# Homology Groups and an embedding

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?

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 $$H_2 = 0$$.