The complement of a circle in $\mathbb R^3$

Question

Is there a direct proof (for analysts; not using results from algebraic topology) that the complement of a circle in the Euclidean space $\mathbb R^3$ is not simply connected? I thought about using the following characterisation: A pathwise connected set $X$ in $\mathbb R^3$ is simply connected if every continuous function $\varphi: \mathbb T\to X$ admits a continuous extension to $\overline{\mathbb D}$ with the same range space $X$.

Answer 1

Without loss of generality assume you are considering a circle that passes through the origin $0\in\Bbb R^3$. Let us write every nonzero vector $V\in\Bbb R^3$ as $V=rv$ where $r=\|V\|$ is its euclidean norm, and $v=\frac{V}{\|V\|}$ is the unit vector that points in the same direction as $V$.

Sphere inversion is the involutive homeomorphism $\Bbb R^3\setminus\lbrace 0\rbrace\longrightarrow \Bbb R^3\setminus\lbrace 0\rbrace$ defined by the formula $V=rv\longmapsto V'=r^{-1}v$. It has the property to convert circles passing through the origin to lines avoiding the origin (and vice versa).

Using sphere inversion, we get a homeomorphism $\Bbb R^3\setminus S^1\simeq\Bbb (\Bbb R^3\setminus\lbrace 0\rbrace)\setminus L$ for some line $L$ not meeting $0$.

Let us prove that $\Bbb R^3$ minus a line and a point isn't simply connected.

Note that if you have a subspace $x_0\in A\subset X$, and a loop $\gamma: S^1\to A$ that defines a non trivial element in $\pi(X,x_0)$, then the same loop defines a non trivial element in $\pi_1(A,x_0)$, for if we could extend it in $A$, i.e. extend it to a map $D^2\to A$, we certainly could extend it in $X$.

So in order to show that $\Bbb R^3-(\lbrace 0\rbrace\cup L)$, three space minus a point and a line, isn't simply connected, it is enough to find a loop drawn in that space that doesn't contract in $\Bbb R^3-L$. For this, note that $\Bbb R^3-L$ is homotopy equivalent to a circle : squish lines that are parallel to $L$, this leaves you with a plane missing a point, then divide by length, to get the circle.

Thus, a small circle that loops around $L$ (and misses the origin) cannot be homotoped to a point in $\Bbb R^3-L$, and hence cannot be homotoped to a point in $\Bbb R^3-(\lbrace 0\rbrace\cup L)$.

Thus, since $\Bbb R^3-(\lbrace 0\rbrace\cup L)$ is homeomorphic to $\Bbb R^3-S^1$, we get that space minus a circle isn't simply connected.

Answer 2

A retract of a simply connected space is still simply connected. This is an elementary fact that can be proved just by composition of mappings.

From rotational symmetry it is easy to see a retraction (not a deformation retraction, but you don't need that) from $\mathbb{R}^3\setminus S^1$ to a punctured half plane. That in turn retracts to a circle, which is not simply connected. Hence $\mathbb{R}^3\setminus S^1$ is not simply connected either.