An issue with homology of topological space obtained by antripodal identification of equatorial $S^2$ in $S^3$ 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. 

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). 
