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.

**Answer**

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.)

**Attribution***Source : Link , Question Author : MGN , Answer Author :
Jason DeVito
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