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.
Best Answer
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,
$$ (x,y_0,t)\sim (x_0,y_0,0);$$ $$ (x_0,y,t)\sim (x_0,y_0,0);$$ $$ (x,y,0)\sim (x_0,y_0,0);$$ $$ (x,y,1)\sim (x_0,y_0,0).$$