Characterisation of longitude and meridian of Knot groups $V = V(k)$ denotes a tubular neighbourhood of the knot $k$ and $C = S^{3}− V$ is called
the complement of the knot. $H_{j}$ will denote the (singular) homology with coefficients
in $\mathbb{Z}$

Choose the homology class of $k$ as a generator of
$H_1(V)$ and represent it by a simple closed curve l on ∂V which is homologous to 0 in
$H_1(C)$. These conditions determine the homology class of l in $∂V$ ; hence, l is unique
up to isotopy on $∂V$ . A generator of $H_1(C)$ can be represented by a curve m on $∂V$
that is homologous to 0 in $V$ . The curves l and m determine a system of generators
of $H_{1}(∂V ) $= $\mathbb{Z} ⊕ \mathbb{Z}$. By a well-known result, we may assume that m is simple and
intersects l in one point. As m
is homologous to 0 in $V$ it is nullhomotopic in $V$ , bounds a disk, and is a meridian of
the solid torus $V $. These properties determine m up to an isotopy of $∂V$ .
A consequence is that l and k bound an annulus $A ⊂ V $.

With respect to the complement $C$ of a knot the longitude l and the meridian
m have quite different properties: The longitude l is determined up to isotopy and
orientation by $C$ ; this follows from the fact that l is a simple closed curve on $∂C$ which
is not homologous to 0 on $∂C$ but homologous to 0 in $C$. The meridian m is a simple
closed curve on $∂C$ that intersects l in one point; hence, l and m represent generators
of $H_{1}(C) $= $\mathbb{Z}^{2}$. The meridian is not determined by $C$ because simple closed curves on
$∂C$ which are homologous to $m^{±1}l^{r}$, r ∈ $\mathbb{Z}$, have the same properties.


My Question is precisely the last statement, why the meridian isn't determined by $C$?
 A: Yes, the homology groups $H_1(\partial V)$ and $H_1(C)$ along with the map $H_1(\partial V)\to H_1(C)$ only characterize the meridian slope up to some number of longitudes.  There is a general principle called "half lives half dies," which is that for a compact oriented $3$-manifold $M$, the kernel of $H_1(\partial M;\mathbb{Q})\to H_1(M;\mathbb{Q})$ is half-dimensional.  In the case of a torus boundary, like a knot complement, then the kernel is one-dimensional so is generated by a well-defined simple closed curve -- the longitude.  There is ambiguity, however, in the meridian.
The second inclusion $H_1(\partial V) \to H_1(V)$ is what pins down the meridian, again by the generator of the kernel.
The meridian problem is this: given $C$, how do we glue in $V$ to yield $S^3$? The core of $V$ then corresponds to a knot, and the meridian of $V$ is the meridian.  A priori, we also can't be sure that there aren't multiple ways to glue in $V$ that yield different knots, though.
The data of gluing in $V$ ends up, up to homeomorphism, only depending on where a meridian of $V$ is glued into the boundary of $C$ (a procedure called Dehn filling).  This cannot be solved purely homologically, in the sense that for each possible meridian $m^{\pm 1}l^r$, if you glue in $V$ according to this loop you get a space whose homology is the same as that of $S^3$.  That is to say, each of them gives you an integer homology $S^3$, often denoted as a $\mathbb{Z}HS^3$.  For example, you can obtain the Poincaré homology sphere by using $m^{\pm 1}l^{\pm 1}$ (sorry, I forget which!) on a trefoil knot.
That's not to say the problem is not solvable.  Gordon and Luecke proved that there is precisely one boundary slope whose Dehn filling yields $S^3$.  Thus, for knots the meridian is completely determined by $C$.  It is a very delicate argument, and my understanding is that through some careful combinatorial analysis, one shows that if there were two different slopes whose Dehn fillings yield $S^3$, then one could find that $S^3$ had a Lens space connect summand, which violates the fact that $H_1(S^3)=0$.
Gordon, C. McA.; Luecke, J., Knots are determined by their complements, Bull. Am. Math. Soc., New Ser. 20, No. 1, 83-87 (1989). ZBL0672.57009.
For links of more than one component, it turns out that $C$ does not determine the meridians, and there are infinitely many links with the same complement.
