Generalizing the example of \(\mathbb{RP}^2\) and \(S^2 \times \mathbb{RP}^\infty\), show that if \(X\) is a connected finite-dimensional CW complex with universal cover \(\tilde{X}\), then \(X\) and \(\tilde{X} \times K(\pi_1(X), 1)\) have isomorphic homotopy groups but are not homotopy equivalent if \(\pi_1(X)\) contains elements of finite order.
Solution
1. \(X\) and \(\tilde X \times K(\pi_1(X), 1)\) have isomorphic homotopy groups.
1.1. \(\pi_1(\tilde X \times K(\pi_1(X),1)) = \pi_1(\tilde X) \times \pi_1(K(\pi_1(X),1)) = 1 \times \pi_1(X) = \pi_1(X)\).
\(\tilde X\) is simply connected and \(K(\pi_1(X),1)\) has fundamental group \(\pi_1(X)\).
1.2. For \(n \ge 2\), \(\pi_n(\tilde X \times K(\pi_1(X),1)) = \pi_n(\tilde X) \times \pi_n(K(\pi_1(X),1)) = \pi_n(\tilde X) \times 0 = \pi_n(\tilde X)\).
\(K(\pi_1(X),1)\) has vanishing higher homotopy groups.
1.3. \(\pi_n(X) = \pi_n(\tilde X)\) for \(n \ge 2\).
the universal cover \(\tilde X \to X\) induces isomorphisms on \(\pi_n\) for \(n \ge 2\).
1.4. Hence \(\pi_n(X) \cong \pi_n(\tilde X \times K(\pi_1(X),1))\) for all \(n\).
<2>1–<2>3.
2. \(X\) and \(\tilde X \times K(\pi_1(X),1)\) are not homotopy equivalent if \(\pi_1(X)\) has an element of finite order.
2.1. \(X\) is finite-dimensional, so \(\pi_n(X) = 0\) for all sufficiently large \(n\).
a finite-dimensional CW complex has finitely many nonzero homotopy groups in the sense that \(\pi_n(X) = 0\) for \(n > \dim X\) (by cellular approximation, since \(S^n\) has no cells below dimension \(n\)… more precisely, \(\pi_n(X) = 0\) for \(n > \dim X\)).
2.2. But \(K(\pi_1(X),1)\) has infinitely many nonzero homotopy groups when \(\pi_1(X)\) has an element of finite order.
if \(\pi_1(X)\) has an element of finite order, then \(K(\pi_1(X),1)\) has nonzero homology (hence nonzero homotopy) in infinitely many dimensions (e.g. \(\ZZ/m\) has \(H_{2k+1}(K(\ZZ/m,1)) = \ZZ/m\) for all \(k\)).
2.3. Hence \(\tilde X \times K(\pi_1(X),1)\) has nonzero homotopy groups in infinitely many dimensions.
2 and the product structure.
2.4. Therefore \(X\) and \(\tilde X \times K(\pi_1(X),1)\) are not homotopy equivalent.
homotopy equivalence preserves homotopy groups, but 1 and 3 contradict.
3. Q.E.D.
1 and 2.