I'm attempting to solve exercise 71.2 of Munkres, and I have only one small roadblock remaining.

Suppose $X$ is a space that is the union of the closed subspaces $X_1, \dots, X_n$; assume there is a point $p$ of $X$ such that $X_i \cap X_j = \{p\}$ for $i \neq j$. Then we call $X$ the wedge of the spaces $X_1, \dots, X_n$, and write $X = X_1 \vee \dots \vee X_n$. 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.

I'm proceeding by modifying the proof of theorem 71.1, which covers the special case when each $X_i$ is homeomorphic to $S^1$. Letting $$ U = X_1 \cup W_2 \cup \dots \cup W_n \textrm{ and } V = W_1 \cup X_2 \cup \dots \cup X_n,$$ I've been able to show that $U$, $V$, and $U \cap V$ are open in $X$ and that $U \cap V$ is simply connected. However, I still need to show that $U$ and $V$ are path connected to apply Seifert-van Kampen and obtain the desired result. Munkres does say on page 332

...it is usual to deal with only path-connected spaces when studying the fundamental group.

But I don't believe he explicitly states that as a convention. Can I simply assume path connectedness, or is there more work for me to do?

  • $\begingroup$ Each X_i is path connected else the result is not true $\endgroup$ – Soumik Apr 10 '19 at 23:22
  • $\begingroup$ The fundamental group is always defined in terms of a basepoint, and the group doesn't depend at all on the components that don't contain the basepoint. Specifically if $X_i(p)$ denotes the component of $X_i$ that contains $p$ then $\pi_1(X_i,p) \cong \pi_1(X_i(p),p)$ (and similarly for $X$). You don't have to assume path-connectedness, but you don't lose any generality by doing so (as long as you are mindful about basepoints). $\endgroup$ – William Apr 11 '19 at 1:36
  • $\begingroup$ In your argument the $W_i$ are already path-connected, so if you replace $X_i$ with $X_i(p)$ then $U$ and $V$ will be path-connected. $\endgroup$ – William Apr 11 '19 at 2:00
  • $\begingroup$ @William Thank you for the comments! They cleared up my confusion and I've submitted my own answer below. $\endgroup$ – Itserpol Apr 11 '19 at 15:58

Okay, so based on William's comment, I've been able to answer the question. Since all loops are paths, any loop in $X$ based at $p$ must lie entirely within the path-component $X(p) \subset X$ which contains $p$, so that $\pi_1(X, p) = \pi_1(X(p), p)$. That is, I don't need to assume path connectedness.

Therefore, if I instead let $$U = X_1(p) \cup W_2 \cup \dots \cup W_n \textrm{ and } V = W_1 \cup X_2(p) \cup \dots \cup X_n(p),$$ then $U$ and $V$ are path connected and I am able to apply Seifert-van Kampen (as well as the rest of the modified proof) to obtain $$\begin{align} \pi_1(X,p) &= \pi_1(X(p),p) \\ &= \pi_1(U,p)*\pi_1(V,p)\\ &= \pi_1(X_1(p),p) * (\pi_1(X_2(p),p) * \ldots * \pi_1(X_n(p),p)) \\ &= \pi_1(X_1,p) * \ldots * \pi_1(X_n,p), \end{align}$$ which is the desired result. (Here I've abused the notation to omit the monomorphisms induced by inclusion.)

Thank you for the help, William!

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