Download Chapter 2 Product and Quotient Spaces

Chapter 2
Product and Quotient Spaces
Let (Xi , Ji ), i = 1, 2, . . . , n be given topological spaces. Also note that it is possible
that (Xi , Ji ) = (Xj , Jj ) even when i 6= j. What do we mean by the cartesian product
n
X = X1 × X2 × · · · × Xn ? We define the cartesian product as X = Π Xj = X1 ×
j=1
X2 × · · · × Xn = {f : J → ∪ Xj : f (i) = xi ∈ Xi ∀ i ∈ J}, where J = {1, 2, . . . , n}.
j∈J
Here we identify an f ∈ X with (x1 , x2 , . . . , xn ), where xi = f (i) for all i = 1, 2, . . . , n.
What is the advantage of defining the cartesian product in this way? Let J
be a nonempty set (finite or infinite) and for each α ∈ J we have a topological space
(Xα , Jα ). Now we define the cartesian product Π Xα = X as X = {f : J → ∪ Xα :
α∈J
α∈J
f (α) = xα ∈ Xα , ∀ α ∈ J}.
2.1 Product Space
Definition 2.1.1. For each α ∈ J, define a function pα : X → Xα , known as α-th
projection or coordinate function, as pα (f ) = f (α) = xα .
Our aim is to define a topology J on Π Xα which will have the following
α∈J
properties:
• Each projection function pα : (X, J ) → (Xα , Jα ) is continuous.
• Whenever J 0 is a topology on X such that each pα : (X, J 0 ) → (Xα , Jα ) is
continuous then J ⊆ J 0 .
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That is J is the smallest (or weakest topology) on X that makes each pα continuous.
−1
For α ∈ J, let Sα = {p−1
α (Uα ) : Uα ∈ Jα }. Then we require that pα (Uα ) is open
in our proposed topological space (X, J ).
Note that
p−1
α (Uα )
Hence we require that ∪ Sα ⊆ J .
α∈J
= {f ∈ X : pα (f ) = f (α) = xα ∈ Xα }. Hence if we fix an
α ∈ J, then p−1
α (Uα ) = Π Aβ , where Aα = Uα and Aβ = Xβ when β 6= α. If
β∈J
α1 , α2 , . . . , αn , n ∈ N and Uαi ∈ Jαi , i = 1, 2, . . . , n, then p−1
αi (Uαi ) ∈ J . Also J
n
is closed under finite intersections means that ∩ p−1
αi (Uαi ) = Π Aα ∈ J , where
i=1
α∈J
Aαi = Uαi , i = 1, 2, . . . , n and Aα = Xα when α 6= α1 , α2 , . . . , αn . Now it is easy
to see that B = { Π Uα : Uα ∈ Jα for all α ∈ J and Uα = Xα , except for finitely
α∈J
many α 6= α1 , α2 , . . . , αn ∈ J} is a basis for a topology on X. The topology J induced
by B is called the product topology on X = Π Xα and the topological space (X, J )
α∈J
is called the product topological space (also known as product space) induced by
the topological spaces (Xα , Jα ), α ∈ J.
Remark 2.1.2. What will happen when J = {1, 2, . . . , n} for some natural number
n? When n = 1, X = {f : {1} → X1 : f (1) = x1 ∈ X1 } = X1 and X =
n
{f : {1, 2, . . . , n} → ∪ Xi : f (i) = xi ∈ Xi }. That is f = (f (1), f (2), . . . , f (n)) =
i=1
n
(x1 , x2 , . . . , xn ) ∈ X = Π Xi . Hence X = {(x1 , x2 , . . . , xn ) : xi ∈ Xi , i = 1, 2, . . . , n}
i=1
= X1 × X2 × · · · × Xn .
In this case, that is when J = {1, 2, . . . , n} is a finite index set containing n
n
elements, B = { Π Ui : each Ui is open in Xi , i = 1, 2, . . . , n} is a basis for the product
i=1
n
topology J on X = Π Xi .
i=1
Now let us prove the following theorem:
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Theorem 2.1.3. Let J 6= φ be an index set and (Xα , Jα ), α ∈ J be a collection of
Hausdorff topological spaces. Then the product space Π Xα , J is also a Hausdorff
α∈J
topological space.
Proof. Our aim is to prove that the product topological space ( Π Xα , J ) is a
α∈J
Hausdorff topological space. So take two distinct elements f, g in Π Xα . Now
α∈J
f, g ∈
Π Xα implies f : J
α∈J
→ ∪ Xα and
α∈J
g : J
→ ∪ Xα such that
α∈J
f (α) = xα ∈ Xα , g(α) = yα ∈ Xα , for each α ∈ J. Also f 6= g ⇒ there exists α0 ∈ J
such that xα0 = f (α0 ) 6= g(α0 ) = yα0 . We have xα0 , yα0 ∈ Xα0 and xα0 6= yα0 . Hence
(Xα0 , Jα0 ) is a Hausdorff topological space implies that there exist Uα0 , Vα0 ∈ Jα0
satisfying:
(i) xα0 ∈ Uα0 , yα0 ∈ Vα0 and (ii) Uα0 ∩ Vα0 = φ.
Now use (i) and (ii) to construct basic open sets U , V in the product space
satisfying f ∈ U , g ∈ V, and U ∩ V = φ. So, let Uα = Xα , Vα = Xα , whenever
α 6= α0 . We already have Uα0 , Vα0 which are open sets in (Xα0 , Jα0 ). Define U, V
as U = Π Uα , V = Π Vα , where U , V are defined as above. We have f (α) ∈ Xα ,
α∈J
α∈J
g(α) ∈ Xα for all α ∈ J and hence f ∈ U , g ∈ V (why?). Also U ∩ V = ( Π Uα ) ∩
α∈J
( Π Vα ) = Π Uα ∩ Vα = φ, since Uα0 ∩ Vα0 = φ. That is for f, g ∈ Π Xα with f 6= g
α∈J
α∈J
α∈J
there exist basic open sets U, V in the product space such that f ∈ U, g ∈ V, and
U ∩ V = φ. This implies that the product space ( Π Xα , J ) is a Hausdorff space. α∈J
Note. Let (X, J ) be a topological space and B be a basis for (X, J ) (or say B is
a basis for J ). Then for a subset A of X, x ∈ A if and only if for each U ∈ B with
x ∈ U, U ∩ A 6= φ. That is x ∈ A if and only if for each basic open set U
containing x, U ∩ A 6= φ.
?
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Theorem 2.1.4. Let (Xα , Jα ), α ∈ J be a collection of topological spaces and
Aα ⊆ Xα for each α ∈ J then Π Aα = Π Aα , with respect to the product space
α∈J
α∈J
( Π Xα , J ).
α∈J
Proof. First let us prove Π Aα ⊆ Π Aα . Let f ∈ Π Aα . Then f : J → ∪ Aα
α∈J
α∈J
α∈J
α∈J
such that f (α) = xα ∈ Aα for all α ∈ J. We aim to prove that f is in the closure
of Π Aα in the product topological space Π Xα . So take a basic open set B in the
α∈J
α∈J
product space Π Xα , J containing f, say B = Π Uα . It is given that f ∈ Π Aα .
α∈J
α∈J
α∈J
Hence f (α) = xα ∈ Aα for each α ∈ J. Now f ∈ B = Π Uα implies f (α) ∈ Uα
α∈J
for all α ∈ J. That is Uα is an open set containing xα and xα ∈ Aα . This implies
that Uα ∩ Aα 6= φ. Let zα ∈ Uα ∩ Aα for all α ∈ J. Define g : J → ∪ Aα as
α∈J
g(α) = zα ∈ Aα then g ∈ B ∩ Π Aα . This implies that for each basic open set B
α∈J
containing f, B ∩ Π Aα 6= φ. This implies f ∈ Π Aα . That is f ∈ Π Aα implies
α∈J
α∈J
α∈J
f ∈ Π Aα and this proves the assertion
α∈J
Π Aα ⊆ Π Aα .
α∈J
(2.1)
α∈J
Now let us prove the converse part namely Π Aα ⊆ Π Aα . So let f ∈ Π Aα ⊆
α∈J
α∈J
α∈J
Π Xα = X. Our aim is to prove: f ∈ Π Aα . But f ∈ Π Xα if and only if f (α) ∈ Xα
α∈J
α∈J
α∈J
for each α ∈ J and f ∈ Π Aα if and only if f (α) ∈ Aα for each α ∈ J. For a fixed
α∈J
α0 ∈ J take an open set Uα0 containing f (α0 ) = xα0 . We will have to use the fact that
f ∈ Π Aα . To use this fact we will have to construct a basic open set containing f .
α∈J
Keeping this in mind, we define B = Π Uα , where Uα = Xα , when α 6= α0 and Uα0
α∈J
is as given above. Now this B is a basic open set containing f and hence f ∈ Π Aα
α∈J
implies B ∩ Π Aα 6= φ. That is Π Uα ∩ Π Aα = Π Uα ∩ Aα 6= φ. This implies
α∈J
α∈J
α∈J
α∈J
that each Uα ∩ Aα 6= φ. In particular Uα0 ∩ Aα0 6= φ implies xα0 ∈ Aα0 . Note that
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though our α0 ∈ J is a fixed element, there is no restriction on α0 ∈ J and the proof
will go through for any α ∈ J. This gives that f (α) = xα ∈ Aα for all α ∈ J and this
implies that f ∈ Π Aα . Hence f ∈ Π Aα ⇒ f ∈ Π Aα . This implies
α∈J
α∈J
α∈J
Π Aα ⊆ Π Aα .
α∈J
(2.2)
α∈J
Now combining Eqs. (2.1) and (2.2) we have Π Aα = Π Aα .
α∈J
α∈J
2.2 The Box Topology
Definition 2.2.1. Let (Xα , Jα ), α ∈ J, be a collection of topological spaces and
Bb = { Π Uα : Uα ∈ Jα for α ∈ J}. Then Bb is a basis for a topology on X = Π Xα
α∈J
α∈J
and Jb , the topology induced by Bb , is called the box topology on X.
Remark 2.2.2. From the definitions of product and box topologies, it is clear that
B ⊆ Bb , where the product topology J on X is induced by B (refer the definition of
product topology) and the box topology Jb is induced by Bb . Now B ⊆ Bb implies
JB = J ⊆ Jb . That is the product topology on Π Xα = X is weaker than the box
α∈J
topology Jb on X.
It is to be noted that if a subset A of X is open with respect to the product
topology on X then it is also open with respect to the box topology on X. Note that
∞
the set A = Π − n1 , n1 is an open set in Rw = R × R × · · · with respect to the box
i=1
topology but not open with respect to the product topology on Rw . Also we have
proved that if (Xα , Jα ), α ∈ J, is a collection of Hausdorff topological spaces then the
product space Π Xα , Jα is also a Hausdorff space. Since J ⊆ Jb it is clear that
α∈J
if Xα , Jα , α ∈ J is a collection of Hausdorff topological spaces then Π Xα , Jb is
α∈J
also a Hausdorff space.
Note that if J is a nonempty finite index set then J = Jb on Π Xα .
α∈J
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Theorem 2.2.3. Let (Xα , Jα ), α ∈ J, be a collection of topological spaces and for
each α ∈ J, let Aα ⊆ Xα . Then Π Aα = Π Aα , where Aα denotes the closure of
α∈J
α∈J
Aα in (Xα , Jα ) and Π Aα denotes the closure of Π Aα in ( Π Xα , Jb ).
α∈J
α∈J
α∈J
Proof. Proof of this theorem is similar to that of theorem 2.1.4.
Theorem 2.2.4. Let
and
(X, J ),
(Y, J 0 ),
(Z, J 00 )
be
topological
spaces
f : (X, J ) → (Y, J 0 ), g : (Y, J 0 ) → (Z, J 00 ) be continuous functions then the
composite function g ◦ f : (X, J ) → (Z, J 00 ) defined as (g ◦ f )(x) = g(f (x)) is also a
continuous function.
Proof. We aim to prove g ◦ f : (X, J ) → (Z, J 00 ) is a continuous function. So start
with an open set W in (Z, J 00 ). Now W is an open set in Z (means W ∈ J 00 ) and
g : (Y, J 0 ) → (Z, J 00 ) is a continuous function implies g −1 (W ) is an open set Y .
Now f : (X, J ) → (Y, J 0 ) is also a continuous function. Hence f −1 (g −1 (W )) is an
open set in X. We define for A ⊆ Y , f −1 (A) = {x ∈ X : f (x) ∈ A}. That is
x ∈ f −1 (A) if and only if f (x) ∈ A. Hence f −1 (g −1 (W )) = {x ∈ X : f (x) ∈ g −1 (W )}
= {x ∈ X : g(f (x)) ∈ W } = (g ◦ f )−1 (W ). That is we have proved: W is an open set
in (Z, J 00 ) implies (g ◦ f )−1 (W ) is an open in (X, J ) implies g ◦ f : (X, J ) → (Z, J 00 )
is a continuous function.
Definition 2.2.5. A sequence {xn } is a topological space (X, J ) is said to converge
an element x in X if for each open set U containing x, there exists a natural number
n0 (that is n0 ∈ N) such that xn ∈ U for all n ≥ n0 .
∞
The product topology R × R × R × · · · = Π Rn = Rw , where Rn = R,
n=1
n = 1, 2, 3 . . . is metrizable. That is, we will have to define a metric say d1 of Rw
w
such that Jd1 = J , the product topology on Rw . For x = (xn )∞
n=1 = (xn ) ∈ R ,
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w
y = (yn ) ∈ R , let d1 (x, y) = sup
n≥1
n
d(xn ,yn )
n
o
, where d(xn , yn ) = min{1, |xn − yn |}.
(Exercise. Let (X, d) be a metric space and d(x, y) = min{1, d(x, y)} for all x, y ∈ X.
Prove that (i) d is a metric on X, (ii) Jd = Jd .)
It is easy to prove that d1 is a metric on Rw . First let us prove that Jd1 ⊆ J . So,
let U ∈ Jd1 . We aim to prove that each point of U is an interior point of U with respect
to the product topology J on Rw . Take x ∈ U . Now x ∈ U , U is an open set in the
metric space (Rw , d1 ) implies there exists r > 0 such that Bd1 (x, r) ⊆ U . Now choose
n0 ∈ N such that
1
n0
< r and B = (x1 − , x1 + ) × · · · × (xn0 − , xn0 + ) × R × R × · · ·
then B is a basic open set in (Rw , J ) containing x = (xn ). Now we leave it as an
exercise to prove that B ⊆ Bd1 (x, ). Hence for each x ∈ U , there exists a basic open
set B in (Rw , J ) such that x ∈ B ⊆ U. This proves that U ∈ J that is
Jd1 ⊆ J .
(2.3)
Now let us prove that J ⊆ Jd1 . To prove this statement it is enough to prove
that every basic open subset V of (Rw , J ) is in Jd1 . Now V is a basic open set in the
product topology implies there exists k ∈ N such that V = V1 × V2 × · · · × Vk × R ×
R × · · · . Let x = (xn )∞
n=1 ∈ V . Hence there exist 1 , 2 . . . k , 0 < i < 1 for i =
1, 2, . . . , k such that (xi − i , xi + i ) ⊆ Vi . Now let = min{ ii : i = 1, 2, . . . , k}. (note:
we have Ui = R for all i > k and hence it is enough to consider 1 , . . . , k ) and we claim
n
o
that Bd1 (x, ) ⊆ V . So, let y ∈ Bd1 (x, ) then d1 (x, y) < implies sup d(xnn,yn ) < n≥1
implies
d(xn ,yn )
n
< for all n = 1, 2, . . . , k implies
1
n
min{1, |xn − yn |} < for all
n = 1, 2, . . . , k implies min{1, |xn − yn | < n < n < 1 for all n = 1, 2, . . . , k implies
|xn − yn | < n for all n = 1, 2, . . . , k implies y = (yn ) ∈ V1 × · · · × Vk × R × · · · = V .
This proves that Bd1 (x, ) ⊆ V . That is for each x ∈ V there exists > 0 such that
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Bd1 (x, ). Hence every point of V is an interior point of V with respect to (Rw , Jd1 ).
Hence
V ∈ Jd1 .
(2.4)
∞
Now if U ∈ J then there exists k ∈ N and B1 , B2 , . . . , Bk ∈ B (B = { Π Un : each
n=1
Un is open in R and Un = R for except finitely many n0 s} is our standard basis for
the product topology J on Rw ) such that U = B1 ∩ · · · ∩ Bk . We have proved that
each basic open set B of J belongs to Jd1 (i.e B ∈ Jd1 ) (refer Eq. (2.4)). Now
B1 , B2 , . . . , Bk ∈ Jd1 and Jd1 is a topology implies B1 ∩ B2 ∩ · · · ∩ Bk ∈ Jd1 . This
proves that U ∈ Jd1 That is
U ∈ J ⇒ U ∈ Jd1 ⇒ J ⊆ Jd1 .
(2.5)
From Eqs. (2.3) and (2.5) we see that J = Jd1 . Hence the product space
Rw = R × R × · · · with product topology J is metrizable. It is interesting to note
that if we consider the box topology say Jb on Rw , then (Rw , Jb ) is not a metrizable
topological space.
How to prove that a given topological space (X, J ) is not metrizable. If (X, J )
is metrizable then we will have to find a metric (finding such a metric is not at all an
easy task and this statement will become meaningful if we have patience to wait and
see the proof of the Urysohn metrization theorem) say d on X such that Jd = J . We
have just proved that (Rw , J ) (with product topology) is metrizable and in this case
we could define a metric d on Rw such that Jd = J .
To prove that a topological space is not metrizable space is comparatively
easier.
For example, if the given topological space (X, J ) is not a Hausdorff
topological space then it is clear that there cannot exist any metric d on X such
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that Jd = J . We know that if d is a metric on X, then (X, Jd ) is a Hausdorff
space. So, what we need here is to find a property a metric space has whereas the
given topological space does not have that particular property. Now let us come back
and prove that (Rw , Jb ) is not metrizable. Suppose there exists a metric say d on
Rw such that Jd = Jb . Then we know that for A ⊆ X, x ∈ A if and only if there
exists a sequence (xn ) in A such that (xn ) → x as n → ∞ (to prove this statement,
observe the following: For x ∈ A, B(x, n1 ) ∩ A 6= φ, for each n ∈ N and hence we have
∞
{B(x, n1 )∩A}∞
n=1 = {An }n=1 = {An }n∈N a collection of nonempty sets. Now by axiom
∞
of choice there exists a choice function say f : N → ∪ An such that xn = f (n) ∈ An .
n=1
So using axiom of choice we have got a sequence (xn ) in A and now it is easy to see
that xn → x as n → ∞).
Note that normally we just say that B(x, n1 ) ∩ A 6= φ implies there exists
xn ∈ B(x, n1 )∩A = An . In such case it is to be understood that in fact we use axiom of
choice to define such a sequence (xn ). Now let us prove that if A = {(x1 , x2 , x3 . . . , ) :
xk > 0 for all k ∈ N}. Then 0 = (0,0,0,. . . ,) ∈ A, but there does not exist any
sequence (x(n) ) in A such that x(n) → (0, 0, . . .) in (Rw , Jb ).
Step 1: Prove that 0 ∈ A.
∞
So take an open set say U containing 0 then there exists a basic open set B = Π Bn =
n=1
B1 × B2 × · · · such that (0, 0, . . . , ) ∈ B1 × B2 × · · · ⊆ U . (Here each Bk is an open set
in R containing 0 ∈ R) 0 ∈ Bk , k = 1, 2, 3 . . . implies there exist ak , bk ∈ R, ak < bk
such that 0 ∈ (ak , bk ) ⊆ Bk , bk > 0 and hence
bk
2
> 0 implies b = ( b2k )∞
k=1 ∈ A ∩ B
implies A ∩ B 6= φ implies A ∩ U 6= φ implies 0 = (0, 0, . . . , ) ∈ A. Now we claim
that there cannot exist any sequence x(n) in A such that x(n) → (0, 0, 0 . . .).
Let x(n) = (a1n , a2n , . . . , ) ∈ A. Then each ain > 0 for all i = 1, 2, . . .. In particular,
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akk > 0 for all k = 1, 2, . . .. Let U = ( (−a211 ) , a211 ) × ( (−a222 ) , a222 ) × · · · then U is an
open set in (Rw , Jb ) containing 0.
What will happen if U ∩ A 6= φ. Note that for each n, ann ∈
/ ( (−a2nn ) , ann
) and
2
hence x(n) = (a1n , a2n , . . . , ) ∈
/ U . If x(n) → (0, 0, . . .) then there exists n0 ∈ N such
that x(n) ∈ U for all n ≥ n0 . But here x(n) ∈
/ U for every n. Hence x(n) does not
converge to (0,0,0, . . .). So (0, 0, . . .) ∈ A but there cannot exist any sequence in A
which converges to (0, 0, . . .) with respect to Jb . This proves that (Rw , Jb ) is not a
metrizable topological space.
2.3 Quotient (Identification) Spaces
We start with a given topological space (X, J ). By identifying some of the
points of X we can produce a new topology on a new set say X ∗ . For example if
we consider the closed unit ball in R2 , then our given topological space is (X, J ),
where X is the closed unit ball in R2 . Here we consider (X, J ) as a subspace of the
Euclidean space R2 . Now we get a new set X ∗ = {(x1 , x2 ) ∈ R2 : x21 + x22 < 1} ∪ {S 0 },
where S 0 is the unit circle (boundary) of the closed disc X. By defining a suitable
topology J ∗ on X ∗ we can show that (X ∗ , J ∗ ) is homeomorphic to the 2-sphere
S 2 = {(x1 , x2 , x3 ) ∈ R3 : x21 + x22 + x23 = 1}. It is to be noted that here we are
considering S 2 as a subspace of R3 (also note that if no topology on Rn , n ≥ 1 is
mentioned then it is understood that we have the usual topology on Rn ).
Now let us see how to construct the quotient topology. Let (X, J ) be a
topological space and X ∗ be a nonempty set. Let p : X → X ∗ be a surjective map.
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Then J ∗ = {A ⊆ Y : p−1 (A) is open in (X, J )} is a topology on X ∗ . This topology
J ∗ on X ∗ is called the quotient topology on X ∗ induced by p.
It is easy to prove that J ∗ is a topology on X and we leave it as an exercise.
Definition 2.3.1. Let (X, J ) be a topological space and X ∗ be a partition of X into
disjoint subsets whose union is X. Let p : X → X ∗ be the natural map satisfying the
condition namely x ∈ p(x), for each x ∈ X. Suppose for a given x ∈ X there exist
A, B ∈ X ∗ such that x ∈ A and x ∈ B. Then x ∈ A ∩ B. This implies B = A. Hence
for each x ∈ X there exists a unique A ∈ X ∗ such that x ∈ A and this A is our p(x).
Also ∪ ∗ A = X implies that p is onto. The quotient topology J ∗ on X ∗ is induced
A∈X
by p and we say that (X ∗ , J ∗ ) is a quotient topology of (X, J ).
Let (X, J ) be a topological space and X ∗ be a partition of X into disjoint
subsets whose union is X. Define a relation R on X as follows:
R = {(x, y) ∈ X × X : x, y ∈ A for some A ∈ X ∗ } then (i) xRx, that is (x, x) ∈ R for
all x ∈ X, (ii) for x, y ∈ X, xRy implies there exists A ∈ A∗ such that x, y ∈ A. Hence
y, x ∈ A and this gives yRx that is for x, y ∈ X xRy ⇒ yRx, (iii) for x, y, z ∈ X, xRy
and yRz implies there exist A, B ∈ X ∗ such that x, y ∈ A and y, z ∈ B. Therefore
y ∈ A∩B and this implies that A = B. From this we have x, z ∈ A. Hence xRz. That
is xRy and yRz implies xRz. From (i), (ii) and (ii) we see that R is an equivalence
relation on X and hence this relation R will partition X into disjoint equivalence
classes.
For each x ∈ X, the equivalence classes determined by x is given by
x = {y ∈ X : yRx}. Hence if x ∈ A, for some A ∈ X ∗ then x = A. Now it is
easy to see that for U ⊆ X ∗ , U ∈ J ∗ if and only if ∪ A is an open subset of X.
A∈U
∗
Let (X, J ) be a topological space and X be a family of disjoint nonempty subsets
45
of X such that X =
∪ A. Define q : X → X ∗ as q(x) = A, where A ∈ X ∗ is
A∈X ∗
such that x ∈ A. Then the topology Jq on X ∗ is the largest topology on X ∗ which
makes q : (X, J ) → (X ∗ , Jq ) a continuous function is called the quotient topology
(or identification) topology on X ∗ induced by q.
Theorem 2.3.2. Let (X ∗ , Jq ) be an identification space (i.e Jq is the identification
topology on X ∗ with respect to q) defined as above and (Y, J 0 ) be an arbitrary
topological space. Then a function f : (X ∗ , Jq ) → (Y, J 0 ) is continuous if and only if
f ◦ q : (X ∗ , Jq ) → (Y, J 0 ) is continuous.
Proof. Let f : (X ∗ , Jq ) → (Y, J 0 ) be a continuous function. We know that by
the definition of identification space, q : (X, J ) → (X, Jq ) is a continuous function.
Hence the composite function f ◦ q : (X ∗ , Jq ) → (Y, J 0 ) is a continuous function.
We will have to prove that f : (X ∗ , Jq ) → (Y, J 0 ) is continuous. So start with an
open set U in Y . That is we will have to prove that f −1 (U ) is open in (X ∗ , Jq ).
But the subset f −1 (U ) is open in the identification space if and only if q −1 (f −1 (U ))
is open in (X, J ). But q −1 (f −1 (U )) = (f ◦ q)−1 (U ) an open set in (X, J ) (since
f ◦ q : (X, J ) → (Y, J 0 ) is a continuous function). This is what we wanted to prove
and hence f : (X, Jq ) → (Y, J 0 ) is a continuous function.
Let (X, J ) be a topological space and Y be a nonempty set. Let f : X → Y
be an onto map. Then X ∗ = {f −1 (y) : y ∈ Y } is a family of disjoint subsets of X
such that ∪ f −1 (y) = X. That is X ∗ is a partition of X. Let q : X → X ∗ be the
y∈Y
map, known as identification map, defined as above. Let J 0 be the largest topology
on Y for which f : (X, J ) → (Y, J 0 ) is continuous. Then it is easy to prove the
following:
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onto
Theorem 2.3.3. Let (X, J ), (Y, J 0 ) be topological spaces and f : (X, J ) −−→ (Y, J 0 )
be a homeomorphism. Further suppose (Z, J1 ) is any topological space. Then a
function g : (Y, J 0 ) → (Z, J1 ) is continuous if and only if g ◦ f : (X, J ) → (Z, J1 ) is
continuous.
Proof. To prove (Y, J 0 ) and (X ∗ , Jq ) are homeomorphic we will have to define a
map say h : X ∗ → Y and prove that this map is a homeomorphism.
Let z ∈ X ∗ = {f −1 (y) : y ∈ Y } be any element. Then z = f −1 (y), for some
y ∈ Y . So let h(z) = h(f −1 (y)) = y. The defined map h : X → Y is such that
(h ◦ q)(x) = h(q(x)) = h(f −1 (y)) (where y ∈ Y is such that x ∈ f −1 (y)) = y = f (x).
That is h ◦ q = f . Let us prove that h is continuous. Let V be an open set in Y .
X
f
q
Y
X*
h
Figure 2.1
The given topology J 0 on Y
is the largest topology on Y for which
f : (X, J ) → (Y, J 0 ) is continuous. Hence V is an open set on Y implies f −1 (V ) is
an open set in X and hence (h◦q)−1 (V ) = q −1 (h−1 (V )) is an open set in X. Therefore
h−1 (V ) is an open set in X ∗ . That is V is an open set in Y implies h−1 (V ) is an open
set in X ∗ . This implies that h is a continuous map.
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• The Torus
Let X = [0, 1] × [0, 1] with the topology J on X induced by the standard
topology on R2 (that is J is the topology on X induced by the Euclidean metric).
Partition X into the subsets of the type:
• the set A = {(0, 0), (0, 1), (1, 0), (1, 1)} consisting of the four corner points,
• all the sets of the form Ax = {(x, 0), (x, 1)} for 0 < x < 1,
• all the sets of the form Ay = {(0, y), (1, y)} for 0 < y < 1,
• all singleton sets of the form {(x, y)}, 0 < x < 1, 0 < y < 1. Then the resulting
identification space is the torus.
Exercise 2.3.4. Let (X, J ), (Y, J 0 ) be topological spaces and f : X → Y be an onto
map. If f maps open sets in X to open sets in Y (that is f is an open map) then
prove that J 0 is the quotient topology on Y induced by f.
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