Two knots $K_0=\varphi_0(S^1)$, $K_1=\varphi_1(S^1)$ are equivalent if the embeddings $\varphi_0$ and $\varphi_1$ are isotopic, that is there exists a continues function
\begin{align*}
&\Phi: S^1 \times [0, 1] \hookrightarrow S^3 \\
&\Phi(x, t) = \Phi_t(x)
\end{align*}
such that $\Phi_t$ is an embedding for any $t \in[0,1]$, $\Phi_0=\varphi_0$ and
$\Phi_1=\varphi_1$.
\end{definition}
\begin{theorem}
Two knots $K_0$ and $K_1$ are isotopic if and only if they are ambient isotopic, i.e. there exists a family of self-diffeomorphisms $\Psi=\{\psi_t: t \in[0, 1]\}$ such that:
\begin{align*}
&\psi(t) = \psi_t \text{ is continius on $t\in[0,1]$}\\
&\psi_t: S^3 \hookrightarrow S^3,\\
&\psi_0 = id ,\\
&\psi_1(K_0) = K_1.
\end{align*}
\end{theorem}
\begin{definition}
A knot is trivial (unknot) if it is equivalent to an embedding $\varphi(t)=(\cos t, \sin t, 0)$, where $t \in[0, 2\pi]$ is a parametrisation of $S^1$.
\end{definition}
\begin{definition}
A link with k - components is a (smooth) embedding of $\overbrace{S^1\sqcup\ldots\sqcup S^1}^k$ in $S^3$
Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing each crossing:
\begin{align*}
\PICorientpluscross\mapsto\PICorientLRsplit\\
\PICorientminuscross\mapsto\PICorientLRsplit
\end{align*}
We smooth all the crossings, so we get a disjoint union of circles on the plane. Each circle bounds a disks in $\mathbb{R}^3$ (we choose disks that don't intersect). For each smoothed crossing we add a twisted band: right-handed for a positive and left-handed for a negative one. We get an orientable surface $\Sigma$ such that $\partial\Sigma= L$.\\
Note: the obtained surface isn't unique and in general doesn't need to be connected, but by taking connected sum of all components we can easily get a connected surface (i.e. we take two disconnected components and cut a disk in each of them: $D_1$ and $D_2$; now we glue both components on the boundaries: $\partial D_1$ and $\partial D_2$.
The three genus $g_3(K)$ ($g(K)$) of a knot $K$ is the minimal genus of a Seifert surface $\Sigma$ for $K$.
\end{definition}
\begin{corollary}
A knot $K$ is trivial if and only $g_3(K)=0$.
\end{corollary}
\noindent
Remark: there are knots that admit non isotopic Seifert surfaces of minimal genus (András Juhász, 2008).
\begin{definition}
Suppose $\alpha$ and $\beta$ are two simple closed curves in $\mathbb{R}^3$.
On a diagram $L$ consider all crossings between $\alpha$ and $\beta$. Let $N_+$ be the number of positive crossings, $N_-$ - negative. Then the linking number: $\Lk(\alpha, \beta)=\frac{1}{2}(N_+- N_-)$.
Let $\alpha$ and $\beta$ be two disjoint simple cross curves in $S^3$.
Let $\nu(\beta)$ be a tubular neighbourhood of $\beta$. The linking number can be interpreted via first homology group, where $\Lk(\alpha, \beta)$ is equal to evaluation of $\alpha$ as element of first homology group of the complement of $\beta$:
Let $L$ be a link and $\Sigma$ be an oriented Seifert surface for $L$. Choose a basis for $H_1(\Sigma, \mathbb{Z})$ consisting of simple closed $\alpha_1, \dots, \alpha_n$.
Let $\alpha_1^+, \dots\alpha_n^+$ be copies of $\alpha_i$ lifted up off the surface (push up along a vector field normal to $\Sigma$). Note that elements $\alpha_i$ are contained in the Seifert surface while all $\alpha_i^+$ are don't intersect the surface.
Let $\Lk(\alpha_i, \alpha_j^+)=\{a_{ij}\}$. Then the matrix $S =\{a_{ij}\}_{i, j =1}^n$ is called a Seifert matrix for $L$. Note that by choosing a different basis we get a different matrix.
The Seifert matrices $S_1$ and $S_2$ for the same link $L$ are S-equivalent, that is, $S_2$ can be obtained from $S_1$ by a sequence of following moves:
\begin{enumerate}[label={(\arabic*)}]
\item
$V \rightarrow AVA^T$, where $A$ is a matrix with integer coefficients,
