So the following comes to me from an old algebraic topology final that got the best of me. I wasn't able to prove it due to a lack of technical confidence, and my topology has only deteriorated since then. But, I'm hoping maybe someone can figure out the proof as I've always been interested in seeing it all at once!

Let $E_\infty$ denote the 2-D analogue of the Hawaiian Earring, i.e.

$$E_\infty = \bigcup_{n=1}^\infty \{(x,y,z) \in \mathbb{R}^3 | \hspace{2mm} (x-1/n)^2 + y^2 + z^2 = 1/n^2\}.$$

The object of the exercise is to show that even though $E_\infty$ is 2-dimensional, $H_3(E_\infty) \neq 0$, which I find interesting even though I know it's not a CW-complex. Parts of the proof were helped along by my professor to serve as a road map for the solution, but sadly I'm still lost in the driveway filling in some of the remaining bits:

Let $h: S^3 \longrightarrow S^2$ denote the Hopf map i.e. the attaching map for the 4-cell in $\mathbb{C}P^2$. One can readily observe that there is a continuous map $\tilde{h}: S^3 \longrightarrow E_\infty$ so that the projection of $\tilde{h}$ to any $S^2$ in $E_\infty$ is homotopic to the Hopf map. Let $C_\tilde{h}$ denote the mapping cone of $\tilde{h}$. Then we have a mapping cones that looks something like a LOT of $\mathbb{C}P^2$'s.

(1) First we must prove that $H^2(C_\tilde{h})$ contains a subgroup $\mathbb{Z}<\zeta_1, \zeta_2, ...>$ and $H^4(C_\tilde{h}) \cong \mathbb{Z}<\eta>$ where

$$\zeta_i \cup \zeta_i = \eta \text{ and } \zeta_i \cup \zeta_j = 0 \text{ for } i \neq j.$$

(I suspect that this comes in part from the cup product structure on $\mathbb{C}P^2$).

Now let $[S^3]$ denote the fundamental class of $S^3$. Assumption: Suppose that $\tilde{h}_* = 0 \in H_3(E_\infty)$. Under this assumption, there is a finite simplicial complex $X$ with boundary $S^3$ so that $\tilde{h}$ extends to a map $k: X \longrightarrow E_\infty$. Let $Y = X \cup \mathbb{D}^4$ where $\mathbb{D}^4$ is glued to $S^3$ in the obvious way. Then extending $k$, we have another map $l: Y \longrightarrow C_\tilde{h}$ sending the $B^4$ to the cone $S^3 \times [0,1]/ S^3 \times 1$.

(2) Here it must be proven that $l^*(\eta)$ is a nontrivial element of $H^4(Y)$.

(I truly do not see how to do this, but it seems like it would be true [here "seems" doesn't really mean anything]).

(3) And now it needs to be shown that the infinitely many $l^*(\zeta_i)$ are all linearly independent.

(This would seem to follow from naturality of the cup product in some way).

If I could prove these things, it would only remain to observe that $Y$ came from gluing to a finite simplicial complex, and therefore it is itself a finite simplicial complex. Hence $H^2(Y)$ is finitely generated, so we arrive at the desired contradiction, which is awesome since we have found a nontrivial element in the third homology group of a 2-D object! Awesome!

Anyways, if anyone could complete this proof in its entirety, I would be supremely grateful. I assure you that I will upvote it a hundred times over! Even though the last 99 won't really do much. Also sorry that there are sort of several questions embedded in one. I thought it would be justified as they lie in the same vein of the single proof.

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  • Sadly this is as close as I could get to an answer. My professor told me to deduce the above contradiction and gave me the trick about considering the mapping cone, but these are the remaining gaps. I guess a hint would be wonderful. The above maps were constructed because I figured I would want to reduce it to a finite complex, so that I could say anything about the homology under the assumption about $H_3$, but this is as far as I can get, with only some vague intuition as to how to proceed. – MGN Jun 26 '12 at 00:26
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    Also it's not homework, just an old exam question from about a year back that I've always been curious about finishing. – MGN Jun 26 '12 at 00:28
  • Alternatively, if anyone sees a different approach (or even modifications on what I think I need), that would be welcome as well! – MGN Jun 26 '12 at 00:29
  • What's up with the bolding? Seems odd to me. – Potato Jun 29 '12 at 04:20
  • Perhaps I got a little carried away with the typesetting. – MGN Jun 29 '12 at 04:56

3 Answers3


Barratt and Milnor were (I think) the first to show that this space has nontrival $H_3$. In fact, they show that $H_{2k+1}$ is uncountable for all $k$. Here is a free .pdf of their paper.

(I've made this CW because a) I did nothing, and b) I haven't read the paper in detail, so I only have a vague notion of what Barratt and Milnor did.)

Jason DeVito
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  • Nitpicking, but actually they show even more, that it has uncountable $H_k$ for $k>1$; the congruence there is modulo $r-1$, not $r$. – Harry Altman Jun 26 '12 at 03:22
  • That's a cool paper, but certainly a much stronger result and at first glance seems a little bit above me, but I'll take a deeper look. – MGN Jun 26 '12 at 04:05
  • @Harry: You're right - I glanced through the paper too quickly. – Jason DeVito Jun 26 '12 at 11:57
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    @MGNewman: I think it's above me as well ;-). I don't really understand Whitehead products and those are the main tool used. – Jason DeVito Jun 26 '12 at 11:57

$\newcommand{\Ch}{ C_{\tilde h}} \newcommand{\CP}{\mathbb{CP}^2}$ We will make use of natural inclusions $\alpha_i\colon S^2\hookrightarrow C_{\tilde h}$ and natural projections $\pi_j\colon \Ch\to \CP$. The composition $\pi_j\circ \alpha_j$ is the standard inclusion of $S^2$ in $\mathbb{CP}^2$ and the compositions $\pi_j\circ\alpha_i$ are the constant map to the base point when $i\neq j$. The composition on the level of $H^2$ is the identity, giving that $\pi_i^*(\zeta)=:\zeta_i$ are all nonzero. (We let $\zeta$ be a generator of $H^2(\CP)$.) In fact, we want to show they are linearly independent. This follows from the fact that the homology-cohomology pairing satisfies $\langle \pi_i^*(\zeta),(\alpha_j)_*([S^2])\rangle =\delta_{ij}$.

To continue, I'm going to cheat slightly and assume that $H^4(E_\infty)=H^5(E_\infty)=0$, since otherwise we would have a higher dimensional class, which would be equally surprising as a $3$ dimensional one. In that case, by the long exact sequence of the pair $$\mathbb Z\cong H^4(\Ch,E_\infty)\cong H^4(\Ch).$$ Note that the generator of $H^4(\Ch,E_\infty)$ is dual to the $4$-cell as is the image of $\zeta^2\in H^4(\CP)$. Thus if we let $\eta$ denote the generator of $H^4(\Ch)$ we have $\zeta_i^2=\eta$ for all $i$. Furthermore $\zeta_i\zeta_j=0$ for $i\neq j$ because they are disjointly supported.

The image $\ell^*(\eta)$ is nonzero as it is dual to the $4$-cell, but $Y$ by construction represents a $4$-cycle on which $\ell^*(\eta)$ evaluates nontrivially.

To see that $\ell^*(\zeta_i)$ are linearly independent, suppose that $\sum_{i\in I} n_i\ell^*(\zeta_i)=0$. Now take the cup product of this element with itself! We have $$\sum_{i\in I}\sum_{j\in I} n_in_j\ell^*(\zeta_i\zeta_j)=\sum_{i\in I} n_i^2\ell^*(\eta)=0,$$ implying $\sum n_i^2=0$ and so $n_i=0$ for all $i$. This contradicts the finite generation of $H^4(Y)$ and we are done!

Cheerful Parsnip
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Here is what I would say:

I'm not sure about the 2-D case, but perhaps you could adapt the results from this paper available online: http://jlms.oxfordjournals.org/content/62/1/305.short

It seems to follow along the lines of what you said in the comments about reducing the question to a finite complex.

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