dehn lemma

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]