Exercise 71.2

Question

Suppose $X$ is a space that is the union of the closed subspaces $X_1,
\ldots, X_n$; assume there is a point $p$ of $X$ such that $X_i \cap X_j =
\{p\}$ for $i 
e j$. Then we call $X$ the \emph{	extbf{wedge}} of the spaces
$X_1, \ldots, X_n$, and write $X = X_1 \vee \cdots \vee X_n$. Show that if
for each $i$, the point $p$ is a deformation retract of an open set $W_i$ of
$X_i$, then $\pi_1(X,p)$ is the external free product of the groups
$\pi_1(X_i,p)$ relative to the monomorphisms induced by inclusion.

Amit Awasthi · Accepted Answer

\begin{proof}
  By induction, it suffices to show when $X = X_1 \vee X_2$; moreover, it
  suffices to consider when the $X_i$ are both path-connected since if $C_i$
  are the path components containing $p$ in $X_i$, then
  $\pi_1(C_i,p) = \pi_1(X_i,p)$ as on p.~332. So, let $U = X_1 \cup
  W_2$ and $V = X_2 \cup W_1$. Then, both $U$ and $V$ are path connected since
  they deformation retract to $X_1,X_2$, respectively, and $U \cap V = W_1 \cup
  W_2$ is moreover simply connected since it deformation retracts to $\{p\}$.
  Thus, by Corollary $70.3$, there is an isomorphism
  $\pi_1(X_1,p) * \pi_1(X_2,p) \simeq \pi_1(X,p)$.
\end{proof}

Exercise 71.2

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Exercise 71.2

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