Homology of 3-sphere minus an embedding of $S^1 \times \mathbb{D}^2$
As suggested by @Lee Mosher, using the Mayer-Vietoris sequence with $A = \phi(S^1 \times \mathbb{D}^2) \cup \{ small \textrm{ } nbhood \}$ and $B = S^3-A$, we get that $A \cup B = S^3$ and $A \cap B \simeq T^2$. If we want to use Mayer-Vietoris to solve for just the first homology group say, then the sequence $$ H_2(A \cup B) \to H_1(A \cap B) \to H_1(A) \oplus H_1(B) \to H_1(A \cup B) $$ becomes $$ 0 \to \mathbb{Z}^2 \to \mathbb{Z} \oplus H_1(S^3-\phi(S^1 \times \mathbb{D}^2)) \to 0, $$ which implies that $H_1(S^3-\phi(S^1 \times \mathbb{D}^2)) = \mathbb{Z}$.
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msteve
Updated on July 25, 2020Comments
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msteve over 3 years
I'm having trouble with the following past qual question: Let $\phi \colon S^1 \times \mathbb{D}^2 \hookrightarrow S^3$ be an embedding, where $\mathbb{D}^2$ is the open unit disk in $\mathbb{R}^2$. Compute the homology groups of $S^3-\phi(S^1 \times \mathbb{D}^2)$.
My thought was to consider $\mathbb{R}^3$ minus some solid torus, and realize the space in question as the 1-point compactification of this space. This, however, has not been terribly successful so far.
Thanks in advance for any help or hints!
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Lee Mosher over 9 yearsHave you tried the Mayer-Vietoris theorem?
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msteve over 9 yearsI haven't actually: just taking $A = \phi(S^1 \times \mathbb{D}^2)$ and $B = S^3-A$ (plus open nbhoods, where required), so that $S^3 = A \cup B$?
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John Hughes over 9 yearsOne cheap answer: since you know that the answer doesn't depend on $\phi$ (by the nature of the question), you can pick $\phi$ to be the standard embedding of the solid torus, whose complement is another (open) solid torus; that retracts onto $S^1$...
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Stefan Hamcke over 9 yearsA theorem says that $\tilde H_i(S^n-h[S^k])$ is $\Bbb Z$ for $i=n-k-1$ and $0$ otherwise, where $h$ is an embedding, see Hatcher, Algebraic Topology, chapter 2.B. I think the $S^3-h[S^1\times \Bbb D^2]$ is a deformation retract of $S^3-h[S^1\times\{0\}]$.
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Daniel Valenzuela over 9 yearsYour question is actually: what is the homology of a knot complement. So is the embedding considered to be tame, e.g. smooth? Or are wild knots allowed?
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John Hughes over 9 yearsI don't see how to embed the $D^2$ factor for a wild knot...I think that the embedding is automatically a tubular neighborhood of a tame knot.
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Daniel Valenzuela over 9 yearsokay that's true but I don't see why your comment above should be true.
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