# 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!

• Uh.. draw a picture to see , every point $x\in \mathbb R^3-S^2$ can be jointed by a straight segment with initial point x and terminal point y ( y is on the $S^1 \text {V}S^1$) [Really , I don't know how to explain it through words] Jan 25, 2015 at 4:56

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 $P=\{x,y\}$.
$$\matrix{\pi_1(W,P) & \to & \pi_1(V,P) \cr \downarrow && \downarrow \cr \pi_1(V,P) & \to & \pi_1(X,P)}$$ 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.