Transcript Construction of Spaces

Construction of Spaces
Comments on Chap. 5 from
Sieradski's book
http://cis.k.hosei.ac.jp/~yukita/
The choice of topology influences
the continuity.
Prop.
Given : An inclusion function i : X  Y . A topology S on the set Y .
Claim : i : ( X ,2 X )  (Y , S ) is continuous .
Proof. U  S ; i 1 (U )  2 X
Is there a smallest t opology T such that i : ( X , T )  (Y , S ) is continuous ?
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1.1 Topological Subspace Construction
Let i : X  Y be an inclusion of a set X into the topologic al space (Y , S ).
Then,
(a) i * S  {i 1 (V )  V  X  X | V  S} is a topology on X .
(b) i : ( X , T )  (Y , S ) is continuous if and only if T  i * S .
Therefore, i * S is the smallest t opology on X for which i is continuous .
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Proof of (a)


By definition U  X ; U  i * S  V  S ;U  i 1 (V ) .
Clearly, we have   i 1 () and X  i 1 (Y ), and thus , X  i * S .
Arbitrary union
1
1
*
i
(
V
)

i
(
V
)

i
 
  S since

Finite intersecti on

1
1
*
i
(
V
)

i
(
V
)

i
 k
 k S since
k
k
 V  S .
V
k
 S.
k
Remark. Taking arbitrary intersecti on does not make sense because
 V  S
is not guaranteed .
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Proof of (b)
i : ( X , T )  (Y , S ) is continuous  V  S ; i 1 (V )  T .
This means i * S  T .
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Transitivity
Def. Let i : X  Y be an inclusion and Y have topology S .
Then, ( X , i * S ) is said to have the subspace topology.
Prop. If W is a subspace of X and X is a subspace of Y ,
then W is a subspace of Y .
Proof. Let j : W  X and i : X  Y be inclusions .
(i  j ) 1 (V )  j 1 (i 1 (V )) for all open sets V  Y .
All open sets in X have the form i 1 (V ), and thus all open sets
in W have the form j 1 (i 1 (V )).
In other word s, all open sets in W have the form (i  j ) 1 (V )
for some open set V  Y .
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Key feature
Let i : X  Y be a subspace inclusion.
k : W  X is continuous  i  k : W  Y is continuous .
Proof.  is trivial.
If i  k : W  Y is continuous , for each open set V  Y ,
(i  k ) 1 (V )  k 1 (i 1 (V ))  k 1 (V  X ) is open in W .
Thus k 1 is continuous since all open subsets of X have the
form V  X for some open set V  Y .
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Restriction
Theorem. Let i : X  Y be an inclusion map
and g : Y  Z be a continuous map.
Then, g | X : X  Z is continuous .
Proof . g | X  g  i.
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Embedding
Any continuous function h : W  Y gives a continuous function
h : W  h(W ), provided that the image h(W )  Y is given
the subspace topology.
The continuous function h : W  Y is called an embedding of
W into Y provided that h : W  h(W ) is a homeomorph ism.
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Distinguishing Spaces
embeddable
non-embeddable
Cylinder
Moebius strip
Sphere
W1 and W2 is non - homeomorph ic if
One has an embedding into a space Y while the other does not.
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Topological Invariants
non-embeddable
Wire handcuff
embeddable
Wire eight
the cross
For a fixed space W , the existence of an embedding W  Y
is a topological invariant of Y .
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Identification Topology
(Dual to the subspace topology)
Let p : X  Y be a surjection and X have topology T .
The indiscrete topology {, Y } is the smallest t opology on Y ,
and it makes the given function p : ( X , T )  (Y , {, Y })
continuous .
Is there a largest to pology S on Y for which
p : ( X , T )  (Y , S ) is continuous .
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1.5 Identification Space
Construction
Let p : ( X , T )  Y be a surjection from the topologic al space
( X , T ) to the set Y . Then,
(a) The collection p*T  {V  Y | p 1 (V )  T } is a topology on Y , and
(b) p : ( X , T )  (Y , S ) is continuous if and only if S is contained in p*T .
Therefore, p*T is the largest to pology on Y for which p is continuous .
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Proof of (a)
By definition , V  p*T  p 1 (V )  T .
Since p 1 ()    T and p 1 (Y )  X  T , we have {, Y }  p*T .
Let V  p*T , namely p 1 (V )  T .
Then, p 1 (V )   p 1 (V )  T , namely


 V  p T .
*
Let Vi  p*T , namely p 1 (Vi )  T .
Then, p 1 (Vi )   p 1 (Vi )  T , namely
i
i
V  p T .
i
*
i
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Proof of (b)
p : ( X , T )  (Y , S ) is continuous  V  S ; p 1 (V )  T .
This means S  p*T .
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Identification Topology
Surjection p : X  Y identifies p 1 ({ y}) to the correspond ing
point y  Y .
p*T on Y is called the identification topology determined by
p and T .
(Y , p*T ) is called an identification space of ( X , p)
and p is called an identification map.
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Ex 4.
X  {1,2,3}, T  {, {1},{1,2},{1,3}, X }, Y  {a, b}.
p : X  Y is defined by p (1)  p(3)  a and p (2)  b.
Then, p*T  {, {a}, Y }.
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1.6 Theorem
A continuous surjection p:XY that is
either open or closed is an identification map.
Proof.
Let p is an open(close d) function.
(i) Let V  Y be a subset wit h p 1 (V ) is open(close d) in X .
Then V  p ( p 1 (V )) since p is surjective , and V is open(close d)
since p is open(close d).
(ii) Let V  Y be open(close d) in Y .
Then p 1 (V ) is open(close d) since p is continuous .
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Ex. 6, 7 Identification Maps
e :  1 is defined by e(t )  (cos 2t , sin 2t ).
p : 2  is defined by p( x, y)  x.
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1.7 Transitivity
Let p : X  Y and q : Y  Z be indentific ation maps.
Then q  p : X  Z is an identifica tion map.
Proof. V  Z is open if and only if q 1 (V )  Y is open,
and q 1 (V )  Y is open if and only if p 1 (q 1 (V ))  X
is open. This proves that q  p : X  Z is an identifica tion map.
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1.8 Key Feature
Let p : X  Y be an identifica tion map.
k : Y  Z is continuous if k  p is continuous .
Proof. For each subset W  Z ,
1
1
1
(k  p ) (W )  p (k (W ))  X is open.
Hence, k 1 (W )  Y is open since p is an identifica tion.
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1.9 Transgression Theorem
Let p : X  Y be an identifica tion map and let g : X  Z be any
continuous function t hat respects the identifica tion of p :
x, x  X ; p( x)  p( x) implies g ( x)  g ( x).
Then there exists a unique continuous function
h : Y  Z such that h  p  g.
Proof. Since y  g ( p 1 ({ y})) is single valued, h is well - defined.
From the key feature, h is continuous .
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1.10 Corollary
Let p : X  Y and q : X  Z be identifica tion maps that respect
one another' s identifica tions. Then, X and Y are homeomorph ic.
Proof. Applying Transgress ion Theorem twice, we have continuous
functions h : Y  Z and k : Z  Y such that h  p  q and k  q  p.
Then k  h  p  k  q  p and h  k  q  h  p  q;
hence, k  h  1Y and h  k  1Z , by the surjectivi ty of p and q.
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Quotient Modulo A Subspace
U
X
U
X/A
A
V
V
[A]
qA : X  X / A
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