Compute \(\pi_1(X)\) where \(X \definedas S^2/\sim\), where \(x\sim -x\) only for \(x\) on the equator \(S^1 \injects S^2\).
- Hint: try cellular homology. Should yield \([\ZZ, \ZZ/2\ZZ, \ZZ, 0, \cdots]\).
Solution
1. The quotient of the equator by \(x\sim -x\) is again a circle; call its generator \(a\).
The antipodal quotient map \(S^1\to S^1/(x\sim -x)\) is the degree-\(2\) covering of a circle by a circle.
2. The northern and southern hemispheres descend to two \(2\)-cells, each attached to this quotient circle by a degree-\(2\) map.
Each hemisphere boundary is the original equator, and after passing to the quotient its attaching map is precisely the antipodal quotient \(S^1\to S^1/(\pm1)\), of degree \(2\) up to orientation.
3. Hence van Kampen gives\[\pi_1(X)\cong\langle a\mid a^2,a^2\rangle\cong\boxed{\mathbb Z/2}.\]
Attaching either \(2\)-cell kills \(a^2\); the second gives the same relation up to inversion.
4. The same CW complex has cellular boundary \(\partial_2:\mathbb Z^2\to\mathbb Z\) given by \((m,n)\mapsto2m\pm2n\), so\[H_2(X)\cong\mathbb Z,\qquad H_1(X)\cong\mathbb Z/2,\]in agreement with the hint.
The image is \(2\mathbb Z\) and the kernel has rank \(1\).