Prove that there is no compact \(4\)-manifold \(M\) (with or without boundary) which is homotopy-equivalent to the suspension \(\Sigma \mathbb{RP}^3\).
Solution
1. Suspension shifts reduced homology, so\[ \widetilde H_i(\Sigma\mathbb{RP}^3;\mathbb Z) \cong \widetilde H_{i-1}(\mathbb{RP}^3;\mathbb Z). \]Hence\[ H_2(\Sigma\mathbb{RP}^3)\cong\mathbb Z/2,\qquad H_4(\Sigma\mathbb{RP}^3)\cong\mathbb Z,\qquad H_1=H_3=0. \]
The integral homology of \(\mathbb{RP}^3\) is \(H_0=H_3=\mathbb Z\), \(H_1=\mathbb Z/2\), and \(H_2=0\).
2. If a compact \(4\)-manifold \(M\) were homotopy equivalent to this suspension, then \(H_4(M;\mathbb Z)\cong\mathbb Z\). Thus \(M\) would be closed and orientable.
A compact connected \(4\)-manifold with nonempty boundary has zero absolute top homology, and a closed nonorientable manifold has zero integral top homology.
3. We would also have \(H_1(M)=0\) and \(H_2(M)=\mathbb Z/2\).
Homotopy equivalence preserves homology; use 1.
4. For a closed oriented \(4\)-manifold with \(H_1=0\), the group \(H_2\) is torsion-free.
Poincaré duality gives \(H_2(M)\cong H^2(M)\). The universal coefficient sequence \[ 0\to\operatorname{Ext}(H_1(M),\mathbb Z)\to H^2(M)\to\operatorname{Hom}(H_2(M),\mathbb Z)\to0 \] has zero left term, so \(H^2(M)\) is free abelian. Hence \(H_2(M)\) is free abelian as well.
5. This contradicts \(H_2(M)\cong\mathbb Z/2\). Therefore\[ \boxed{\Sigma\mathbb{RP}^3\text{ is not homotopy equivalent to any compact }4\text{-manifold}.} \]
Combine <1>2–<1>4.