A space obtained by $S^3$ by removing a Hopf link If we remove a Hopf link from $3$-dimensional sphere $S^3$, can we obtain a space homotopy equivalent to (or deformation retract to) an annulus?
If the answer yes, can we write it explicitly? 
EDIT: Let $H^+$ denote the positive Hopf link. If we remove $H^+$ from $S^3$, we obtain a Milnor fibration $\pi_+: S^3-H^+ \to S^1$ given by the rule $(r_1,\theta_1,r_2,\theta_2) \to \theta_1+\theta_2$, see Etnyre's note. This gives an open book decomposition of $S^3$ with a page annulus. 
 A: There seem to be a couple of confusions here. 
The Hopf fibration is a presentation of $S^3$ as an (oriented) circle bundle over $S^2$. In particular, when you delete the north and south pole, you have a circle bundle over $\Bbb R \times S^1$; projecting once more to the circle factor, you have an annulus bundle over $S^1$. This is the open book decomposition you cite. 
You can not conclude from this that the total space is homotopy equivalent to a generic fiber of this bundle: it's simply not true! You're ignoring the homotopy type of the base. Instead, what you are able to actually say is that the total space is homotopy equivalent to a circle bundle over $S^1$ (start with the circle bundle over $S^1 \times \Bbb R$, and then crush the $\Bbb R$ factor to a point). In fact, the whole bundle over $S^1 \times \Bbb R$ is trivial (exercise: write down a global trivialization), and hence $S^3 \setminus H \cong S^1 \times \Bbb R \times S^1$. As a corollary, you find the description in one of the comments that $\pi_1(S^3 \setminus H) = \pi_1 T^2 = \Bbb Z^2$. 
As a brief comment on the difference between $S^3$ and $\Bbb R^3$, here are a few general principles. First, deleting a set of codimension larger than $2$ does not change the fundamental group. Second, you can explicitly understand the homotopy type of $\Bbb R^3 \setminus L$ given that of $S^3 \setminus L$: in fact, we have a homotopy equivalence $$\Bbb R^3 \setminus L \simeq (S^3 \setminus L) \vee S^2.$$ (This is a general fact: if $M$ is a noncompact $n$-manifold, then $M \setminus \{pt\} \simeq M \vee S^{n-1}$. Argue by showing that $M$ deformation retracts onto an $(n-1)$-dimensional subcomplex, but even better, that it deformation retracts onto something of the form $M' \vee B^n$, where $M'$ is a subcomplex of dimension $(n-1)$; then puncturing in the interior of $B^n$, we obtain a deformation retraction onto $M' \vee S^{n-1}.$)
A: The Hopf link is a torus link, it's complement (as the one of any algebraic link) fibers over the circle with fiber a minimal genus Seifert surface. More explicitly $S^3 \setminus \nu (L)$ is diffeomorphic to $A\times [0,1]/\sim$, where $A=S^1 \times [0,1]$ denotes the annulus, and $\sim$  the equivalence relation identifying $(x,1) \sim (\tau_\gamma(x), 0)$ where $\tau_\gamma:A \to A$ denotes a Dhen twist along $\gamma= S^1 \times \{1/2\}$. Thus $S^3 \setminus \nu (L)$ is homotopic equivalent to the two-torus $S^1 \times S^1$
In general $S^3 \setminus \nu (K_{p,q})$, where $K_{p,q}$ denotes the $(p,q)$ torus knot, is homotopic equivalent to $F_g \times S^1$ where $F_g$ denotes a bouquet of $g=(p-1) (q-1)$ circles.    
