**Hatcher Exercise 2.2.11**

Let be the space obtained from the cube by identifying each face with its opposite face via a right-handed one-quarter twist. This is a cell-complex structure on the resulting space:

This is a structure with two 0-cells (the marked vertices represent one equivalence class and the rest represent an equivalence class ), four 1-cells (the labeled edges), three 2-cells (which is clear since we identify the faces in pairs) – call the front/back , left/right , and bottom/top – and one 3-cell . The cellular homology is:

The maps are:

To determine , we compute the degrees of each map in the cellular boundary formula by counting local degrees. First, consider how is mapped onto the face . The image of the center point of has a preimage consisting of two points – the center of the front and back face, respectively. Neighborhoods of these points are mapped homeomorphically onto a neighborhood of the image, so the local degree at each is . They must have opposite sign, because one neighborhood differs from the other by a reflection and a rotation, and this would multiply the degree by and respectively, yielding an opposite sign. Then, the degree of the map is . We get the same thing for the other faces, so each coefficient in the sum in the cellular boundary formula is zero, and .

Let’s compute homology. We have:

Now that we have the maps explicitly and their Smith forms, the homology is just a matter of tabulation: