dehn lemma

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Maria Marchwicka 2019-06-07 16:20:17 +02:00
parent 94e0077612
commit 129b2608be

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@ -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 $H_2$)
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\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]