,t)\sim (x_0,y_0,t);$$
$$ (x_0,y,t)\sim (x_0,y_0,t);$$
$$ (x,y,1)\sim (x_0,y_0,1);$$
$$ (x,y,0)\sim (x_0,y_0,0).$$
The reduced suspension adds the relation
$$ (x_0,y_0,t)\sim (x_0,y_0,0).$$
Now it is not hard to see you are quotienting out by the same relations for both constructions. Namely,
$$ (x,y_0,t)\sim (x_0,y_0,0);$$
$$ (x_0,y,t)\sim (x_0,y_0,0);$$
$$ (x,y,0)\sim (x_0,y_0,0);$$
$$ (x,y,1)\sim (x_0,y_0,0).$$
A: Some pictures from "Topology and Groupoids",  Chapter 5,  which you may find  helpful are   the join $X * Y$ as 
and the subspace to be collapsed  to a point to give the suspension of the smash product is 
 where the two vertices on the mid line are the base points.  
