Alexander对偶和纽结补空间的同调群

Poincare duality with boundary

Theorem. Let M be an n-dimensional compact, oriented manifold with boundary, and $\[M\]\in H_n(M,\partial M)$ be its natural orientation class. Then
$$ H^i(M,\partial M)\overset{\cdot\cap\[M\]}{\longrightarrow} H_{n-i}(M) $$
and
$$ H^i(M)\overset{\cdot\cap\[M\]}{\longrightarrow} H_{n-i}(M,\partial M) $$
are both isomorphisms.

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Alexander duality

Corollary. (Alexander duality in Sⁿ) Let K ⊂ Sⁿ be a subcomplex of Sⁿ under some finite cell structure. Then there is a canonical isomorphism
H̃^q(K) ≅ H̃_{n − q − 1}(Sⁿ − K)
Proof. Suppose that N(K) is a regular neighborhood of K. Then ∀q > 0, using Poincare duality with boundary:
$$ \begin{aligned} \tilde{H}^q(K) &\cong H^q(K)\\\\ &\cong H^q(N(K))\\\\ &\cong H_{n-q}(N(K),\partial N(K))\\\\ &\cong H_{n-q}(S^n,S^n-K)\quad\\\\ &\cong H_{n-q-1}(S^n-K) =\tilde{H}_{n-q-1}(S^n-K). \end{aligned} $$
The last equality is because of the long exact sequence of homology induced by the pair (Sⁿ, Sⁿ − K), and for q ≠ 0, we have H_n − q(Sⁿ) = 0. Therefore, the case for q ≠ 0 has been established. For q = 0, notice that
H_n − 1(Sⁿ − K) ⊕ ℤ ≅ H_n(Sⁿ, Sⁿ − K)
and
H̃⁰(K) ⊕ ℤ ≅ H⁰(K)
Still using the above result: H⁰(K) ≅ H_n(Sⁿ, Sⁿ − K), we have
H̃_n − 1(Sⁿ − K) ≅ H̃⁰(K).
This completes the case for q = 0. ▫

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Knot complement

Using Alexander duality, the computation of homology groups of knot complement in S³ is simple.

e.g. Suppose that K ⊂ ℝ³ is a knot. Then by Alexander duality in S³,
H₁(S³ − K) ≅ H¹(K) ≅ H¹(S¹) ≅ ℤ
and
H₂(S³ − K) ≅ H̃⁰(K) = 0.