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)?



Homology boundary links
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).


