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.
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.