# Alexander duality and homology equivalence

While reading the paper of Kauffman and Taylor "Signature of links" I found the following situation.

In the proof of Theorem 2.6 they suppose that two links $$L_1, L_2\in \mathbb{S}^3$$ are topologically concordant, i.e. that there exists a proper embedding (not necessarily locally flat) of a finite number of cylinders $$F\subset \mathbb{S}^3\times I$$ such that $$F\cap \partial({\mathbb{S}^3}\times I)$$ is $$L_1 \sqcup L_2$$. They want to study the manifold $$M=\mathbb{S}^3\times I\setminus F$$ and they claim that the inclusion in $$M$$ of each of its two boundary components (that are $$\mathbb{S}^3\setminus{L_1}$$ and $$\mathbb{S}^3\setminus{L_2}$$) is a homology equivalence mod $$2$$.

They say that this is a corollary of the Alexander Duality, which in this context states that $$H_i(\mathbb{S}^3\times I, M)\cong H^{4-i}(F, \partial F)$$, but I have no idea of how to show this implication.

• The assertion also holds with $\mathbb{Z}$ coefficients and the groups of interest are i=1,2. Here is an alternative to Alexander duality. Using the long exact sequence of the pair and excision, $H_1(S^3 \times I \setminus \nu F)=H_{2}(S^3 \times I,S^3 \times I \setminus \nu F)=H_{2}(\nu F,\partial \nu F)$ is generated by the meridians of the cylinders. These meridians are images of the meridians of the link under the inclusion $S^3\setminus \nu L_k \hookrightarrow S^3 \times I \setminus \nu F$. For $i=2$, the same type of argument works. – Anthony Conway Jan 28 at 7:30
• Thank you for your answer. One question: are you supposing that $F$ admits a tubular neighbourhood inside $S^3\times I$? Because I do not know whether it is true, since $F$ could be embedded in a non-locally flat way. – Diego95 Jan 28 at 9:51

Alexander Duality says that if $$X$$ is a nice enough subspace of $$S^n$$ then $$\tilde H^i(X)\cong \tilde H_{n-i-1}(S^n-X)$$. In this case, the top and bottom of the cobordism, $$(S^3,L_1)$$ and $$(S^3,L_2)$$ can be coned off, giving $$S^4$$ as the ambient space and $$X=F\cup c(L_1)\cup c(L_2)$$ as the subspace, with $$c(Z)$$ indicating the cone on $$Z$$. Since the cones are contractible, $$\tilde H^i(X)\cong H^i(F,\partial F)$$ from the long exact sequence of the pair. Putting together the pieces so far gives $$H^i(F,\partial F)\cong \tilde H_{4-i-1}(S^4-X).$$ Next, since $$\tilde H_{j}(S^4)=0$$ for all $$i<4$$, the long exact sequence of the pair gives $$\tilde H_j(S^4,S^4-X)\cong \tilde H_{j-1}(S^4-X)$$ for $$j\leq 3$$. So $$\tilde H_{4-i-1}(S^4-X)\cong \tilde H_{4-i}(S^4,S^4-X)$$ for $$4-i\leq 3$$, and $$S^4-X$$ is homotopy equivalent to your $$M$$. This gives the isomorphism you asked about since I believe the Kauffman-Taylor paper only needs it in this range of dimensions.

• Thank you for your answer. I have a doubt and a comment. 1) it seems to me that the isomorphism $\tilde{H}^i(X)\cong \tilde{H}^i(F,\partial F)$ holds when $i\ne 1$. In fact, a posteriori, I know that $\tilde{H}_{2}(S^4-X)$ has rank equal to $\mu-1$, where $\mu$ is the number of components of the link, but from your isomorphism I get that it has rank equal to $\mu$. 2) It is not clear to me of how from this argument I can show that the inclusion in $M$ of each of the two boundary components induces the required homology equivalence. – Diego95 Jan 27 at 9:38
• Oops, you're right about that isomorphism being off for the i=1 case, but looking at the Kauffman-Taylor paper, I think you only need it for i=2, at least if your question is about the isomorphism in the diagram at the top of page 353. I don't see where they talk about a homology equivalence, though. Can you provide a more precise reference? – Greg Friedman Jan 27 at 19:32
• Of course, I am referring to the proof of Theorem 2.6 at page 355. In any case, your idea of the cones is very useful and I managed to prove the homology equivalence with it. I will probably post an answer with the solution in a while. Thank you very much :) – Diego95 Jan 28 at 9:44

I found a solution for my problem. I post here a brief proof, in case anyone needs it.

I use the version of Alexander Duality, as stated in Bredon's book "Topology and Geometry" that states that if $$M$$ is a closed oriented $$n$$-manifold and $$K\subset L$$ are nice compact subspaces, then there is an isomorphism $$H_i(M\setminus{K}, M\setminus{L})\cong H^{n-i}(L, K)$$

Step 1: If $$F\subset D^4$$ is a proper (not necessarily locally flat) surface in the $$4$$-disc, then $$H_i(D^4\setminus{F}, S^3\setminus{\partial F})\cong H^{4-i}(D^4, F)$$.

Coning off $$(S^3, \partial F)$$ we obtain $$S^4\supset X$$, with $$X=F\cup c(\partial F)$$, where $$c(\partial F)$$ denotes the cone on $$\partial F$$. At this point we can use Alexander Duality to get $$H_i(S^4\setminus{X}, S^4\setminus{(D \cup X)})\cong H^{4-i}(D\cup X, X)$$ where $$D$$ is the complement of a small open collar of $$(S^3, \partial F)$$ inside $$D^4$$. To prove Step 1 one simply notes that the inclusion $$(D^4\setminus F, S^3\setminus \partial F)\hookrightarrow (S^4\setminus{X}, S^4\setminus{(X\cup D)})$$ induces isomorphisms in homology, and that by excision we have $$H^{4-i}(D\cup X, X)\cong H^{4-i}(D^4, F)$$

Step 2: If $$L_1, L_2$$ are two links in $$S^3$$ and $$F\subset S^3\times I$$ defines a (non necessarily locally flat) concordance between $$L_1$$ and $$L_2$$, then the inclusion of $$\partial_-M=S^3\setminus L_1$$ in $$M=(S^3\times I)\setminus F$$ induces a homology equivalence.

Consider the cone on $$(S^3, L_2)$$ so to obtain $$D^4\supset X$$, where $$X=F\cup c(\partial_+F)$$. Notice that $$\partial D^4\setminus \partial X=\partial_-M$$. Now apply step 1 to obtain $$H_i(D^4\setminus X, \partial_-M)\cong H^{4-i}(D^4,X).$$ One obtains the thesis observing that $$D^4\setminus X$$ is homotopy equivalent to M, and that since both $$D^4$$ and $$X$$ are contractibles, the homology groups $$H^{4-i}(D^4, X)$$ all vanish.