Use the Mayer–Vietoris sequence to show that a nonorientable closed surface, or more generally a finite simplicial complex \(X\) for which \(H_1(X)\) contains torsion, cannot be embedded as a subspace of \(\mathbb{R}^3\) in such a way as to have a neighborhood homeomorphic to the mapping cylinder of some map from a closed orientable surface to \(X\). [This assumption on a neighborhood is in fact not needed if one deduces the result from Alexander duality in §3.3.]
Solution
1. Suppose \(X\) embeds in \(\RR^3\) with a neighborhood \(N\) homeomorphic to the mapping cylinder of a map \(f: S \to X\) from a closed orientable surface \(S\).
assume the contrary.
2. \(N\) is a compact \(3\)-manifold with boundary \(S\) (the closed orientable surface).
the mapping cylinder of \(f: S \to X\) has boundary \(S\) (the “top” of the cylinder), and \(N\) is a regular neighborhood of \(X\).
3. Apply Mayer–Vietoris to \(\RR^3 = N \cup (\RR^3 \setminus \operatorname{int} N)\) with intersection \(S\).
decompose \(\RR^3\) into the neighborhood \(N\) and its complement, meeting along the boundary surface \(S\).
4. The relevant part of the Mayer–Vietoris sequence is \[H_2(\RR^3) \to H_1(S) \to H_1(N) \oplus H_1(\RR^3 \setminus \operatorname{int} N) \to H_1(\RR^3).\] Proof: the Mayer–Vietoris sequence in low degrees.
5. \(H_2(\RR^3) = 0\) and \(H_1(\RR^3) = 0\).
\(\RR^3\) is contractible.
6. Hence \(H_1(S) \to H_1(N) \oplus H_1(\RR^3 \setminus \operatorname{int} N)\) is injective.
4 and 5 (the map is injective since its kernel is the image of \(H_2(\RR^3) = 0\)).
7. \(H_1(S)\) is free abelian (since \(S\) is a closed orientable surface).
\(H_1\) of a closed orientable surface of genus \(g\) is \(\ZZ^{2g}\).
8. \(H_1(N) \cong H_1(X)\) (the mapping cylinder deformation retracts onto \(X\)).
a mapping cylinder deformation retracts onto its base.
9. The map \(H_1(S) \to H_1(N) \oplus H_1(\RR^3 \setminus \operatorname{int} N)\) is an isomorphism.
it is injective (<1>6), and its cokernel maps into \(H_1(\RR^3) = 0\) (<1>5), so it is also surjective.
10. Hence \(H_1(N) = H_1(X)\) is a direct summand of the free abelian group \(H_1(S)\).
9 and 7; a direct summand of a free abelian group is free abelian.
11. But \(H_1(X)\) contains torsion, so it is not free abelian.
hypothesis.
12. Contradiction.
10 and 11.
13. Hence no such embedding exists.
<1>1–<1>12.
14. Q.E.D.
13.