Homotopy equivalent prove I want to know how to prove the next homotopy equivalent: between the set $\mathbb{R}^3\setminus ((\{0\}\times\{0\}\times\mathbb{R})\cup (\mathcal{S}^1\times\{0\}))$ and the torus $\mathcal{S}^1\times\mathcal{S}^1$. The homotopy equivalent has sense but i'm not capable to find the homotopy function. If anyone could help me i'll be very grateful! Thank you very much!
 A: Do it as a composition of homotopy equivalences:
First shrink the "outside" to $S^1\times D$ (with $D$ a disc), then expand the circle $S^1\times \{0\}$ missing inside to $S^1\times D$. The composition then gives a homotopy with the surface of $S^1\times D$, i.e. $S^1\times S^1$.
The second homotopy should be clear, but how do we construct the first one?
For a fixed point $x$ in $S^1\times \{0\}$ take for each point $y$ in $\{0\}\times\{0\}\times\mathbb R$ the line going through $x$ and $y$. The collection of these lines together with the line parallel to the $z$-axis span a plane containing $x$.
Do the homotopy along these lines. For each $x$ you will shrink a halfplane of $\mathbb R^2\setminus (\{0\} \times \mathbb R)$ to a disc $D$. (These lines are the same ones along which we will do the second homotopy)
Doing this for each $x$ in $S^1\times \{0\}$ simultaneously will give you a homotopy $\mathbb{R}^3\setminus ((\{0\}\times\{0\}\times\mathbb{R})\cup (\mathcal{S}^1\times\{0\}))\approx (S^1\times D)\setminus (\mathcal{S}^1\times\{0\}))$.
Explicitely writing down the homotopy might be cumbersome, but from the constuction it should be clear how to do this. I had some trouble writing down the geometric picture I had in mind, I can upload a scetch if you have trouble understanding it.
