Theorem. <\/strong>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 Corollary. <\/strong>(Alexander duality in $S^n$) Let $K\\subset S^n$ be a subcomplex of $S^n$ under some finite cell structure. Then there is a canonical isomorphism Using Alexander duality, the computation of homology groups of knot complement in $S^3$ is simple.<\/p>\n<\/blockquote>\n\n\n\n e.g.<\/strong> Suppose that $K\\subset \\mathbb{R}^3$ is a knot. Then by Alexander duality in $S^3$, Using Poincare duality on the manifolds with boudary, prove the Alexander duality on n-sphere and calculate the homology group of knot complement. \u7ee7\u7eed\u9605\u8bfb Alexander\u5bf9\u5076\u548c\u7ebd\u7ed3\u8865\u7a7a\u95f4\u7684\u540c\u8c03\u7fa4<\/span>
$$
H^i(M,\\partial M)\\overset{\\cdot\\cap[M]}{\\longrightarrow} H_{n-i}(M)
$$
and
$$
H^i(M)\\overset{\\cdot\\cap[M]}{\\longrightarrow} H_{n-i}(M,\\partial M)
$$
are both isomorphisms.<\/p>\n\n\n\n
\n\n\n\nAlexander duality<\/h2>\n\n\n\n
$$
\\tilde{H}^q(K)\\cong\\tilde{H}_{n-q-1}(S^n-K)
$$
Proof.<\/strong> Suppose that $N(K)$ is a regular neighborhood of $K$. Then $\\forall q>0$, 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 $(S^n,S^n-K)$, and for $q\\neq 0$, we have $H_{n-q}(S^n)=0$. Therefore, the case for $q\\neq 0$ has been established. For $q=0$, notice that
$$
H_{n-1}(S^n-K)\\oplus\\mathbb{Z}\\cong H_n(S^n,S^n-K)
$$
and
$$
\\tilde{H}^0(K)\\oplus\\mathbb{Z}\\cong H^0(K)
$$
Still using the above result: $H^0(K)\\cong H_n(S^n,S^n-K)$, we have
$$
\\tilde{H}_{n-1}(S^n-K)\\cong \\tilde{H}^0(K).
$$
This completes the case for $q=0$. $\\square$<\/p>\n\n\n\n
\n\n\n\nKnot complement<\/h2>\n\n\n\n
\n
$$
H_1(S^3-K)\\cong H^1(K)\\cong H^1(S^1)\\cong\\mathbb{Z}
$$
and
$$
H_2(S^3-K)\\cong \\tilde{H}^0(K)=0.
$$<\/p>\n","protected":false},"excerpt":{"rendered":"