Hatcher Algebraic Topology 0.24

Question

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.

Answer 1

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,

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

Answer 2

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.