fundamental group of the complement of a circle This should be a quick question. I am reading Hatcher's Algebraic Topology book. At page 46, in the example of computing the fundamental group of the complement in $\mathbb{R^3}$ of a single circle, he says the point inside $S^2$ and not in the circle can be pushed away from the circle towards $S^2$ or the diameter, which I don't understand. I understand the case where the point is outside $S^2$ deformation retract onto $S^2$ .Can anyone please explain this to me? Thanks!
 A: This seems an example where the natural solution uses  the fundamental groupoid on two base points, and groupoid methods, see the books Topology and Groupoids (first edition, differently named,   1968) and
Categories and Groupoids (first published, 1971). See also https://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one
We can represent  $\mathbb R^3\setminus S^1 $ as the union  $U \cup V$ whose intersection $W$ is the plane through the circle, 
but without the circle. Then $U,V$ are contractible.  Since  $W$ is the union of two  components, choose  base points $x,y$, one  in each component, and let $P=\{x,y\}$.   
We then have a  pushout, not of  the usual groups, but of groupoids:   
$$ \matrix{\pi_1(W,P) & \to & \pi_1(V,P) \cr
\downarrow && \downarrow \cr 
\pi_1(V,P) & \to & \pi_1(X,P)} $$
This is now a standard  situation for the algebra of groupoids, which enables one to show that $\pi_1(X,x)$, and also $\pi_1(X,y)$, is isomorphic to  the integers. See also the paper Brouwer. 
This type of method also works for the fundamental group of the complement of a knot or graph in $\mathbb R^3$ - see again Topology and Groupoids, section 9.1. 
