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Any hint is appreciated.

Let $S^3 = \{x=(x_1,x_2,x_3,x_4) \in \mathbb{R}^4 : |x|=1\}$, $A = \{(x_1,x_2,x_3,x_4) \in S^3 : {x_1}^2+{x_2}^2 \leq {x_3}^2+{x_4}^2 \}$ and $B = \{(x_1,x_2,x_3,x_4) \in S^3 : {x_3}^2+{x_4}^2 \leq {x_1}^2+{x_2}^2 \}$. We know that $S^3 =A \cup B$ and $A \cap B$ is a torus, and both $A$ and $B$ are solid tori. Let $K = \{(x_1,x_2,x_3,x_4) \in A : x_1=x_2=0 \}$ be the "unknotted circle", we see the boundary of $A$ is a deformation retract of $A \setminus K$. Why $B$ is the deformation retract of $S^3 \setminus K$?

Thanks.

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It follows from what you just said: removing the center circle from solid torus $A$, one deformation retracts to $\partial A$. So $S^3 \setminus K = (A \setminus K) \cup B$ deformation retracts to $\partial A \cup B$. Because $B$ is glued to $A$ by a homeomorphism $\partial A \to \partial B$, this is just literally $B$.

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  • $\begingroup$ But the boundary of $A$ is a torus, right? $\endgroup$ – Aleph-null Dec 23 '16 at 16:48
  • $\begingroup$ Did you mean $S^3 \setminus K$ deformation retracts to $\partial A \cup B$? $\endgroup$ – Aleph-null Dec 23 '16 at 17:24
  • $\begingroup$ @AlPinky Edited, thanks. Also, +1 on your own summary because it's absolutely correct. $\endgroup$ – Balarka Sen Dec 24 '16 at 0:20
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I think I know now what's going on:

We see that $K \subset A$ and $K \cap B =\emptyset $, so $S^3 \setminus K=(A \setminus K) \cup B$. Obviously, $A \setminus K$ deformation retracts to $\partial A = \partial B = \{(x_1,x_2,x_3,x_4) \in S^3 : {x_1}^2+{x_2}^2 = {x_3}^2+{x_4}^2=1/2 \}=A \cap B $. Therefore $S^3 \setminus K$ deformation retracts to $\partial A \cup B = (A \cap B) \cup B=B$. We note that the solid torus $B$ deformation retracts to its central circle, so $\pi_1(S^3 \setminus K)=\mathbb{Z}$.

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