From 129b2608be2ca810b3dae76f764bfaca5a38fbf8 Mon Sep 17 00:00:00 2001 From: Maria Marchwicka Date: Fri, 7 Jun 2019 16:20:17 +0200 Subject: [PATCH] dehn lemma --- lectures_on_knot_theory.tex | 14 ++++++++++++-- 1 file changed, 12 insertions(+), 2 deletions(-) diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex index 750d8ad..d759edf 100644 --- a/lectures_on_knot_theory.tex +++ b/lectures_on_knot_theory.tex @@ -558,6 +558,16 @@ There are not trivial knots with Alexander polynomial equal $1$, for example: $\Delta_{11n34} \equiv 1$. \end{example} %removing one disk from surface doesn't change $H_1$ (only $H_2$) +% +% +% +\begin{lemma}[Dehn] +Let $M$ be a $3$-manifold and $D^2 \overset{f} \rightarrow M^3$ be a map of a disk such that $f_{\big|\partial D^2}$ is an embedding. Then there exists an embedding +${D^2 \overset{g}\hookrightarrow M}$ such that: +\[ +g_{\big| \partial D^2} = f_{\big| \partial D^2.} +\] +\end{lemma} \section{} \begin{example} \begin{align*} @@ -580,8 +590,8 @@ A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$. \end{definition} \begin{definition} -Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists an annulus $A$ that is smoothly embedded in $S^3 \times [0, 1]$ such that -$\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\} $. +Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists an annulus $A$ that is smoothly embedded in ${S^3 \times [0, 1]}$ such that +${\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\}}$. \end{definition} \begin{figure}[h]