$\mathbb{R}^3 \setminus A$ deformation retracts onto $S^1 \vee S^2$

Hatcher says the following:

The complement $\mathbb{R}^3 \setminus A$ of a single circle $A$ deformation retracts onto a wedge sum $S^1 \vee S^2$.

He then goes on to explain it, but I do not understand the explainations (pages 46/47, example 1.23). I'm staring at the following picture:

The first is $S^1 \vee S^2$ and the second is homotopy equivalent to the first. So the question is, why is the second space homotopy equivalent to $\mathbb{R}^3 \setminus A$?

Hatcher says, we push points from outside the sphere onto the sphere. So far, so good. Points that are outside of $A$ can also be pushed onto the sphere. What about points inside $A$? Can we contract them to a point and stretch it from one end of the sphere to the other to get the desired space?

Solution 1:

Say the circle has radius $1$, center at the origin, and lies in the $xz$-plane. Now let $S^2$ be embedded with center in the origin and radius $2$. Everything outside the sphere deformation retracts onto the sphere.

For every point $x$ inside the sphere, if it's on the $y$-axis, leave it. Otherwise, there is a point $p$ on the circle that is closest to it. Now let $x$ move directly away from $p$ until it either hits the $y$-axis or the sphere, whichever comes first (this is continuous because the line between $p$ and $x$ is coplanar to the $y$-axis, so there are no points that "just misses" the $y$-axis and goes past it). You've now created the second space.