Complement of a circle in $\mathbb R^2$ So this is the exercise in Saul Stahl and Catherine Stenson's Introduction to Topology and Geometry (2nd edition), Chapter 9.1. Let $S$ be the complement of a circle in the plane $\mathbb R^2$, and let $X$ be a point of $S$. Describe $\pi(S, X)$, namely the set of homotopy classes of all the circuits of S that are based at $X$. A circuit is a path with identical initial and terminal points, and the circuit is said to be based at that point.
My thoughts:
For $X$ outside the circle, there would be the circuits homotopic to the point $X$ and the circuits homotopic to the circle itself. For $X$ inside the circle, there would be the circiuts homotopic to the point $X$ in $S$. I don't know how to put this formally though.
I know that $\pi(\mathbb R^2, X)$ is homotopic to the point $X$ (constant circuit $1_x$), namely $\pi(\mathbb R^2, X)\cong <1>$ so does any $X$ on the unit sphere, namely $\pi(S_0, X)\cong <1>$.
It would be great if you could also decribe how you arrive at the conclusion, or general strategy to solve similar questions. Thanks a lot.
 A: I'm going to discuss the case where "circle" means "$\{p\ :\ |p-p_0| = r\}$ for some $p_o, r$." You should work to understand the "why?" I throw in, and then address the case where "circle" means "image of $S^1$ under a continuous map to $\mathbb{R}^2."
Our general strategy is going to be: divide and standardize. We're going to split the problem into several simpler cases, then in each case we're going to search for a nice representative of each homotopy class of circuits that we can use to understand what the set $\pi(S,X)$ looks like.
Without loss of generality (why?) we may assume the circle $C$ is of radius $r$ and is centered at $0$. Then $S$ has two components: $D = \{p\ :\ |p| < r\}$ and $O = \{p\ :\ |p| > r\}$. ("D" for "disc" and "O" for "outside".)
If $X$ is an element of the first component, since the disc of radius $r$ has no "holes", intuition suggests that the set of homotopy classes should have one element. In fact this is the case: let's use the "spaghetti slurp" to prove this. Let $\gamma: [0,1]\to S$ be an arbitrary circuit based at $X$. Because $X$ is in the image of $\gamma$, we must in fact have $\gamma:[0,1]\to D$ (why?). 
Because $D$ is convex, for each $t\in [0,1]$ there exists a line segment connecting $\gamma(t)$ to $X$. That line segment can be parametrized by $s\mapsto (1-s)\gamma(t) + sX$. Define the map $H(s,t) = (1-s)\gamma(t) + sX$. This is a continuous (why?) homotopy (why?) from $\gamma$ to the constant map $[0,1]\to X$. Therefore $\pi(S,X) = \{ [\mbox{constant map}] \}$ if $X\in D$.
If $X\in O$, then because there's one hole, intuition suggests that homotopy classes should be in 1-1 correspondence with "numbers of times we can wrap around the hole." The question to make this precise is: how can we standardize "wrap around the hole"? We'll do this in two steps: first, homotope a circuit onto an honest-to-god circle; second, homotope every circuit on a circle to a map of the form $t\mapsto Re^{2\pi i k t}$.
For the first step, an obvious choice of circle is $C_{|X|} = \{ p\ :\ |p| = |X|\}$. Let's use it. Let $\gamma:[0,1]\to O$ be a circuit based at $X$ (why can we say "$O$" instead of "$S$"?). For each point $\gamma(t)$ there's a radial line segment connecting $\gamma(t)$ to $C_{|X|}$. We can parametrize it and as before define a homotopy 
$$H(s,t) = \bigg((1-s) + \frac{s|X|}{|\gamma(t)|}\bigg)\gamma(t).$$
(Why is $H$ continuous? Why is it well-defined, e.g. doesn't cross $C$?)
We have shown that every circuit in $O$ is homotopic to a circuit on a circle. The set of homotopy classes of circuits on a circle is in bijection with the integers (why?). Therefore, if $X\in O$, we have $\pi(S,X)$ is a countably infinite set.
