Let \(X\) be the topological space obtained as the quotient of the sphere \(S^2 = \theset{\vector x \in \RR^3 \suchthat \norm{\vector x} = 1}\) under the equivalence relation \(\vector x \sim -\vector x\) for \(\vector x\) in the equatorial circle, i.e. for \(\vector x = (x_1, x_2, 0)\).
Calculate \(H_* (X; \ZZ)\) from a CW complex description of \(X\).
Solution
Write \[ E=\{(x_1,x_2,0)\in S^2\} \cong S^1 \] for the equator, and let \(D_+\) and \(D_-\) be the closed upper and lower hemispheres.
1. The image of the equator in \(X\) is\[ E/(x\sim -x)\cong\RP^1\cong S^1. \]Give it a CW structure with one \(0\)-cell \(v\) and one \(1\)-cell \(e\).
The antipodal action on the circle is free, and the quotient map \[ q_E:S^1\longrightarrow S^1/(z\sim -z) \] is the standard two-sheeted covering. Identifying the quotient circle with \(S^1\) via \[ [z]\longmapsto z^2, \] the quotient map becomes \[ z\longmapsto z^2. \] Thus the quotient equator is a circle and has the stated one-vertex CW structure.
2. The interiors of \(D_+\) and \(D_-\) descend to two open \(2\)-cells \(e_+^2\) and \(e_-^2\) of \(X\).Hence \(X\) has a CW structure with\[ \#\{0\text{-cells}\}=1, \qquad \#\{1\text{-cells}\}=1, \qquad \#\{2\text{-cells}\}=2. \]
The equivalence relation identifies points only on the equator. Therefore the quotient map is injective on each open hemisphere, so each hemisphere interior maps homeomorphically onto an open disk in \(X\).
1.Together with the cells \(v,e\), these two hemisphere interiors exhaust \(X\).
3. After choosing orientations of the two \(2\)-cells suitably, both attaching maps have degree \(2\) on the \(1\)-skeleton.Thus\[ d_2:\ZZ e_+^2\oplus\ZZ e_-^2\longrightarrow\ZZ e \]is given by\[ d_2(a,b)=2a+2b. \]
The boundary circle of either hemisphere is the original equator \(E\). By 1, its attaching map is exactly the antipodal quotient \[ S^1\longrightarrow E/(x\sim -x)\cong S^1, \] which, under the coordinate \(z\mapsto z^2\), has degree \(2\).
The cellular boundary formula says that the coefficient of the unique \(1\)-cell is this degree. Reversing the chosen orientation of a \(2\)-cell reverses the sign of its coefficient, so we may orient both cells so that the two coefficients are \(+2\). Therefore \[ d_2=(2\ \ 2). \]
4. The cellular differential\[ d_1:\ZZ e\longrightarrow\ZZ v \]is zero.
The unique \(1\)-cell is a loop whose two endpoints are both the unique \(0\)-cell \(v\). Hence its cellular boundary is \[ v-v=0. \]
5. The cellular chain complex is therefore\[ 0 \longrightarrow \ZZ^2 \xrightarrow{\ (2\ \ 2)\ } \ZZ \xrightarrow{\ 0\ } \ZZ \longrightarrow0. \]
This is immediate from the cell counts in 2 and the boundary computations in 3 and 4.
6. The second homology group is\[ \boxed{H_2(X;\ZZ)\cong\ZZ.} \]
There are no \(3\)-cells, so \[ H_2(X)=\ker d_2. \] From 5, \[ \ker d_2 =\{(a,b)\in\ZZ^2:2a+2b=0\} =\{(a,-a):a\in\ZZ\} \cong\ZZ. \]
7. The first homology group is\[ \boxed{H_1(X;\ZZ)\cong\ZZ/2\ZZ.} \]
Since \(d_1=0\), \[ H_1(X) =\frac{\ker d_1}{\operatorname{im}d_2} =\frac{\ZZ}{2\ZZ}. \] Indeed, \[ \operatorname{im}d_2 =\{2a+2b:a,b\in\ZZ\} =2\ZZ. \]
8. Finally,\[ \boxed{ H_i(X;\ZZ) \cong \begin{cases} \ZZ, & i=0,2,\\ \ZZ/2\ZZ, & i=1,\\ 0, & i\ge3. \end{cases}} \]
The space has one \(0\)-cell and \(d_1=0\), so \[ H_0(X)\cong\ZZ. \] Steps 6 and 7 compute the remaining nonzero groups. There are no cells in dimensions at least \(3\), so the higher homology groups vanish.