\subsection{Existence of a Seifert surface - second proof}
%\begin{theorem}
%For any knot $K \subset S^3$ there exists a connected, compact and orientable surface $\Sigma(K)$ such that $\partial \Sigma(K) = K$
%\end{theorem}
\begin{proof}(\Cref{theo:Seifert})\\
Let $K \in S^3$ be a knot and $N = \nu(K)$ be its tubular neighbourhood. Because $K$ and $N$ are homotopy equivalent, we get:
\begin{align*}
H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K).
\end{align*}
Let us consider a long exact sequence of cohomology of a pair $(S^3, S^3 \setminus N)$  with integer coefficients:

\begin{center}
\begin{tikzcd}
[
    column sep=0cm, fill=none, 
    row sep=small,
    ar symbol/.style =%
        {draw=none,"\textstyle#1" description,sloped},
         isomorphic/.style = {ar symbol={\cong}},
]
&\mathbb{Z}
\\

& H^0(S^3) \ar[u,isomorphic]  \to 
&H^0(S^3 \setminus N) \to 
\\
\to H^1(S^3, S^3 \setminus N) \to
 & H^1(S^3) \to
 & H^1(S^3\setminus N)  \to
 \\
& 0  \ar[u,isomorphic]&
 \\
 \to H^2(S^3, S^3 \setminus N) \to
 & H^2(S^3) \ar[u,isomorphic] \to
 & H^2(S^3\setminus N) \to
 \\
\to H^3(S^3, S^3\setminus N)\to
& H^3(S) \to
& 0
\\
&  \mathbb{Z} \ar[u,isomorphic] &\\
 \end{tikzcd}
\end{center}
The tubular neighbourhood of the knot is homomorphic to 
$D^2 \times S^1$. 
So its boundary 
$\partial N \cong \  S^1 \times S^1$ and therefore:
$H^1(N, \partial N) \cong \ \mathbb{Z} \oplus \mathbb{Z}$. By excision theorem we have:
\begin{align*}
H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N).
\end{align*}
Therefore:
\begin{align*}
H^	1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z}.
\end{align*}
Let us consider the following diagram: 
\begin{equation*}
\begin{tikzcd}[row sep=huge]
H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] & 
H^1(N \setminus K) \arrow[d,"\Theta"] \\ 
{[S^3 \setminus K, S^1]} \arrow[r,]&  
{[N \setminus K, S^1]}
\end{tikzcd}
\end{equation*}
\noindent
$\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface. 
%
% picture for excision theorem
% Thom isomorphism, 
\end{proof}
%$S$ - equivalence $\Sigma$\\
%simple closed curves $\alpha_1, ... \alpha_n \in H_1(\Sigma, \mathbb{Z})$ basis for $H_1$
\subsection{Alexander polynomial}
\begin{definition}
Let $S$ be a Seifert matrix for a knot $K$. The Alexander polynomial $\Delta_K(t)$ is a Laurent polynomial:
\[ 
\Delta_K(t) := \det (tS - S^T)  \in 
\mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}]
\]
\end{definition}

\begin{theorem}
$\Delta_K(t)$ is well defined up to multiplication by $\pm t^k$, for $k \in \mathbb{Z}$.
\end{theorem}
\begin{proof}
We need to show that $\Delta_K(t)$ doesn't depend on $S$-equivalence relation.
\begin{enumerate}[label={(\arabic*)}]
\item Suppose $S\prime = CSC^T$, $C \in \Gl(n, \mathbb{Z})$ (matrices invertible over $\mathbb{Z}$). Then $\det C = 1$ and:
\begin{align*}
&\det(tS\prime - S\prime^T) =
\det(tCSC^T - (CSC^T)^T) =\\
&\det(tCSC^T - CS^TC^T) = 
\det C(tS - S^T)C^T = 
\det(tS - S^T)
\end{align*}
\item
Let \\
$ A := t
\begin{pmatrix}
    \begin{array}{c|c}
           S &  
           \begin{matrix}
                 \ast & 0  \\
                 \sdots & \sdots\\
                 \ast & 0
           \end{matrix} \\
    \hline
           \begin{matrix}
                \ast & \dots & \ast\\
                0      & \dots & 0
           \end{matrix}
           &
           \begin{matrix}
               0 & 0\\
               1 & 0
           \end{matrix}
    \end{array}
\end{pmatrix}
-
\begin{pmatrix}
    \begin{array}{c|c}
           S^T &  
           \begin{matrix}
                 \ast & 0  \\
                 \sdots & \sdots\\
                 \ast & 0
           \end{matrix} \\
    \hline
           \begin{matrix}
                \ast & \dots & \ast\\
                0      & \dots & 0
           \end{matrix}
           &
           \begin{matrix}
               0 & 1\\
               0 & 0
           \end{matrix}
    \end{array}
\end{pmatrix}
=
\begin{pmatrix}
    \begin{array}{c|c}
           tS - S^T &  
           \begin{matrix}
                 \ast & 0  \\
                 \sdots & \sdots\\
                 \ast & 0
           \end{matrix} \\
    \hline
           \begin{matrix}
                \ast & \dots & \ast\\
                0      & \dots & 0
           \end{matrix}
           &
           \begin{matrix}
               0 & -1\\
               t & 0
           \end{matrix}
    \end{array}
\end{pmatrix}
$
\\
\\
Using the Laplace expansion we get $\det A = \pm t \det(tS - S^T)$. 
\end{enumerate} 
\end{proof}
%
%
%
\begin{example}
If $K$ is a trefoil then we can take
$S = \begin{pmatrix}
-1 & -1 \\
0 & -1
\end{pmatrix}$. Then
\[
\Delta_K(t) = \det 
\begin{pmatrix}
-t + 1 & -t\\
1 & -t +1
\end{pmatrix}
= (t -1)^2 + t = t^2 - t +1 \ne 1
\Rightarrow \text{trefoil is not trivial.}
\]
\end{example}
\begin{lemma}
$\Delta_K(t)$ is symmetric.
\end{lemma}
\begin{proof}
Let $S$ be an $n \times n$ matrix. 
\begin{align*}
&\Delta_K(t^{-1}) = \det (t^{-1}S - S^T) = (-t)^{-n} \det(tS^T - S) = \\
&(-t)^{-n} \det (tS - S^T) = (-t)^{-n} \Delta_K(t)
\end{align*}
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 $2g \times 2g$ 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}
$\Delta_{11n34} \equiv 1$.
\end{example}

\subsection{Decomposition of \texorpdfstring{
$3$-sphere}{3-sphere}}
We know that $3$ - sphere can be obtained by gluing two solid tori:
\[
S^3 = \partial D^4 = \partial (D^2 \times D^2) =  (D^2 \times S^1) \cup (S^1 \times D^2).
\]
So the complement of solid torus in $S^3$ is another solid torus.\\
Analytically it can be describes as follow. \\
Take $(z_1, z_2) \in \mathbb{C}$ such  that ${\max(\vert z_1 \vert, \vert z_2\vert) = 1.}
$
Define following sets: 
\begin{align*}
S_1 = \{ (z_1, z_2) \in S^3: \vert z_1 \vert = 0\} \cong S^1 \times D^2 ,\\ 
S_2 = \{(z_1, z_2) \in S ^3: \vert z_2 \vert = 1 \} \cong  D^2 \times S^1.
\end{align*}
The intersection 
$S_1 \cap S_2 = \{(z_1, z_2): \vert z_1 \vert = \vert z_2 \vert = 1 \} \cong S^1 \times S^1$.
\begin{figure}[h]
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.3\textwidth}{!}{\includegraphics[width=0.3\textwidth]{sphere_as_torus.png}}
\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 $H_2$)
%
%
%
\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 $\pi_1(M) \ne 0$. 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 $f : \pi_1(\partial M ) \longrightarrow \pi_1(M)$. 
If $\pi_1(M)$ is infinite cyclic group then the map $f$ is non-trivial. Suppose ${\lambda \in \ker (\pi_1(S^1 \times S^1) \longrightarrow \pi_1(M))}$.  
There is a map $g: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $g(\partial D^2) = \lambda$.\\
 By Dehn's lemma there exists an embedding ${h: (D^2, \partial D^2) \longhookrightarrow (M, \partial M)}$ such that 
$h\big|_{\partial D^2} = f \big|_{\partial D^2}$ and  $h(\partial D^2) = \lambda$.
Let $\Sigma$ be a union of the annulus and the image of $\partial D^2$. 
%$g_3$
If $g_3(\Sigma) = 0$, then $K$ is trivial. \\
Now we should proof that: 
\[
H_1(M) \cong \mathbb{Z} \Longrightarrow \lambda \in \ker ( \pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)).
\]
\begin{figure}[h]
\fontsize{40}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.4\textwidth}{!}{\input{images/torus_lambda.pdf_tex}}
}
\caption{$\mu$ is a meridian and $\lambda$ is a longitude.}
\label{fig:meridian_and_longitude}
\end{figure}
Choose a meridian $\mu$ such that $\Lk (\mu, K) = 1$.  Recall the definition of linking number via homology group (Definition \ref{def:lk_via_homo}).
$[\mu]$ 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 $H_1(M)$. 
\end{proof}
