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.

  • 2
    $\begingroup$ Unfortunately, both spaces are connected, so you can't use connectedness to distinguish them. So, here's a hint for showing that the two spaces are not homeomorphic. First, for the subspace of $\mathbb{R}^2$, note that the topology is first countable, that is, around every point, there is a neighborhood basis consisting of countable many elements. This follows because it's true for $\mathbb{R}^2$ and first countability is always inherited by subspaces. On the other hand, show the wedge point in the wedge sum construction doesn't have such a countable basis of open sets. $\endgroup$ – Jason DeVito Nov 23 '15 at 0:50

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.

  • $\begingroup$ Thank you, that makes sense. But I am still unsure as to whether, for example, $(I_{\alpha_1},1_{\alpha_1})\cup\bigvee_{\alpha\in{\mathcal{A}\backslash\alpha_1}}(I_{\alpha},1_{\alpha})$ would be a valid decomposition to show whether or not the wedge sum is disconnected? Or is this union not disjoint, since we have the identification of 1? (This notation may not be correct - I'm trying to single out one of the 'spokes' in order to decompose the wedge product into two disjoint pieces) $\endgroup$ – jl2 Nov 22 '15 at 20:36
  • 2
    $\begingroup$ Right, those are not disjoint, because (the equivalence class of) $1_{\alpha_1}$ is in the second set, because it is equal to (the equivalence class of) all the other $1_\alpha$s. $\endgroup$ – Eric Wofsey Nov 22 '15 at 20:44
  • $\begingroup$ Just one final thing - is it correct to think that an open ball (in the wedge sum) centred at $1_\alpha$ will 'seep out' into all intervals? So for example an open ball of radius 1/2 centred at $1_\alpha$ for some $\alpha\in\mathcal{A}$ is $\cup_{\alpha\in\mathcal{A}}(1/2,1_\alpha)_\alpha$ where $(1/2,1_\alpha)_\alpha$ is the subinterval of the pointed interval $(I_\alpha,1_\alpha)$? $\endgroup$ – jl2 Nov 22 '15 at 23:14
  • 1
    $\begingroup$ The topology on the wedge sum is not metrizable (and in any case certainly doesn't have a canonical metric), so I don't know what "open ball of radius $1/2$" means. To understand open sets in the wedge sum, you have to look to the definition of the quotient topology. $\endgroup$ – Eric Wofsey Nov 22 '15 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.