If $K$ is a knot, then $n$ is necessarily even, and so $\Delta_K(t^{-1})= t^{-n}\Delta_K(t)$.
\end{proof}
\begin{lemma}
\begin{align*}
\frac{1}{2}\deg\Delta_K(t) \leq g_3(K),
\text{ where } deg (a_n t^n + \dots + a_1 t^l )= k - l.
\end{align*}
\end{lemma}
\begin{proof}
If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma)=\mathbb{Z}^{2g}$, so $S$ is an $2g \times2g$ matrix. Therefore $\det(tS - S^T)$ is a polynomial of degree at most $2g$.
\end{proof}
\begin{example}
There are not trivial knots with Alexander polynomial equal $1$, for example:
\includegraphics[width=0.3\textwidth]{11n34.png}
$\Delta_{11n34}\equiv1$.
\end{example}
\subsection{Decomposition of $3$-sphere}
We know that $3$ - sphere can be obtained by gluing two solid tori:
\caption{The complement of solid torus in $S^3$ is another solid torus.}
\label{fig:sphere_as_tori}
}
\end{figure}
\subsection{Dehn lemma and sphere theorem}
%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
${D^2\overset{g}\longhookrightarrow M}$ such that:
\[
g\big|_{\partial D^2} = f\big|_{\partial D^2.}
\]
\end{lemma}
\noindent
Remark: Dehn lemma doesn't hold for dimension four.\\
Let $M$ be connected, compact three manifold with boundary.
Suppose $\pi_1(\partial M)\longrightarrow\pi_1(M)$ has non-trivial kernel. Then there exists a map $f: (D^2, \partial D^2)\longrightarrow(M, \partial M)$ such that $f\big|_{\partial D^2}$ is non-trivial loop in $\partial M$.
\begin{theorem}[Sphere theorem]
Suppose $\pi_1(M)\ne0$. Then there exists an embedding $f: S^2\hookrightarrow M$ that is homotopy non-trivial.
\end{theorem}
\begin{problem}
Prove that $S^3\ K$ is Eilenberg–MacLane space of type $K(\pi, 1)$.
\end{problem}
\begin{corollary}
Suppose $K \subset S^3$ and $\pi_1(S^3\setminus K)$ is infinite cyclic ($\mathbb{Z})$. Then $K$ is trivial.
\end{corollary}
\begin{proof}
Let $N$ be a tubular neighbourhood of a knot $K$ and $M = S^3\setminus N$ its complement. Then $\partial M = S^1\times S^1$. Let $f : \pi_1(\partial M )\longrightarrow\pi_1(M)$.
$[\mu]$ represents the generator of $H_1(S^3\setminus K, \mathbb{Z})$. From definition of $\lambda$ we know that $\lambda$ is trivial in $H_1(M)$ ($\Lk(\lambda, K)=0$, therefore $[\lambda]$ was trivial in $pi_1(M)$). If $K$ is non-trivial then $\lambda$ is non-trivial in $\pi_1(M)$, but it is trivial in $H_1(M)$.
