The first homology group $ H_1(E(K); Z) $ of a knot exterior is an infinite cyclic group which is generated by the class of the meridian. I'm trying to solve the following exercice : 
Prove that the first homology group $H_1(E(K); Z)$ of a knot 
exterior is an infinite cyclic group which is generated by the class of the meridian. 
With $ E(K) $ is the closure of the complement of a tubular neigborhood $ N(K) $ of $ K $ and the merdian is just the boundary of a meridien disk of $ K $ .
Here is the adequate formulation :
Let $K$ denote a knot in $S^3$.
$ N(K) $ is the image of an embedding $ S^1 \times D^2 \longrightarrow S^3  $ whose inverse $ f : N(K) \longrightarrow S^1 \times D^2 $ is called a trivialization. A meridien disk of $K$ is $ f^{-1} (D^2 \times \lbrace y \rbrace ) $ for a point $ y \in K $ .
$ E(K) = cl(S^3 \setminus N(K)) $ .
I proved that $  E(K) $ is a submanifold with boundary of $ S^3 $ and that $ E(K) \cup N(K) = S^3 $. Then I used the known fact that the Mayer-Vietoris sequence still relevant for a cover of a manifold consisting of two submanifold with boundary to conclude that indeed $ H_1(E(K); \mathbb{Z}) = \mathbb{Z} $. 
The issue is that I was not able to get the information on the generator by applying this method.
So my question is how can I exhibit the generator ( knowing that normally there is two generators since the boundary of $ N(K) $ is a torus, a meridien and a longitude ).
Thanks in advance.
 A: This can be proved using the Mayer-Vietoris sequence, taking advantage of two facts: $T = N(K) \cap E(K)$ is a torus; and the induced map on homology 
$$\langle l \rangle \oplus \langle m\rangle \approx \mathbb{Z} \oplus \mathbb{Z} \approx H_1(T) \xrightarrow{f_1} H_1(N(K)) \approx H_1(K) \approx \mathbb{Z}
$$ 
is the quotient homomorphism that kills the meridian generator $m$ of $H_1(T)$ and that takes the longitude generator $l$ to a generator of $H_1(K)$ (subject to this restriction there is some freedom to choosing the longitude generator, and for purposes of this problem it does not matter how that choice is made).  
From the Mayer-Vietoris sequence we have:
$$\underbrace{H_2(S^3)}_0 \to H_1(N(K) \cap E(K)) \approx \underbrace{\langle l \rangle \oplus \langle m \rangle}_{\mathbb{Z} \oplus \mathbb{Z}} \xrightarrow{f \, = \, f_1 \, \oplus \, f_2} \underbrace{H_1(N(K))}_{\mathbb{Z}} \oplus \underbrace{H_1(E(K))}_{\text{?}} \to \underbrace{H_1(S^3)}_0
$$
The map $f : \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \text{?}$ is therefore an isomorphism. Combining this with what we know about the first coordinate map $f_1 : \langle l \rangle \oplus \langle m \rangle \to H_1(N(K))$, we may form the following conclusions, in order, for the second coordinate map $f_2 : \langle l \rangle \oplus \langle m \rangle \to H_1(E(K))$: 


*

*$H_1(E(K))$ is infinite cyclic (or else $\mathbb{Z} \oplus \mathbb{Z} \not\approx \mathbb{Z} \oplus \text{?}$);

*$f_2$ restricts to an isomorphism $\langle m \rangle \mapsto H_1(E(K))$ (or else $f$ is not an isomorphism). 


The homomorphism $f_2$ is, of course, induced by inclusion $T \hookrightarrow E(K)$, and hence the meridian curve of $T$ generates $H_1(E(K))$.
A: As @MikeMiller mentioned,  use a Wirtinger presentation. The Hurewicz homomorphism is sufficient to finish the problem after a tiny bit of algebra.  
Say $\Gamma = \pi_1 E(K) = \langle a_1, \ldots, a_n \mid r_1 \ldots r_n \rangle$. Define a function $\phi \colon \{a_i\} \to \mathbb{Z}$ by sending each $a_i$ to 1. Then $\phi$ extends to a homomorphism, since the image of each Wirtinger relation is 0. Notice $\phi$ is onto. 
If $\Gamma'$ is the derived subgroup of $\Gamma$, then $\Gamma' \subseteq \ker \phi$, so $\phi$ factors through a surjective homomorphism $\psi \colon \Gamma / \Gamma' \to \mathbb{Z}$ and the quotient homomorphism $\rho \colon \Gamma \to \Gamma / \Gamma'$ (ie $\phi = \psi \rho$).  Wirtinger generators are all conjugate, so they must have the same image in $\Gamma / \Gamma'$, as it is abelian. This shows $\Gamma / \Gamma '$ is cyclic, so $\psi$ must be an isomorphism and $\ker \phi = \Gamma'$.  
The Hurewicz homomorphism gives $H_1 E(K) \cong \Gamma / \Gamma'$ and shows that $H_1E(K)$ is generated by a meridian. 
A: just commute the relations in the fundamental group of K and since each relation has the form $x_i x_j=x_j x_k$   then if you commute you get $x_i=x_k=$generator of the first homology of K
