Homology of knot complement

Question

I was told in a topology class that if $Y$ is a closed $3$-manifold and $K$ is a null-homologous knot in $Y$, then $H_1(Y- \nu(K)) \cong H_1(Y) \oplus \mathbb{Z}$.

I'm trying to prove this assertion, but have been unsuccessful so far. I'm trying to work with the Mayer-Vietoris sequence applied to $Y-\nu(K)$ and a neighborhood of $K$, which will be isomorphic to $S^1 \times D^2$. The sequence gives

$$ \ldots \rightarrow H_2(Y) \rightarrow H_1(T^2) \rightarrow H_1(S^1 \times D^2) \oplus H_1(Y-\nu(K)) \rightarrow H_1(Y) \rightarrow \ldots $$

and the null-homologous condition tells us that $H_1(S^1 \times D^2) \rightarrow H_1(Y)$ is the zero map. If one could show that the connecting homomorphism were $0$, and there was a splitting of the resulting short exact sequence, then things would start to look good, but I'm not making any progressing towards that.

Answer 1

Note, as $K$ is null-homologous there exists some $3$-disk $D\subset Y$ which bounds the tubular neighourhood $\nu(K)$ of the knot $K$ (I think this is the case anyway - I could be wrong).This is incorrect! There exists null-homologous curves in $Y$ which are not bounded by a disk. The rest of this answer only applies if such a disk exists unfortunately.

Now you only need to apply Mayer-Vietoris on the two subsets $\nu(Y\setminus D)$ and $D\setminus K$ whose intersection is homeomorphic to the sphere $S^2$.

It is well known that the first homology group of a knot complement $\mathbb{R}\setminus K$ is isomorphic to the integers $\mathbb{Z}$ (which can also be computed using a Mayer-Vietoris sequence), and so the calculation falls out quite nicely.

Answer 2

I will developp my comment here: Assume that $Y$ is a closed $3$-manifold with rational sphere homology and let $K$ be a null homologous knot in $Y$.

Let $\nu(K)$ be a neighborhood of $K$ homeomorphic to a solid torus $T=\mathbb D^2\times S^1$. I will denote by $X_K$ the knot complement : $X_K=Y\backslash \stackrel{\ \circ}{\nu(K)}$.

There is a long exact sequence corresponding to the pair $(Y,X_K)$ over $\mathbb Z$ coefficients: $$\cdots \rightarrow H_2(Y) \rightarrow H_2(Y,X_K)\rightarrow H_1(X_K)\rightarrow H_1(Y)\rightarrow H_1(Y,X_K) \rightarrow\cdots$$ and we will show that $H_2(Y)=0$ while $H_1(Y,X_K)\cong \mathbb Z$.

• By Excision Theorem, there is an isomorphism $$H_*(Y,X_K) \cong H_*(\nu(K),\partial \nu(K)).$$ Since $\nu(K)$ is a solid torus, we can use the Lefschetz duality that tells us : $$H_k(T,\partial T)\cong H^{3-k}(T)\cong\left\{\begin{array}{ll} 0 & \text{ if } k=0,1\\ \mathbb Z & \text{ if } k=2,3\end{array}\right. .$$

• Assuming that $H_1(Y,\mathbb Q)=0$ implies now that $H_1(Y)$ is torsion and thus $$H_2(Y)\cong H^1(Y)\cong Hom(H_1(Y),\mathbb Z)=0.$$

Going back to the exact sequence we get : $$\cdots 0 \rightarrow \mathbb Z \rightarrow H_1(X_K) \rightarrow H_1(Y) \rightarrow 0 \rightarrow\cdots$$

Morover, $K$ being null homologous in $Y$ implies that the knot $K$ bounds a disk $D$ in $Y$. Let's $s$ be the map $$s:H_1(X_K)\rightarrow \mathbb Z$$ given by the intersection of curves in $X_K$ with the disk $D$.

It is a morphism and looking at meridian and longitude of $X_K$ and $\nu(K)$ you can see that this is a section of the exact sequence hence giving you the isomorphism $$H_1(X_k)\cong H_1(Y)\oplus \mathbb Z.$$

Anyway with the above observations, you should also be able to understand the connecting homomorphism and the desired section you were looking for in the case of using Mayer-Vietoris exact sequence.