Let \(X\) be \(S^1\) with two 2-cells attached via \(z\mapsto z^5\) and \(z\mapsto z^3\).
Compute \(\pi_1(X)\) and \(H^*(X)\) as a ring.
Show that \(X\) is not homeomorphic to \(S^2\).
Solution
\(X\) is path-connected, so \(H_0(X) = \ZZ\).
\(X\) is a CW complex with only one 1-cell, so \(\pi_1(X^{(1)}) = \ZZ\) for the 1-skeleton.
- Use that \(3\ZZ + 5\ZZ = 1\ZZ\) as a ring, so the attached cells kill off all 1-homotopy and \(\pi_1(X) = 1\).
By Hurewicz, \(H_1(X) = 1\).
By UCT, \(H^1(X) = \dualof{H_1(X)} = \Hom_\ZZ(1, \ZZ) = 1\)
By UCT, \(H^2(X) = \dualof{H_2(X)}\)
We can compute \(H_2\) directly: \[ C^*(X) = (0 \to C_2(X) \mapsvia{d_2} C_1(X) \mapsvia{d_1} C_0(X) ) = (0 \to \ZZ\adjoin{e_1, e_2} \mapsvia{\substack{e_1\mapsto 3e \\ e_2 \mapsto 5e} } \ZZ\adjoin{e} \mapsvia{d_1} \ZZ\adjoin{\pt} ) .\]
\(\ker d_2 = \gens{ 5e_1, -3e_2 }\), so \(H_2 = \ZZ\).
Thus \(H^*(X) = \Extalgebra_\ZZ(x)\) where \(\abs{x} = 2\).
\(X\not\cong S^2\): delete a point \(p\) in the interior of the 2-cell corresponding to \(z^3\), then use that \(S^2\sm\ts{\pt} \cong \RR^2\) is contractible but \(\pi_1 X\sm\ts{\pt} = \ZZ/5\).