Lecture notes for Friday, Mar. 2

In this part we will assume that the complement of any arc (i.e., a subset homeomorphic to a closed interval) on \(S^2\) is connected. This will be proved a little later.

Throughout this section we’ll let \(C\) be a subset of \(S^2\) homeomorphic to \(S^1\), and let \(C = D \cup E\) where \(D\) and \(E\) are arcs that meet in exactly two points \(a\) and \(b\). We’ll also let \(U = S^2 \setminus D\), \(V = S^2 \setminus E\), so that \(U \cap V = S^2 \setminus C\) and \(U \cup V = S^2 \setminus \{a, b\} =: X\).

Note that, since \(C, D\) and \(E\) are all compact, they are closed subsets of \(S^2\). Moreover, since \(U\), \(V\) and \(X\) are open subsets of \(S^2\) (and \(S^2\) is locally path-connected), we won’t need to distinguish between connectedness and path-connectedness for these subspaces.

Proposition. The complement of a simple closed curve is disconnected.

Proof. Assume it were connected. Then we can use van Kampen’s theorem (even the fundamental group version!) to get that

\(\pi_1(U \cap V)\)	\(\to\)	\(\pi_1(U)\)
\(\downarrow\)		\(\downarrow\)
\(\pi_1(V)\)	\(\to\)	\(\pi_1(X)\)

is a pushout square of groups. The lower right corner we know: it is just \(\mathbb{Z}\), since \(X\) is (homeomorphic to) an (open) annulus, and thus equivalent to a circle. We’ll get a contradiction by showing that both \(\pi_1(U) \to \pi_1(X)\) and \(\pi_1(V) \to \pi_1(X)\) are trivial morphisms sending everything to zero.

It should be intuitive that these morphisms are indeed zero: we’re saying that if you have a loop on a sphere that avoids some arc, then the loop can be contracted to a point without going through the endpoints of the arc. To prove it, let’s just put one endpoint off limits to begin with: \(S^2 \setminus \{b\}\) is homemorphic to \(\mathbb{R}^2\) (by stereographic projection, for example), and we can even pick a homemorphism for which \(a\) maps to the origin in the plane. Then the arc \(D\) corresponds to some curve starting at the origin and going off to infinity, and if we pick a parametrization \(\alpha : [0,\infty) \to \mathbb{R}^2\) for this curve, what we’re trying to show is that any loop \(\gamma\) in the plane avoiding the image of \(\alpha\) can be contracted to a point while avoiding the origin. Our strategy for that is to translate \(\gamma\) by the vector \(- \alpha(t)\): when \(t=0\) we just get \(\gamma\), but as \(t \to \infty\), \(\gamma\) gets pushed away from the origin until it “looks tiny when viewed from the origin” and then be contracted.

More formally, since the image of \(\gamma\) is compact, it lies inside some big ball of radius \(R\) around the origin and there is a \(t_0\) such that \(\left|\alpha(t_0)\right| > R\). Consider the homotopy \(H_t(s) = H(t,s) = \gamma(s) - \alpha(t)\). Because the loop \(\gamma\) avoid the image of \(\alpha\), \(H\) never passes through the origin. Also, we have \(H_0 = \gamma\) and \(H_{t_0}\) is a loop that lies inside the ball \(B\) of radius \(R\) with center \(\alpha(t_0)\). Since \(\left|\alpha(t_0)\right| > R\), \(B\) does not contain the origin and the loop \(H_{t_0}\) can be safely contracted to a point inside of \(B\). This shows that the inclusion \(U \to X\), induces the zero map on \(\pi_1\), as required.

Proposition. The complement of a simple closed curve has exactly two components.

Proof. We’ve seen it’s disconnected, so it has at least two components; we need only show it can’t have three or more components. To do this we’ll apply van Kampen to \(U\) and \(V\) again (but this time we do need the groupoid version). Take a set \(A\) of base points that consists of exactly one point from each component of \(S^2 \setminus C = U \cap V\). Notice that since \(U\), \(V\) and \(X\) are connected we don’t have to worry about \(A\) failing to meet some component of them. Now, van Kampen gives us a pushout of groupoids:

\(\pi_{\le 1}(U \cap V,A)\)	\(\to\)	\(\pi_{\le 1}(U,A)\)
\(\downarrow\)		\(\downarrow\)
\(\pi_{\le 1}(V,A)\)	\(\to\)	\(\pi_{\le 1}(X,A)\)

Again, we know all about the lower right corner. In particular, if we take a point \(p \in A\), we know that the fundamental group of \(X\) based at \(p\) is just \(mathbb{Z}\). But here is a heristic argument indicating that this is not consistent with the pushout square: we get \(|A|-1\) independent loops at \(p\) coming from following some path from \(p\) to \(q \in A\) in \(U\) and coming back along some path in \(V\). The fact that in \(U \cap V\) there are no paths from \(p\) to \(q\) tells us these loops in \(X\) are non-trivial and have no relations between them. This (correctly) suggests we actually get the free group on \(|A|-1\) generators sitting inside \(\pi_1(X,p)\), which shows \(|A|=2\).

All that remains is make the heuristic precise, which is the next section.

Lemma. Consider a pushout square of groupoids,

\(A\)	\(\to\)	\(B\)
\(\downarrow\)		\(\downarrow\)
\(C\)	\(\to\)	\(G\)

where

• \(A\), \(B\), \(C\), \(G\) all have the same objects and the morphisms in the square are all the identity on objects,
• \(A\) is skeletal, i.e., a disjoint union of groups, and
• \(B\) and \(C\) are connected.

In this situation, for any object \(p\) in \(G\), the automorphism group of \(p\) in \(G\) contains a free group on \(n-1\) generators, where \(n\) is the common number of objects of \(A\), \(B\), \(C\) and \(G\).

Proof. Let \(D\) be the discrete groupoid on the common set of objects of \(A\), \(B\), \(C\) and \(G\), and let \(I\) be the indiscrete groupoid on the same set of objects. The unique morphisms \(D \to A \to D\) which are the identity on objects show that \(D\) is a retract of \(A\). We also have that \(I\) is a retract of both \(B\) and \(C\) and indeed the retractions \(B \to I\) and \(C \to I\) are again determined uniquely by being the identity on objects. The inclusions \(I \to B\) and \(I \to C\) are not unique; we will choose them as follows: for each object \(q \neq p\), choose a morphism \(\beta_q : p \to q\) in \(B\) and a morphism \(\gamma_q : p \to q\) in \(C\), then \(I \to B\) is determined uniquely by asking that the unique morphism \(p \to q\) in \(I\) is sent to \(\beta_q\), and similarly for the morphism \(I \to C\).

These retractions are compatible in the sense that they make the diagram [1]

\(D\)	\(\to\)	\(I\)		\(A\)	\(\to\)	\(B\)
\(\downarrow\)			a retract of	\(\downarrow\)
\(I\)				\(C\)

and one can easily check that this induces a retraction of the pushouts. So, if \(F\) is the pushout of the first of the two diagrams above, then \({\text{Aut}}_G(p)\) has \({\text{Aut}}_F(p)\) as a retract; we will show that this group, \({\text{Aut}}_F(p)\), is free on \(n-1\) generators. To do so, notice that both \(D\) and \(I\) are free groupoids on a directed graph: \(D\) is free on the empty graph \(E\) with \(n\) vertices and no edges, and \(I\) is free on any tree with \(n\) vertices, but to fix ideas let’s pick one: let \(T\) be the “star” with center \(p\), i.e., a directed graph which one edge going from \(p\) to every other vertex \(q\). Since the free groupoid functor is a left adjoint, the pushout of \(I \leftarrow D \to I\) is the free groupoid on the pushout of the directed graphs \(T \leftarrow E \to T\). This pushout of directed graphs is easy to describe: it simply has two edges, say \(a_q\) and \(b_q\), going from \(p\) to any other vertex \(q\). The automorphism group of \(p\) in the free groupoid on this doubled tree is clearly free on the \(n-1\) generators of the form \(b_q^{-1} a_q\).

Notice that in the setting where we used this lemma in the previous (where, in particular, the pushout square of groupoids we’re talking about comes from van Kampen) the images of the generators \(b_q^{-1} a_q\) of \({\text{Aut}}_F(p)\) in \({\text{Aut}}_G(p)\) are precisely the morphisms \(\beta_q^{-1} \alpha_q\) described in the heuristic argument at the end of previous section (“going from \(p\) to \(q\) along some path in \(U\) and coming back along some path from \(q\) to \(p\) in \(V\)”).

By an arc \(A \subset S^2\) we just mean a subset of \(S^2\) which is homeomorphic to \(S^1\).

curve

[1] In the diagram below all the morphisms are identities on objects.