Fundamental Group of Wedge Sum: Does Munkres Assume Path Connectedness

algebraic-topology

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?

Best Answer

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!