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I have a problem computing the homology of a certain space $X$, where my intuition and my answer don't coincide.

The space $X$ is given by identifying antipodal points $x \sim -x$ on the equator $S^2$ of $S^3$, $X=S^3/\sim$.

Since the space is easily obtained by attaching two copies of $D^3$ and attaching them along their $S^2$ boundary with the antipodal map, by the homology-effect of attaching cells, I get an exact sequence for reduced homology modules: \begin{align*} 0 \to H_3(D^3) \to H_3(X) \to H_2 (S^2) \to H_2 (D^3) \to H_2 (X)\to 0. \end{align*} from which I conclude $H_3=\mathbb{Z}$, $H_2=0$. However, the result for $H_3$ seems wrong. First, it does not agree with my intition, I should get 2, $3$d-holes via the identification, not one, and the second, it is in contradiction with the answer obtained in

Homology of quotient of 3-sphere by identifying antipodal points on equator

So I am asking, where am I going wrong? Note that I don't really want an explanation of how to do this via Mayer-Vietoris as in the question I linked but via the "theoremy" method I use, as I could then just as well copy the question linked. All advice to this end is well appreciated.

EDIT: As I guess a lot of people have read Hatcher I'll state what I meant by attaching in my third paragraf:

By attaching I mean $X=D^3\amalg D^3 / \sim$ where $\sim$ is given by the antipodal map from $\partial D^3 \to \partial D^3$. As I stated in the comments I'm not familiar with Hatcher so I'm not sure what he means by attaching.

afightingchance
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    Are you sure this space consists of a CW complex with a $3$-disc attached to another $3$-disc? Identification is not quite the same thing as attaching. If it had the CW structure you describe, it would follow from Hatcher, page 11 ("If $(X,A)$ is a CW pair...") that your space is a $3$-sphere. – Peter Franek Dec 10 '17 at 20:17
  • By attaching I mean $X=D^3\amalg D^3 / \sim$ where $\sim$ is given by the antipodal map from $\partial D^3 \to \partial D^3$. Im not sure whether this conflicts with the notion of attaching in Hatcher or not, I'll look into it, though. ($\amalg$ is the disjoint union) – afightingchance Dec 10 '17 at 20:50

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The space $X=S^3/\sim$ where $\sim$ identifies antipodal pair on the equatorial $S^2$ is not equal to two copies of $D^3$ glued along their $S^2$ boundary via the antipodal map.

Note that such space $D^3\sqcup D^3/\sim$ is the ordinary $S^3$ (you just identify two $3$-hemispheres on their boundary $2$-spheres via an orientation-reversing homeomorphism).

Peter Franek
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  • Oh, so that's where my mistake lies. Thank you. Is there a way to construct the space in terms of $D^3$s? – afightingchance Dec 10 '17 at 22:16
  • I think it has a CW structure of $RP^3$ with another $3$-cell attached to $RP^2$ in a similar way as the first $3$-cell. As I outlined in the comment, it cannot be just $D^3\sqcup D^3$ glued via some $\varphi: \partial D^3\to \partial D^3$ because this would always result in a homotopy sphere. – Peter Franek Dec 10 '17 at 22:23
  • Thank you, I'm accepting your answer. Seems messy, so Maybe I'll try to compute it via Mayer-Vietoris after all.. – afightingchance Dec 10 '17 at 22:46