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According the the original definition by Smythe, a homology boundary link $L\subseteq S^3$ with $m$ components is a link which sarisfies one of the following equivalent conditions:

(1): The fundamental group $\pi_1 (S^3\setminus L)$ of the link complement surjects onto the free group $F_m$ of rank $m$;

(2): There exists a collection $S_1,\ldots,S_m$ of disjoint oriented surfaces properly embedded in the complement $M$ of a tubular neighbourhood of $L$ such that for every $i=1,\ldots,m$, the boundary of $S_i$ is homologous in $\partial M$ to the $i$-th longitude of $L$.

It is not difficult to show that (2) implies (1): the homological condition implies that $S_1,\ldots,S_m$ are linearly independent as elements of $H_2(M;\partial M;\mathbb{Z})$, so the complement of the collection of surfaces is connected, and we may construct a map on the bouquet of $m$ copies of $S^1$ inducing an epimorphism on fundamental groups.

On the other hand, if (1) holds one may construct a family of disjoint surfaces that represent a basis of $H_2 (M;\partial M;\mathbb{Z})$. However, it could happen that these surfaces do not satisfy (2): for example, if $m=2$ it could happen that $\partial S_1$ is equal to 3 times the first longitude plus 2 times the second one, while $\partial S_24 is equal to 2 times the first longitude plus the second one. How could I trade such a system of surfaces for a system satisfying (2)?

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I think this is a version of Brown representability: mathoverflow.net/questions/11458/… The construction Ryan points out there, which Serre and Thom used, looks to me like it carries through to this context with no change. –  Daniel Moskovich Dec 22 '10 at 12:23

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This is easy.

if (1) holds one may construct a family of disjoint surfaces that represent a basis of $H_2(M;\partial M;\Bbb Z)$

Indeed; but in order to do so, one has to identify the free group $F_m$ with the fundamental group of the wedge of $m$ circles. This identification depends on a choice of basis of $F_m$. Given one choice $(x_1,\dots,x_m)$, other choices can be obtained by Nielsen transformations:

a$_{ij}$) $x_i\mapsto x_{ij}$, $x_k\mapsto x_k$ for $k\ne i$;

b$_i$) $x_i\mapsto x_i^{-1}$, $x_j\mapsto x_j$ for $j\ne i$.

One may also renumber the $x_i$'s.

Now the given epimorphism $\pi_1(S^3\setminus L)\to F_m$ descends to an isomorphism $h$ between the abelianizations. The first abelianization, $H_1(S^3\setminus L)$, has a basis given by a choice of meridians of the link; and a basis for the second is given by our basis $(x_i)$ of the free group. Then $h$ corresponds to an $m\times m$ matrix over $\Bbb Z$ with determinant $\pm 1$. Nielsen transformation of the $x_i$'s induce elementary row transformations of this matrix. They suffice to diagonalize the matrix, with $\pm 1$ on the diagonal. The result is that, geometrically, the $i$th meridian goes homologically onto the $i$th circle of the wedge with some orientation. Apply the Pontryagin construction now, and the surfaces will satisfy (2).

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