The closure of a braid connects the ends of the braid as shown to create a knot or knotted links. Mathematically, for a braid \(b\) the resulting closure is denoted by \(\beta(b)\)

We can determine if the closure of a braid is a knot rather than links by considering the "canonical epimorphism" - which to my understanding essentially means a mapping of sorts - of the braid group \(B_{n}\) onto the permutation group \(S_{n}\). Consider the braid

The mapping \(B_{n} \mapsto S_{n}\) is thus a mapping of \(b \mapsto \sigma(b)\) where \(b_{i} \mapsto s_{i}\) and \(s_{i}\) denotes the switching of the \(i\)th and \((i+1)\)th numbers (or elements at said indices) in the sequence. Thus, for the braid b above, we have \(\sigma(b) = s_{1}s_{2}s_{1}s_{2}\) or we can reduce \(b\) to \(b = b_{2}b_{1}\) and get \(\sigma(b) = s_{2}s_{1}\). In either case, \(\sigma(b) = (3, 1, 2)\). Since this permutation can be obtained from the original point by moving each digit to the right once (w/ the last digit cycling to the front), this is a cyclic permutation. This means that the braid's closure yields a knot because the cyclic nature of the permutation ensures that we will pass through all points with a single strand when drawing the closure (this is much easier to make sense of if one draws the closure and follows the flow of the lines). More precisely, "the closure of \(\beta(b)\) of the braid \(b\) is a knot if and only if the permutation \(\sigma(b)\) associated to the braid generates the cyclic subgroup of order \(n\), \(\mathbb{Z}/n\mathbb{Z}\), in the permutation group \(S_{n}\)" (pg. 55).

The permutation of \((1, 2, 3)\) to \((3, 1, 2)\) is denoted by \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix} The identity permutation is essentially the non-permutation of \((1, 2, 3)\) and the order of a permutation \(\sigma\) is the number \(n\) such that \(\sigma^{n} = 1\) or the identity. The multiplication of permutations is performed like so: \begin{equation} \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \end{equation} \begin{equation} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} \end{equation} Thus, for our example the permutation has an order \(n\). The cyclic nature comes from the fact that we pass through all points in the process: \begin{align} 1 \mapsto 3 \mapsto 2 \mapsto 1 \\ 2 \mapsto 1 \mapsto 3 \mapsto 2 \\ 3 \mapsto 2 \mapsto 1 \mapsto 3 \end{align}

The closure of braids is not injective, meaning that there isn't always a 1-to-1 map between braids and their closures. However, Alexander's Braiding Theorem does state that the closure operation is surjective meaning that all knots and links constitute the closure of some braid.
One way to prove this is via the Vogel algorithm which is as follows:

which is illustrated in the examples below.

The algorithm always yields a set of nested Seifert circles in a finite number of steps thus proving that any knot is the closure of some braid (also note that the final change infinity move appears to take whatever single un-nested Seifert circle there is and braid it around so as to create a final outside Seifert circle around the already nested ones hence the lack of a loop back to the while loop). The act of drawing the steps as well as various braid closures should give one an intuitive understanding of this fact although I do hope to update this post with a proof eventually (time permitting or should it become necessary for further work).

The Markov Theorem states that the closures of two braids can be proven to be isotopic if and only if one braid can be transformed to the other by a finite sequence of the Markov moves which are defined as:
1. \(b \leftrightarrow aba^{-1}\) with \(a, b \in B_{n}\)
2. \(b \leftrightarrow bb_{n}^{\pm 1}\) with \(b \in B_{n}\) and \(b_{n} \in B_{n + 1}\)

1. \(\langle L \rangle = a\langle L_{A} \rangle + b\langle L_{B} \rangle\)
2. \(\langle L \sqcup \bigcirc \rangle = c \langle L \rangle\)
3. \(\langle \bigcirc \rangle = 1\)

\[\langle L \rangle = \sum_{s}a^{\alpha(s)}b^{\beta(s)}c^{\gamma(s)-1} = \sum_{s}a^{\alpha(s)-\beta(s)}(-a^{2}-a^{-2})^{\gamma(s)-1}\] where \(s\) is the identifier of each of the \(2^{n}\) states and \(\alpha(s)\) is the exponent on \(a\), \(\beta(s)\) is the exponent on \(b\) and \(\gamma(s)\) is the number of circles created.
For the example above, we get
\begin{split} \langle L \rangle & = a^{3}(-a^{2}-a^{-2})^{2} + (a^{-3} + 3a)(-a^{2}-a^{-2}) + 3a^{-1}\\ & = a^{7} - a^{3} - a^{-5} \end{split}

\[w(L) = \sum_{i}\epsilon_{i}\] where \(i\) is the sum over the \(n\) crossings and \(\epsilon\) is defined as + or -1 as per the diagram below

\[X(L) = (-a)^{-3w(L)}\langle |L| \rangle\]

\[a \mapsto q^{-\frac{1}{4}}\]
The Jones polynomial satisfies the relations
1. \(q^{-1}V(L^{+}) - qV(L^{-}) = (q^{\frac{1}{2}} - q^{-\frac{1}{2}})V(L^{o})\) ...note, this is the skein relation
2. \(V(L \sqcup \bigcirc) = -(q^{-\frac{1}{2}} + q^{\frac{1}{2}})V(L)\)
3. \(V(\bigcirc) = 1\)

\[\frac{1}{\sqrt{\lambda}\sqrt{q}}\chi(L^{+}) - \sqrt{\lambda}\sqrt{q}\chi(L^{-}) = \frac{\sqrt{q}-1}{\sqrt{q}}\chi(L^{o})\]

Conway polynomial: \(\mapsto \nabla(L^{+}) - \nabla(L^{-}) = z\nabla(L^{o})\)
Alexander polynomial: \(\mapsto \Delta(L^{+}) - \Delta(L^{-}) = \frac{1}{\sqrt{1 - t^{2}}}\Delta(L^{o}) \)
HOMFLY polynomial: \(\mapsto x\mathcal{P}(L^{+}) - t\mathcal{P}(L^{-}) = \mathcal{P}(L^{o})\)