# Fundamental Group of Wedge Sum: Does Munkres Assume Path Connectedness?

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?

• Each X_i is path connected else the result is not true – Soumik Apr 10 '19 at 23:22
• 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). – William Apr 11 '19 at 1:36
• 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. – William Apr 11 '19 at 2:00
• @William Thank you for the comments! They cleared up my confusion and I've submitted my own answer below. – 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.)