INTRODUCTION
A braid theory introduced by E. Artin at 1925 and the concept of braid
theory was found to have applications in other fields after the 1950s
and this gave fresh impetus to the study of braids. More studies on braid
theory are studied by many researches by this braid theory has gradually
been prospected, refined and polished. In mathematics it is, now, recognized
from one of the basic theories and is of benefit in such branches as topology
and algebraic geometry. Also, it is of profound use in other areas of
the sciences physics, statistical mechanics, chemistry and biology. The
braid group was took an important role in this field. The iridescent hue
of this concepts flowering into full bloom and activity occurred in 1984,
when V. Jones put into action with inordinate success the original aim
of Artin. i.e. the application of braids to knot theory. In (ElGhoul
et al., 2006, 2007) we introduced a new direction on knot theory
called folding and retraction of knot. More studies of retraction in ElGhoul
(1985, 1995, 1998, 2002). In this study our intention introduce the concepts
of retraction on the braid theory and braid group, continuations to the
two articles above, we study the effect of retraction and conditional
retraction on braid and braid group and singular braid.
DEFINITIONS
Here we will show some definitions and basic concepts which we will use
it in the main results.
Definition 1
Let D be a unit cube, so D = {(x, y, z): 0 ≤ x, y, z ≤ 1} on the
top face of cube place n points, a_{1}, a_{2}, ..., a_{n}
and similarly, place n points on the bottom face b_{1}, b_{2},
..., b_{n} Now,join the points a_{1}, a_{2}, ...,
a_{n} with b_{1}, b_{2}, ..., b_{n} by
means of n arcs d_{1}, d_{2}, ..., d_{n} (as smooth
curves), this arcs are mutually disjoint and each d_{1} connects
some a_{j} to b_{k} (J ≠K or J = K) not connect a_{j}
to a_{K} or b_{j} to b_{K}. Each plane E_{s},
such that z = s, 0 ≤ s ≤ 1 (parallel to xyplane intersections each
arc d_{i} at one and only one point.
A configuration of n arcs d_{1}, d_{2}, ..., d_{n}
with end points a_{1}, a_{2}, ..., a_{n} and b_{1},
b_{2}, ..., b_{n} is called nbraid or a braid with n
strings Fig. 1 (Murasugi, 1996).

Fig. 1: 
Representation of braid in unit cube 

Fig. 2: 
Product of β_{1} and β_{2} 
Definition 2
Let B_{n} be a set of all nbraids and β_{1}, β_{2}
ε B_{n}, we may create a third nbraid from them which
we shall call their product and denoted by β_{1}β_{2}
as follows:
Glue the bottom arcs of β_{1} with the top arcs of β_{2}
(Murasugi and Kurpita, 1999) Fig. 2.
Remark 1
• 
For β_{1} ε B_{n} there
is β_{1}^{1} such that β_{1}β_{1}^{1}
= e ε B_{n} Fig. 3. 
• 
β_{1}β_{2} ≠β_{2}β_{1} 
• 
The product of braids is associative, i.e., (β_{1}β_{2})β_{3}
≈_{1}(β_{2}β_{2}) 
Theorem 1
The set of all nbraids, B_{n} forms a group. This group is usually
called the nbraid group or Artin's nbraid group (Murasugi and Kurpita,
1999; Gemein, 2001).
Theorem 2
For any n ≤ 1 the nbraid group B_{n} has the following presentation:
Where, σ_{i} denotes the standard generator of the braid
group (Kauffman, 1991; Murasugi and Kurpita, 1999; Gemein, 2001) Fig.
4.

Fig. 3: 
Product of β_{1} and β_{1}^{1}
and β_{1} ° β_{2} ≠β_{2} °
β_{1} 

Fig. 4: 
Standard generator σ_{i} 
Definition 3
An ndimensional manifold is a Housdorff topological space M, such that
every point of M has a neighborhood homeomorphic to open set U ⊂ R^{n}
(Munkers, 1975).
Definition 4
Let A be a subset of a topological space X. A continuous map r: X →A
is said to be retraction if r(a) = a for all a ε A (Massay, 1967;
Munkers, 1975).
Definition 5
A subset A ⊂ X is deformations retract of X if there is a retraction
r: X →A such that i ° r homotopic to the identity map. That is,
there exists a continuous function f: X x [0, 1] →X such that for
x ε X, f (x, 0) = x and f (x, 1) = r (x) and for all a ε
A and all t ε [0, 1], f (a, t) = a (Massay, 1967).
Definition 6
Let β_{1}, β_{2} be two nbraids in a cube
D, we say β_{1} ambient isotopic to β_{2}, denoted
by β_{1} ≈ β_{2}, if there exists a homeomorphism
H: Dx[0,1]→Dx [0,1], such that H(x,t) = h_{t}((x),t), tε[0,1],
h_{t}: D→D and h_{0}(β_{1}) = β_{1},
h_{t}(β_{2}) = β_{2} (Murasugi and Kurpita,
1999).
Definition 7
Let β be a nbraid and suppose the ith string d_{i} of β
joins a_{i} to b_{j(i)} for i = 1, 2, 3, ….n. Define g:
B_{n} →S_{n} (the set af all permutations of the set {1,
2, 3, 4, ….., n} as,
the permutation for a given braid is called a braid permutation and denoted
by π(β), if
then β is called a pure nbraid (Murasugi and Kurpita, 1999).
Remark 2
If β_{1} ≈ β_{2} ⇒ π(β_{1})
= π(β_{2}).
Theorem 3
Let B_{n} be a nbraid group and S_{n} the symmetry
group of n elements, then there exist a natural surjective homomorphism
f from B_{n} onto S_{n}, takes any braid β to the
permutation determined by β, f (β) = π (β).
The kernel of f is a pure nbraid group, denoted by P_{n} and
B_{n}/P_{n} is isomorphic to S_{n}, the index
of P_{n} in B_{n} is a finite, [B_{n}:P_{n}]
= n! (Murasugi and Kurpita, 1999).
THE MAIN RESULTS
Let β = σ_{i} be a nbraid and r a retraction from
β{a} onto β', where a is a point on one arc of β, then
we have two cases:
• 
β' is a trivial n1braid, if a ε d_{i} or
a ε d_{i+1}. 
• 
β' = σ_{i} as a n1braid if a ε d_{j},
j ≠i, i + 1 Fig. 5. 
If the retraction r from β{a_{1}, a_{2}, ..., a_{n}}
onto β' by remove all points from top level or bottom level or together,
then we have two cases:
• 
β' is a trivial nbraid. 
• 
β' = σ_{i} as a nbraid Fig. 6. 
Theorem 4
A retract of any braid by remove a point or (points) from any arc
or (arcs) is a braid.
Proof
Let β be a nbraid, β = {d_{1}, d_{2}, ..., d_{n}}
where d_{i} is a string and r be a continuous map from β{a}
onto β' where a is a point on d_{i}, i = 1, 2, ..., n defined by
,
then r(d_{i}) = d_{i} ∀d_{i} ∈ β' hence
r is a retraction and β' is a n1braid.
Theorem 5
Let β = σ_{1}σ_{2}.....σ_{k}
be a nbraid and r: (βa) →β' be a retraction then β'
takes three cases:
• 
Trivial n1braid, if a ε d_{1}. 
• 
β' = σ_{1}σ_{2}....σ_{k1}
as a n1braid, if a εd_{i}, 1 < i ≤ k + 1. 
• 
β' = σ_{1}σ_{2}....σ_{k}
as a n1braid, if a ε d_{i}, i > k + 1. 

Fig. 5: 
Two cases of retraction of σ_{i} by remove one
point 

Fig. 6: 
Two cases of retraction of σ_{i} by remove more
then one point 
Proof
• 
Since d' _{1} of σ_{1} glue with
d_{1}'' of σ_{2}, also d_{1}'' glue with
d_{1}''' of σ_{3}…., d_{1}^{k}^{1}
of σ_{k1} glue with d_{1}^{k} of σ_{k}
and since d_{1} = d_{1}'d_{1}''d_{1}'''....d_{1}^{k}^{1}d_{1}^{k}
hence if we remove d_{1}, then all σ_{1}, σ_{2},
..., σ_{k} are finished. 
• 
Since d_{i}, 1 < i ≤ k + 1 is not glue with any arc
of any σ, then the only σ which finished is σ_{i1}
and σ_{i} takes its place and represents σ_{i}
in the retract. 
• 
Since all d_{i}, i > k + 1 are straight strings then
remove it not finish any σ, hence the retract β' = σ_{1}σ_{2}...σ_{k}. 
Corollary 1
The limit of retractions of i) in Theorem 5 is a 1braid group, also
the limit of the retraction of ii) is a 1braid, but the limit of the
retractions of iii) is σ_{1}σ_{2}....σ_{k}.
Lemma 1
Let β, β_{1}, β_{2} be a three elements of a
nbraid group B_{n} and r be a retraction from B_{n}{a} onto
,
then not necessary
• 
r(β_{1} ° β_{2}) = r(β_{1})
° r(β_{2}), 
• 
r(β^{1}) = (r(β))^{1}, β^{1}
be the inverse element of β. 
We can show the lemma by the following examples:
• 
Let β_{1} = σ_{3}σ_{1}
and β_{2} = σ_{1}σ_{2}σ_{3}σ_{1}
be a two elements of 4braid group B_{4}, r be a retraction from
B_{4}{a} onto
a ∈ d_{1}, then β_{1} ° β_{2}
= σ_{3}σ_{1}^{2}σ_{2}σ_{3}σ_{1},
r(β_{1}) = σ_{2} is a 3braidas, r(β_{2})
= σ_{1} is a 3braid, r(β_{1} ° β_{2})
= σ_{2}σ_{1}σ_{2} is a 3braid and
r(β_{1}) ° r(β_{2}) = σ_{2}σ_{1}
as a 3braid, hence r(β_{1} ° β_{2}) ≠r(β_{1})
° r(β_{2}) Fig. 7. 
• 
Let β = σ_{2}σ_{1}σ_{3}
be an element of 4braid group B_{4} and β^{1} = σ_{3}^{1}σ_{1}^{1}σ_{2}^{1}
be its inverse, r be a retraction from B_{4}{a} onto
a ∈ d_{1}, then r(β) = σ_{2}σ_{3}
as a 3braid, hence (r(β))^{1} = σ_{3}^{1}σ_{2}^{1}
and r(β^{1}) = σ_{1}^{1} as a 3braid,
hence r(β^{1}) ≠(r(β))^{1} Fig.
8. 

Fig. 7: 
r(β_{1} ° β_{2}) ≠r(β_{1})
° r(β_{2}) 
Theorem 6
The retraction of a nbraid group B_{n} is not necessary a
braid group.
Proof
From the lemma above we can say easy there is not exist for all element
in r(B_{n}{a}) an inverse element.
Theorem 7
Let B_{n} be a nbraid group and M = {σ_{1}, σ_{2},
..., σ_{n1}} be a set of its generators, r be a retraction from
B_{n}{a} onto ,
then i) if a ε d_{1} or a ε d_{n} then r(M)
is a set of the generators of B_{n1}, ii) if a ε d_{i},
1 < i < n, then r(M) = {σ_{1}, σ_{2}, ..., σ_{i1},
σ_{i+1}, σ_{n1}}
Proof
• 
If we remove an arc d_{1} from all the generators,
then σ_{1} was vanished, also if we remove d_{n},
then σ_{n1} was vanished and σ_{2} becomes
σ_{1}, etc. 
• 
Proof ii) came directory from the proof i). 
Theorem 8
Let β = σ_{2}σ_{4}σ_{6}...σ_{2k2}
be a nbraid, then β' = σ_{1}σ_{3}...σ_{2k1}
as a n1braid is a retract of β.
Proof
Let β = σ_{2}σ_{4}σ_{6}...σ_{2k2}
be a nbraid, {d_{1}, d_{2}, ..., d_{n}} be an arcs
of β and r be a continuous map from β{a} onto β' where a is
a point on d_{i}.i =1, 2, ..., n defined, by ,
then r(d_{i}) = d_{i} ∀ d_{i} ε β'
hence r is a retraction and β' = σ_{1}σ_{3}...σ_{2k1}
is a n1braid.
Theorem 9
Let β = σ_{1}σ_{3}...σ_{2k1}
be a nbraid, then β = σ_{2}σ_{4}σ_{6}...σ_{2k2}
as a n1braid is a retract of β.

Fig. 8: 
r(β^{1}) ≠(r(β))^{1} 
Proof
The proof is similar to the proof of Theorem 8.
Theorem 10
Let B_{n} be a pure nbraid group generated by a_{1}, a_{2},
..., a_{n1}, where, a_{i} = (σ_{n1}σ_{n2},
..., σ_{i+1})
and r:(a_{i}a) →b_{i} be a retraction then
• 
b_{i} is a trivial n1braid if a ε
d_{i} or a ε d_{n}. 
• 
if a ε d_{j}, i < j < n. 
• 
b_{i} = a_{i1} in n1braid, if a ε d_{j},
j < i 
Proof
The proof is clear.
Theorem 11
Let B_{n} be a nbraid group and r be a retraction from B_{n}{a}
onto ,
then is
a monoid.
Proof
Let B_{n} be a nbraid group and r be a retraction from B_{n}{a}
onto ,
then the all elements of
are n1braids, hence by definition the product of any two elements is also
an element in ,
also the associative law is holds and we can see easy the identity element is
exists, hence
is a monoid.
Theorem 12
A retract of a singular braid not necessary a singular braid.
Proof
We can show that by the following Example:
Example 1
Let β = σ_{1}σ_{3}σ_{2} be a singular
4braid and r: (β{a}) →
be a retraction and {a} be a singular point then r(β{a}) = σ_{1}
which is not singular braid Fig. 9.

Fig. 9: 
Singular braid not necessary a singular braid 

Fig. 10: 
r(β_{1}) = r(β_{2}) but β_{1}
β_{2}/td>


Fig. 11: 
β_{1}, β_{2}, β_{3}
are equivalents and has the same r(β_{i}{a}), i = 1,
2, 3 
Theorem 13
A retraction of any isotopic braids is an invariant.
Proof
Let β_{1}, β_{2} be a two nbraids such that β_{1}
≈ β_{2} and a Γ be a set of all nbraids and r be a
retraction from Γ{a} onto ,
we want to prove r(β_{1}) = r(β_{2}), since β_{1}
≈ β_{2} then π(β_{1}) = π(β_{2}),
this means the ends points of the arcs d_{1}, d_{2}, ..., d_{n}
of β_{1} and β_{2} have the same location, by elementary
move Ω or Ω^{1} on the arcs the ends points can not be changed,
but the σ`s which components of β_{1} and β_{2}
are exchanged the places, hence by remove any arc from β_{1} and
β_{2} gives the same result which means r(β_{1}) =
r(β_{2}).
Remark 1
If r(β_{1}) = r(β_{2}), then not necessary
β_{1} ≈ β_{2}.
By the following example we show that:
Example 2
Let ,
β_{2} = σ_{1}σ_{2}σ_{3}
be a two 4braids, then r(β_{1}) = r(β_{2}) but β_{1}
β_{2} Fig.10.
Example 3
Let β_{1}, β_{2}, β_{3} be a three 4braids,
easy we can get one from the another by some elementary moves, hence this three
4braids are equivalents, if we apply the retraction on the three 4braids,
r:(β_{i}{a}) → ,
a ε d_{1} we get =
(σ_{2}σ_{1})^{3}, (Fig. 11).