Doubt in the Proof on Dehn-Nielsen-Baer theorem

Question

I'm reading the proof of Dehn-Nielsen-Baer theorem given in the book 'A primer on mapping class groups' by Margalit and Farb. I'm having trouble understanding the following line from the proof of lemma 8.5. "Assume for the purposes of contradiction that $\Phi(\gamma)$ and $\Phi(\delta)$ are linked at infinity. It follows that the polygonal paths $\{\Phi(\alpha_i)\}$ and $\{\Phi(\beta_i)\}$ have to cross." Since $\Phi(\gamma)$ and $\Phi(\delta)$ are linked at infinity thier axes intersect. But what is the exact relation between the axes of $\Phi(\gamma)$ and $\Phi(\delta)$ and the polygonal paths that forces them to intersect?

Answer 1

To review the setting in more detail, we have two embeddings of $[0,1]$ in the closed disc $D^2$ with images $A$ and $B$:

• $A$ is the closure of $\Phi(\gamma) = \text{interior}(A) = A \cap \text{interior}(D^2)$ with endpoints $\partial A = \{a_0,a_1\} = A \cap S^1$;
• $B$ is the closure of $\Phi(\delta) = \text{interior}(B) = B \cap \text{interior}(D^2)$ with endpoints $\partial B = \{b_0,b_1\} = B \cap S^1$.

To say that "$\Phi(\gamma)$ and $\Phi(\delta)$ are linked at infinity" means that the two sets $\partial A,\partial B \subset S^1$ are linked in $S^1$, which in turn means that there exists a parameterization $[0,2\pi] \mapsto S^1$ in the usual fashion (continuous, surjective, and injective except that $0,2\pi$ are mapped to the same point of $S^1$), and there exist parameters $0 < s_0 < t_0 < s_1 < t_1 < 2\pi$, such that $$s_0 \mapsto a_0, \quad t_0 \mapsto b_0, \quad s_1 \mapsto a_1, \quad t_1 \mapsto b_1 $$ Another way to say this without parameterizations is that each of the two components of the set $S^1 - \{a_1,a_2\}$ contains a point of $\{b_1,b_2\}$.

That answers your question about what the exact relation is.

The way that you then get the desired conclusion is by applying a theorem of topology:

For any two embeddings of $[0,1]$ in $D^2$ denoted $A,B \subset D^2$, if $A \cap S^1 = \partial A$ and $B \cap S^1 = \partial B$, and if $\partial A$ and $\partial B$ are linked in $S^1$ (as defined above), then $A \cap B \ne \emptyset$.

This can be proved by doubling $D^2$ along $S^1$ to obtain a 2-sphere, in which there is an embedded circle obtained by doubling $A$ along $\partial A$, and then applying the Jordan Curve Theorem.