Exercise 0.24

Question

Let $X$ and $Y$ be CW complexes with $0$-cells $x_0$ and $y_0$. Show that the quotient spaces $X \ast Y/(X \ast \{y_0\} \cup \{x_0\} \ast Y)$ and $S(X \vee Y)/S(\{x_0\} \wedge \{y_0\})$ are homeomorphic, and deduce that $X \ast Y \simeq S(X \wedge Y)$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Recall that
  \begin{equation*}
    X \ast Y = \frac{X 	imes Y 	imes I}{(x,y_1,0) \sim (x,y_2,0), (x_1,y,1)\sim(x_2,y,1)}.
  \end{equation*}
  Under this identification, $X 	imes \{y_0\} 	imes I \cup \{x_0\} 	imes Y 	imes I$ maps to $X \ast \{y_0\} \cup \{x_0\} \ast Y$, and so we have
  \begin{equation*}
    \frac{X \ast Y}{X \ast \{y_0\} \cup \{x_0\} \ast Y} \cong \frac{X 	imes Y 	imes I}{(X 	imes \{y_0\} \cup \{x_0\} 	imes Y) 	imes I}.
  \end{equation*}
  \par Note we also have
  \begin{align*}
    \frac{X 	imes Y 	imes I}{(X 	imes \{y_0\} \cup \{x_0\} 	imes Y) 	imes I} &\cong \frac{X 	imes Y 	imes I}{(X \vee Y) 	imes I}
    \cong \frac{X 	imes Y}{X \vee Y} 	imes I \cong (X \wedge Y) 	imes I\
    &\cong \frac{X \wedge Y}{\{x_0\} \wedge \{y_0\}} 	imes I \cong \frac{(X \wedge Y) 	imes I}{(\{x_0\} \wedge \{y_0\}) 	imes I}.
  \end{align*}
  Recall that
  \begin{equation*}
    S(X) = \frac{X 	imes I}{(x,0) \sim (x',0), (x,1) \sim (x',1)}.
  \end{equation*}
  Under this identification for $X \wedge Y$, we see that $(X \wedge Y) 	imes \{0\}$ and $(X \wedge Y) 	imes \{1\}$ collapse to two points. Thus, $(\{x_0\} \wedge \{y_0\}) 	imes I$ maps to $S(\{x_0\} \wedge \{y_0\})$. By transitivity of homeomorphisms, we see
  \begin{equation*}
    \frac{X \ast Y}{X \ast \{y_0\} \cup \{x_0\} \ast Y} \cong \frac{S(X \wedge Y)}{S(\{x_0\} \wedge \{y_0\})}.
  \end{equation*}
  \par Finally, both $(X \ast Y,X \ast \{y_0\} \cup \{x_0\} \ast Y)$ and $(S(X \vee Y),S(\{x_0\} \wedge \{y_0\}))$ are CW pairs, and so they satisfy the homotopy extension property by Proposition 0.16, and therefore are homotopy equivalent to their quotient spaces by the subcomplex by Proposition 0.17. This gives that
  \begin{equation*}
    X \ast Y \simeq \frac{X \ast Y}{X \ast \{y_0\} \cup \{x_0\} \ast Y} \cong \frac{S(X \wedge Y)}{S(\{x_0\} \wedge \{y_0\})} \simeq S(X \wedge Y),
  \end{equation*}
  and so $X \ast Y \simeq S(X \wedge Y)$.
\end{proof}

Exercise 0.24

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Exercise 0.24

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