I'll ask my question using the specific example of the wedge sum of unit intervals (pointed at 1), motivated by the actual exercise I'm trying to answer. So we have a countable collection of pointed unit intervals (which we'll index by $\alpha\in\mathcal{A}$), whose wedge sum is:
\begin{equation}
\bigvee_{\alpha\in\mathcal{A}}(I_\alpha,1_\alpha)=\bigg{(}\frac{\bigsqcup_{\alpha\in\mathcal{A}}I_\alpha}{1_\alpha\sim1_\beta},[1_\alpha]\bigg{)}
\end{equation}
I understand this as the disjoint union of intervals with 1 identified for all of them. But then surely the unions are not disjoint, if they all meet at 1? (I'm visualising this as infinite number of spokes, i.e. the unit intervals, emanating from a centre point, i.e. the point 1 for all them).
I have been asked to show that this wedge sum is not homeomorphic to $D_n=\cup_{n\in\mathbb{Z}}I_n$ equipped with the subspace topology, where $I_n\subset\mathbb{R}^2$ is the line segment joining $(0,1)$ to $(n,0)$. It is clear to me that this latter space is connected. However, my lack of understanding over the wedge product is preventing me from saying anything about its connectedness. If it really is a disjoint union, then it is not connected, and since connectedness is a topological invariant, I am done. Otherwise I will need to rethink.
Best Answer
The wedge sum is a quotient space of the disjoint union. That is, we take the disjoint union $X=\bigsqcup_{\alpha\in A} I_\alpha$, define an equivalence relation $\sim$ on $X$ by $x\sim y$ iff either $x=y$ or there exist $\alpha,\beta\in A$ such that $x=1_\alpha$ and $y=1_\beta$, and define the wedge sum to be the quotient space of $X$ by this equivalence relation.
In general, in topology, when one speaks of "identifying points", that means that one takes the original space in which the points were not equal (in this case, the disjoint union), and considers the quotient space by the equivalence relation that says that those points are equivalent.