# CW complexes exercise from Hatcher

I am having some trouble with this. Any help would be very appreciated. Thanks.

Exercise 24. Let $X$ and $Y$ be $CW$ complexes with $0$ cells $x_{0}$ and $y_{0}$. Show that the quotient spaces $X * Y / (X*\left\{y_{0}\right\} \cup \left\{x_{0}\right\} * Y)$ and $S(X \wedge Y)/S(\left\{y_{0}\right\} \wedge \left\{x_{0}\right\})$ are homeomorphic, and deduce that $X * Y$ and $S(X \wedge Y)$ are homotopy equivalent.

Best, Anna

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You might want to construct a map from $X\star Y$ into $I\times X\times Y$ and make sure it passes through all the appropriate quotients. – Joe Johnson 126 Oct 2 '11 at 18:01
More details, please? That's what I've been trying to do. – Anna Oct 2 '11 at 18:35
Maybe even start one higher: Make a map from $X \times Y \times I$ to $X \times Y \times I$, then pass through all the quotients. Hint: Don't think too hard. – Dylan Wilson Oct 2 '11 at 19:28

We show that these two spaces are both homeomorphic to $I \times X \times Y / \left(I \times ( X \times \{y_0\} \cup \{x_0\} \times Y)\right)$ as follows:
1. Passing from $I \times X \times Y$ to $X * Y$, we identify the points $(0,x,y) \sim(0,x,y')$ and $(1,x,y) \sim (1,x',y)$. Under these identifications, $I \times X \times \{y_0\} \cup I \times \{x_0\} \times Y$ becomes $X *\{y_0\} \cup \{x_0\}*Y$. Hence $$\frac{I \times X \times Y } {I \times ( X \times \{y_0\} \cup \{x_0\} \times Y)} \simeq \frac{X * Y}{X * \{y_0\} \cup \{x_0\}*Y}$$
2. Notice that $$\frac{I \times X \times Y } {I \times ( X \times \{y_0\} \cup \{x_0\} \times Y)} \simeq \frac{I \times X \times Y } {I \times ( X \vee Y)} \simeq I \times \frac{ X \times Y } { X \vee Y} \\ \simeq I \times(X \wedge Y) \simeq I \times \frac{ X \wedge Y } {\{x_0\}\wedge \{y_0\}} \simeq \frac{I\times ( X \wedge Y) } {I \times (\{x_0\}\wedge \{y_0\})} .$$ When passing form $I\times ( X \wedge Y)$ to $S(X \wedge Y)$, we collapse $\{0\} \times (X \wedge Y)$ and $\{1\} \times (X \wedge Y)$ to two points. With these identifications, $I \times (\{x_0\}\wedge \{y_0\})$ becomes $S(\{x_0\}\wedge \{y_0\})$. Whence $$\frac{I \times X \times Y } {I \times ( X \times \{y_0\} \cup \{x_0\} \times Y)} \simeq \frac{S(X\wedge Y)}{S(\{x_0\}\wedge \{y_0\})}.$$
Since $({X * Y},{X * \{y_0\} \cup \{x_0\}*Y})$ and $({S(X\wedge Y)},{S(\{x_0\}\wedge \{y_0\})})$ are CW-pairs, they both have homotopy extension property (Hatcher prop. 0.16), so the complex is homotopy equivalent to the quotient space by the subcomplex (Hatcher prop. 0.17). Therefore $${X * Y}\simeq {S(X\wedge Y)}.$$