Take the quotient space of the cube $I^3$ obtained by identifying each square face with opposite square via the right handed screw motion consisting of a translation by 1 unit perpendicular to the face, combined with a one-quarter twist of its face about it's center point.

I am trying to calculate the homology of this space.

It is not too hard to see that the CW decomposition of this space has 2 0-cells, 4 1-cells, 3 2-cells and 1 3-cell.

We end up (drawings would help here, but my MS-Paint skills are poor!) with the 2 0-cells ($P$ and $Q$) connected by the 4 1-cells $a,b,c,d$ with $a,c$ from $P$ to $Q$ and $b,d$ from $Q$ to $P$. Thus we have the closed loops $ab,ad,cb,cd$. They also satisfy the relations $abcd=1,dca^{-1}b^{-1}=1,c^{-1}adb^{-1}=1$ via the identification of opposite 2-cells (top/bottom, left/right, up/down). (There is a relationship between the generator loops - the fundamental group is the quaternion group).

From the CW decomposition we get the cellular chain complex

$0 \to \mathbb{Z} \stackrel{d_3}{\to} \mathbb{Z}^3 \stackrel{d_2}{\to} \mathbb{Z}^4 \stackrel{d_1}{\to} \mathbb{Z}^2 \to 0$

I'm struggling to work out the boundary maps. Can it be 'seen' easily from the relations above?

I tried to use the cellular boundary formula. $d_1$ must be a 2 x 4 matrix. The cellular boundary formula gives the relation

$$d_1(e^1_\alpha) = \sum_{\beta=1}^2 d_{\alpha \beta} e^0_\beta$$

Are the entries of the matrix $d_1$ then given by $$\left(\begin{array}{cccc} d_{11} & d_{21} & d_{31} & d_{41} \\ d_{12} & d_{22} & d_{32} & d_{42} \\ \end{array}\right)?$$

I am pretty sure that $d_{\alpha \beta}$ must be $-1$ or $1$ as the attaching map is a homeomorphism (and is not 0), and is dependent on orientation. Therefore I get that $$d_1 = \left(\begin{array}{cccc} 1 & 1 & 1 & 1\\ -1 & -1 & -1 & -1\\ \end{array}\right).$$

Similar logic says that $d_2$ is a 4x3 matrix. Again all entries must be 1 or -1. I'm struggling to see exactly what the boundary map should be here?

Any thoughts on the best approach are appreciated.