I'm dealing with the following question:
Let $V$ be the the image of an embedding $f: S^{1} \times D^{2} \rightarrow S^{3} $. Let $X = S^{3}-\textit{Int}(V)$ be the complement of its interior. Compute the homology groups $H_{*}(X ; \mathbb{Z})$ and $H_{*} (X, \partial X ; \mathbb{Z}).$
My work:
The Mayer Vietoris sequence with
$$ A = f(S^1 \times D^2) \cup { \textit{small neighborhood} } $$
$$ B = S^3 – A $$
we get that $$ A \cup B = S^1 $$
$$ A \cap B \simeq T^2 $$
If we want to use Mayer-Vietoris to solve for just the homology group, then the sequence:
$$ H_{2}(A \cup B) \rightarrow H_1 (A \cap B) \rightarrow H_1 (A) \oplus H_1 (B) \rightarrow H_1 (A \cup B) $$
becomes
$$ 0 \rightarrow \mathbb{Z}^{2} \rightarrow \mathbb{Z} \oplus H_1 (S^3 – f(S^1 \times D^2)) \rightarrow 0 $$
which implies that
$$ H_1 (S^3 – f(S^1 \times D^2 ) = \mathbb{Z} $$
using the Mayer-Vietoris sequence, we get:
$$H_2 (S^3) \rightarrow H_1 (T^2) \rightarrow H_1 (A) \oplus H_2 (Y) \rightarrow H_1 (S^3) $$
the first and last groups are trivial and
$$ H_1 (T^2) \simeq \mathbb{Z}^2, $$ so $$H_1 (Y) \simeq \mathbb{Z} $$
I know that my proof is inadequate in many ways, so I would appreciate any correction and guidance that can be provided. Where am I going wrong and in what ways have I left the problem incomplete?
Best Answer
Your notation is a bit overcomplicated (for example I think that $X$ changed to $Y$ at some point), but the computation of $H_1(X)$ looks fine to me . Mayer-Vietoris will also give $b_{2}(X)=0$, since the relavant terms are $$H_3(S^3) =\mathbb{Z} \rightarrow H_{2}(T^2)=\mathbb{Z} \rightarrow H_{2}(A) \oplus H_2(B) \rightarrow H_{2}(S^3) = 0 $$all of the higher Betti numbers vanish by standard facts since $X$ is an open $3$-manifold.
For a reality check you may consider the Hopf fibration $\pi:S^3 \rightarrow S^2$, and consider open sets $U_{1} , U_{2}$ which are neighbourhoods of north and south hemispheres , so that they are both homeomorphic to $D^2$ and intersect in an open annulus.
Then $\pi^{-1}(U_{1}) \cap \pi^{-1}(U_{2})$ is homotopy equivelant to $T^2$ and $S^3 = \pi^{-1}(U_{1}) \cup \pi^{-1}(U_{2})$. Also clearly $$\pi^{-1}(U_{i}) \cong D^2 \times S^1 $$ for $i=1,2$. So both of the trivialisations $\pi^{-1}(U_{i})$ have $H_{1} = \mathbb{Z}$ and $H_2 = 0$.