Understanding why collapsing $CX$ and $CY$ gives us $\sum (X \times Y)$?

Question

Here is the proposition I am trying to understand in Hatcher:

But I do not understand:

1- why collapsing $CX$ and $CY$ gives us $\sum (X \times Y),$ could someone explain this to me please? as far as I understand suspension means that the space $X \times Y$ will be suspended between 2 points, one point below it and another one above it, how will this happen here? where are the two points?

2- What is the point of intersection of the cones $x_0 \star Y$ and $X \star y_0$?

3- Lastly, why in the last line of the proof, $X \star Y$ will be converted into $\sum (X \wedge Y)$ and not $ \sum X \vee Y$ for example, also why the suspension sign is there?

Could someone clarify to me the answers of these questions please?

Answer 1

This is an extra-long proof of $1$. My hope is that after reading it thoroughly several times, you'll be equipped to answer question 2 and 3 yourself.

The definition of the (reduced) join $(X,x_0)\ast (Y,y_0)$ is the quotient space of $X\times Y\times [0,1]$ where we make three identifications:

1. For each $x\in X$, $(x,y,0)\sim (x,y',0)$ for all $y,y'\in Y$.
2. For each $y\in Y$, $(x,y,1)\sim (x',y,1)$ for all $x,x'\in X$.
3. $(x_0,y_0, t) \sim (x_0, y_0, t')$ for all $t,t'\in [0,1]$.

Now, on the end with $t=0$, we're going to attach the reduced cone $CX$. For concreteness, I'm going to take $CX$ to be the quotient of $X\times [-1,0]$ where we identify:

4. $\{x\} \times \{-1\}\sim \{x'\}\times \{-1\}$ for all $x,x'\in X$

5. $\{x_0\}\times \{t\} \sim \{x_0\}\times \{t'\}$ for all $t,t'\in [0,1]$.

The gluing map gives an extra quotient on $CX \cup X\ast Y$ where we identify

6. $CX\ni (x,0)\sim [(x,y_0,0)]\in X\ast Y$ for all $x\in X$. (Note that by $1$ above, we can use any $y$ in place of $y_0$. That's why I'm using an equivalence class for $[(x,y_0,0)]$.)

Similarly, on the end with $t=1$, we're going to attach $CY$. I'll leave this part to you: $CY$ should be a quotient of $Y\times [1,2]$ by two more conditions (7. and 8.), and the gluing gives another condition (9.).

Now, let's put this all together. We can obtain $CX\cup X\ast Y\cup CY$ as the quotient of $X\times Y\times [-1,2]$ where we make identifications 1-9 above. (I spelled out 1-6 explicitly).

Lemma: Consider $CX\cup X\ast Y \cup CY$. If we add an additional relation where we quotient $CX$ to a point, then all points of the form $(x,y,t)$ with $t\leq 0$ are identified.

Proof: Every such point is contained in $CX$ (if $t< 0)$, or equivalent to one contained in $CX$ (if $t=0$). So the result follows since everything in $CX$ is being identified to a point. $\square$.

Of course, the analogous thing is true if we collapse $CY$ to a point as well.

Proposition: The space $Z$ obtained from $CX\cup X\ast Y\cup CY$ by collapsing $CX$ and $CY$ to a point is homeomorphic to $\Sigma(X\times Y)$.

Proof: By definition, $Z$ is a quotient of $X\times Y \times [-1,2]$ by relations $1.-9.$ together with some more relations coming from the fact that $CX$ and $CY$ are being collapsed to points. From the Lemma, every point with $t\leq 0$ is equivalent to one with $t=0$, and similarly for $t\geq 1$. Thus, $Z$ may be viewed as a quotient of $X\times Y\times [0,1]$. For these $t$, relations $1,2,3$ above apply. However, by the Lemma, quotienting $CX$ identified all points with $t=0$ to a single point. Similarly, all points with $t=1$ are identified.

On the other hand, since $CX$ and $CY$ are only attached at $t=0$ and $t=1$, quotienting them cannot change the identifications for $t\in (0,1)$.

Thus, $Z$ is the quotient of $X\times Y\times [0,1]$ where

1' All points with $t=0$ are identified to a point.

2' All points with $t=1$ are identified to a point.

3'. $(x_0,y_0,t)\sim (x_0,y_0)$ for all $t,t'\in [0,1]$. (The same $3.$ from above.)

But this is precisely the identification which gives $\Sigma(X\times Y)$. Thus, $Z$ is homeomorphic to $\Sigma(X\times Y)$. $\square$