I am trying to complete exercise 3.18(a) in Lee's Introduction to Topological Manifolds. The exercise is as follows:

Let $A \subseteq \mathbb{R}$ be the set of integers and let $X$ be the quotient space $\mathbb{R}/A$, where $A$ is collapsed down to a point.

(a) Show that $X$ is homeomorphic to a wedge sum of countably infinitely many circles.

Lee gives the hint that one should express both spaces as quotients of a disjoint union of intervals, implying that one will make use of the uniqueness of quotient spaces.

The wedge sum of countably infinitely many circles is the space $B=\bigvee_{i=0} ^\infty S_i ^1=\amalg_{i=0} ^\infty S_i ^1/\{p_i\}$, where $p_i \in S_i ^1$ and $\{p_i\}$ denotes the relation $p_i\sim p_j$ for all $i,j$. Since $S^1$ is the quotient space $I/(0\sim1)$, we can express $B$ as $B=\amalg_{i=0} ^\infty I_i/\sim$, where $\sim$ is the relationship $p_i \sim p_j$ for all $i,j$ and $0_i\sim 1_i$ for all $i$. With this one defines the quotient map $q:\amalg_{i=0} ^\infty I_i \rightarrow B$ by sending elements to their equivalence classes.

I am stuck trying come up with a similar formulation for $\mathbb{R}$. Graphically I could imagine $\mathbb{R}$ being constructed by attaching countably infinitely many of the unit intervals $I_i$ together with the identification $1_i\sim 0_{i+1}$. Formally this would be $\mathbb{R}=\amalg_{i=0} ^\infty I_i/(1_i\sim 0_{i+1})$. Using this idea though I can only see myself being able to construct the positive real numbers $\mathbb{R}^+$ which is not homeomorphic to $\mathbb{R}$. Is their some way to save this idea?

If I could figure this out, then I would have to find a quotient map $r:\amalg_{i=0} ^\infty I_i\rightarrow \mathbb{R}/A$ that makes the same identifications as $q$ to prove that the two spaces are homeomorphic by the uniqueness of the quotient space.

  • $\begingroup$ Just an idea (maybe, too trivial) - what about starting with $i=0$ interval being $[0,1]\subset \Bbb R$ and attaching $2i$-th interval to the left and $2i+1$-th interval to the right? $\endgroup$ – Ilya Aug 22 '12 at 15:50
  • $\begingroup$ I thought about that, but I was trying to find a way that would make make the identifications easier. Maybe I should go back to this one. $\endgroup$ – Holdsworth88 Aug 22 '12 at 15:51
  • $\begingroup$ I see. I think Brian's idea of using integers is better as it allows avoiding such mess. $\endgroup$ – Ilya Aug 22 '12 at 15:56

Your idea is fine with just a small modification: $\Bbb R=\amalg_{n\in A}I_n/(1_n\sim 0_{n+1})$. To make life easier, you should also use $A$ as the index set for $\amalg_n S_n^1/P$, where $P$ is the set of distinguished points.

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  • $\begingroup$ That seems much more natural. Thank you for clearing that up. $\endgroup$ – Holdsworth88 Aug 22 '12 at 15:59

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