Alexander duality and homology equivalence

Question

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.

Answer 1

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.

Answer 2

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.