Path components of Wedge Sum

Question

I couldn't find this anywhere else, so I decided to post it here.

I suspect that the wedge sum $⋁X_α$ of pointed spaces $X_α$ has as path components all components of the topological sum $\oplus X_α$ not containing any base point, plus a single component which is the wedge of all components containing the base points.
In particular, if we have $n$ spaces with $m$ components, the wedge will have $m-n+1$ components.

This should follow from

Theorem: If $p$ is a path in $(X,x_0)=\bigvee_α(X_α,x_α)$ between distinct $X_α$ and $X_β$, then $p(0)$ must lie in the path component of $x_α$.
Proof: Case 1: Assume $m=\min\{s\in I\mid p(s)\notin X_α\}$. In that case, we can assume that $m=1$ and $p(1)\in X_β$, so $p([0,1))\subseteq X_α$. If we define $$q(s)=\begin{cases}p(s), & s<1\\ x_0, & s=1\end{cases}$$ we only need to check continuity of $q$ at $1$. Let $U$ be an open neighbourhood of $x_0$ in the subspace $X_α$ of $X$. Then the neighbourhood $U\vee\bigvee_{\gamma\neα} X_\gamma$ of $p(1)$ contains the image of some $(1-\delta,1]$ under $p$ since that path is continuous. Thus $p((1-\delta,1))=q((1-\delta,1))⊆U$. Hence $q$ is a path in $X_α$ from $p(0)$ to $x_α$.
Case 2: Assume $m=\max\{s\in I\mid p([0,s])⊆X_α\}$. Then $p(m)$ is in the closure of $⋁_{γ≠α}X_γ$, so it must be in the closure of $\{x_α\}$. There is thus a path $q:I→\{p(m),x_α\}$ which we can glue to $p|_{[0,m]}$ to get a path within $X_α$ from $p(0)$ to $x_α$.

From this is follows that points in different summands of the wedge can be joined only if they are both in their respective path components containing the base points.

Do you see any flaws? Or do you know if the proof for this fact can be found in literature?

Thanks.

Answer 1

N.B. This answer is flawed. The reasoning below shows that a path-component of $X$ is either the path-component of $x_0$ or the image of a path-component of some $X_\alpha$. It does not follow immediately that the path-component of $x_0$ is the union of the path-components of the $x_\alpha$. (It does follow if $\{x_0\}$ is closed.)

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I don't see anything wrong in your proof, but maybe I can put a different spin on it. The following line of reasoning seems more intuitive to me.

1. Let $q$ be the quotient map taking the disjoint union to the wedge sum. The restriction of $q$ to $q^{-1}(X \setminus \overline{\{x_0\}})$ is a homeomorphic embedding. (it's injective, continuous and open) It follows that any path not passing through $\overline{\{x_0\}}$ is already a path in the disjoint union.
2. The closure of a point in any topological space is path-connected.

The result you want then seems obvious.