@ Appl. Gen. Topol. 15, no. 1 (2014), 43-54doi:10.4995/agt.2014.2144
© AGT, UPV, 2014

On some properties of T0−ordered reflection

Sami Lazaar
∗
and Abdelwaheb Mhemdi

Department of Mathematics, Faculty of Sciences of Tunis. University Tunis-El Manar. “Campus

Universitaire” 2092 Tunis, TUNISIA. (salazaar72@yahoo.fr, mhemdiabd@gmail.com)

Abstract

In [12], the authors give an explicit construction of the T0−ordered
reflection of an ordered topological space (X, τ, ≤) . All ordered topo-
logical spaces such that whose T0−ordered reflections are T1−ordered

spaces are characterized. In this paper, some properties of the

T0−ordered reflection of a given ordered topological space (X, τ, ≤)
are studies. The class of morphisms in ORDTOP orthogonal to all

T0−ordered topological space is characterized.

2010 MSC: 54F05; 18B30; 54G20; 54C99; 06F30; 18A05.

Keywords: ordered topological space; T2−ordered; T1−ordered;

T0−ordered; ordered reflection; ordered quotient; category

and functor.

1. Introduction

Among the oldest separation axioms in topology, there are three famous
ones T0, T1 and T2.

The T0−, T1− and T2−reflections of a topological space have long been of
interest to categorical topologist. The construction of these reflections in the
category TOP of all topological spaces are given in [10].

In [2], the authors introduced some new separation axioms using the Ti−reflections
Ti(X) i ∈ {0, 1, 2} as follow:
Definition 1.1. Let i, j be two integers such that 0 ≤ i < j ≤ 2. A topological
space X is said to be a T(i,j) − space if Ti(X) is a Tj−space.

∗Corresponding author.

Received May 2013 – Accepted February 2014

http://dx.doi.org/10.4995/agt.2014.2144


S. Lazaar and A. Mhemdi

Precisely, there are two new types of separation axioms namely T(0,1)− and
T(0,2)−spaces.

The characterization of these spaces are completely stated in [2, Theorem
3.5.] and [2, Theorem 3.12.].

After this, in [12], H. Künzi and T. A. Richmond are interested in the
corresponding concepts of Ti−ordered reflections in the category ORDTOP
with ordered topological spaces (X, τ, ≤) as objects and continuous increasing
maps as morphisms (or arrows).

The authors show that, given an ordered topological space (X, τ, ≤), the
Ti−ordered reflection, for i ∈ {0, 1, 2} of (X, τ, ≤) is obtained as a quotient of
X ( for more information see [12, Theorem 2.5.]).

Let (X, τ, ≤) be an ordered topological space. For A ⊆ X, the increasing
hull of A is i (A) = {y ∈ X : ∃ x ∈ A x ≤ y}. A subset A of X is an increasing
set if i (A) = A and we denote by I (A) the closed increasing hull of A, that is,
the smallest closed increasing set containing A. Decreasing set, decreasing hull
d(A) and the closed decreasing hull D(A) are defined dually. A subset A which
satisfy A = I(A) ∩ D(A) is called a c − set. We denote by C(A) the smallest
c − set containing A.

An ordered topological space (X, τ, ≤) is said to be T0−ordered if the fol-
lowing equivalent conditions hold.

(1) [I(x) = I(y) and D(x) = D(y)] =⇒ x = y.
(2) C(x) = C(y) =⇒ x = y.
(3) If x 6= y, there exist a monotone open neighborhood of one of the points

which does not contain the other point.

Let (X, τ, ≤) be an ordered topological space. For x, y ∈ X, define an
equivalence relation on X by x ≈ y if and only if [I (x) = I (y) and D (x) =
D (y)], which is equivalent to C (x) = C (y) .

Order the set X/≈ by the finite step order defined by :

z0 ≤0 zn ⇐⇒ ∃ z1, ..., zn−1 and ∃ z
′

i, z
∗
i ∈ zi (i = 0, 1, ..., n)

with z
′

i ≤ z∗i+1 ∀ i = 0, 1, ..., n − 1
.

T. A. Richmond and H-P. A. Künzi show that
(

X/≈, τ/≈, ≤0
)

is the T0−ordered
reflection of X.

This paper consists of some investigations into the T0−ordered reflection of
an ordered topological space (X, τ, ≤) .

In the first section we give the characterization of an ordered topologi-
cal space (X, τ, ≤) such that its T0−ordered reflection is T K1 −ordered and
we characterize ordered topological spaces whose T0−ordered reflections are
T2−ordered. [2, Theorem 3.5] and [2, Theorem 3.12] are recovered.

The second investigation deals with some categorical properties of the cat-
egory ORDTOP0, of T0−ordered topological spaces. More precisely, a char-
acterization of the class of morphisms in ORDTOP rendered invertible, by
the T0−ordered reflection functor, is given. [2, Theorem 2.4] is seen to be a
particular case of our result.

© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 44



On some properties of T0−ordered reflection

2. separation axioms

Given an ordered topological space (X, τ, ≤), the construction of its T0−ordered
reflection denoted by

(

X/≈, τ/≈, ≤0
)

satisfies some categorical properties:
For each ordered topological space (Y, γ, ⊑) and each continuous increasing

map f from (X, τ, ≤) to (Y, γ, ⊑) , there exists a unique continuous increasing
map

≈

f :
(

X/≈, τ/≈, ≤0
)

−→
(

Y/≈, γ/≈, ⊑0
)

such that the following diagram
commutes:

(X, τ, �)
f

//

qX

��

	

(Y, γ, ⊑)

qY

��
(

X/≈, τ/≈, ≤0
)

≈

f

//
(

Y/≈, γ/≈, ⊑0
)

where qX is the canonical surjection map.
From the above properties, it is clear that we have a covariant functor from

the category of ordered topological spaces ORDTOP into the full subcategory
ORDTOP0 of ORDTOP whose objects are T0−ordered topological spaces.

In [12], the authors characterize those ordered topological spaces whose
T0−ordered reflections are T1−ordered as follows:

Theorem 2.1 ([12, Theorem 3.2]). The following statements are equivalent:

(1) The T0 − ordered reflection X/≈ of X is T1 − ordered.
(2) x �0 y in X/≈ implies there exists an open increasing neighborhood of

x not containing y and there exists an open decreasing neighborhood of
y not containing x.

(3)

i (x) =
⋂

{N : N is an open increasing neigborhood of x}

and
d (x) =

⋂

{N : N is an open decreasing neigborhood of x}
∀x ∈ X.

On the other hand, recall that an ordered topological space (X, τ, ≤) is said
to be a T K1 −ordered space if, for any point x in X, we have C(x) = {x}
(for more information see [13]). The following theorem characterizes ordered
topological spaces whose T0−ordered reflections are T K1 −ordered.

Theorem 2.2. Let (X, τ, ≤) be an ordered topological space. The following
statements are equivalent:

(i) The T0 − ordered reflection X/≈ of X is T K1 − ordered;
(ii) For each x ∈ X and each monotone closed subset F of X such that

F ∩ C(x) 6= ∅, we have x ∈ F ;
(iii) For each monotone open subset O of X containing x, we have C(x) ⊆

O.

© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 45



S. Lazaar and A. Mhemdi

Proof.

• (i) ⇒ (ii)
Suppose that X/≈ is T K1 − ordered.
Let F be a closed monotone subset of X such that F ∩ C (x) 6= ∅

and y ∈ F ∩ C (x) .
Clearly, qX(y) ∈ qX (F)∩C (qX(x)). Thus, we can see that qX(x) =

qX(y). Now, since F is monotone and consequently a saturated subset
of X, qX (x) = qX (y) ⊆ F. Therefore, x ∈ F.

• (ii) ⇒ (i)
Let y ∈ X be such that qX(y) ∈ C (qX(x)) . Clearly C (y) ⊆ C (x) .

Conversely, since I (y) ∩ C (x) is non empty then by (ii) x ∈ I (y) and
by the same way we say that x ∈ D (y) . Therefore x ∈ C (y) and
C (x) ⊆ C (y) .

Finally we can see that C (x) = C (y), so we have qX(x) = qX(y).
• (ii) ⇒ (iii)

Let x ∈ X and O be a monotone open subset of X such that x ∈ O.
If C (x) * O then by (ii) x /∈ O which is false.

• (iii) ⇒ (ii)
Let F be a closed monotone subset of X such that F ∩ C (x) 6= ∅.

If x /∈ F then x ∈ F C, since F C is a monotone open subset of X then
by (iii) we obtain C (x) ⊆ F C which is false.

�

As an immediate consequence of Theorem 2.2, we have the following corollary.

Corollary 2.3 ([2, Theorem 3.5]). Let (X, τ) be a topological space. Then the
following statements are equivalent:

(i) X is a T(0,1)-space;

(ii) For each open subset U of X and each x ∈ U, we have {x} ⊆ U;
(iii) For each x ∈ X and each closed subset C of X such that {x} ∩ C 6= ∅,

we have x ∈ C.
Proof. Let (X, τ) be a topological space. It is enough to consider the ordered
topological space (X, τ, =) in Theorem 2.2.

�

Now, let us introduce the following notation and definition:

Notation 2.4. Let (X, τ, ≤) be an ordered topological space and z in X. We
denote by:

T (z) := {x ∈ X : C (x) = C (z)} .
Definition 2.5. Let (X, τ, ≤) be an ordered topological space. Defines on
X the finite step preorder �(X,≤) related to ≤, by x �(X,≤) y if there exists
z0, ..., zn and ∃ z

′

i, z
∗
i ∈ T (zi) (i = 0, 1, ..., n) such that z0 = x, zn = y and

z
′

i ≤ z∗i+1 ∀ i = 0, 1, ..., n − 1.
for short, we denote �(X,≤) also by �≤ .

© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 46



On some properties of T0−ordered reflection

Remarks 2.6.

• It is clear that G (≤) ⊆ G (�≤) .
• If X is a T0−ordered space, we have ≤=�≤.
• For each x, y ∈ X, we have equivalence between x �≤ y and qX (x) ≤0
qX (y) .

Recall that an ordered topological space (X, τ, ≤) is said to be T2−ordered
if there is an increasing neighborhood of x disjoint form some decreasing neigh-
borhood of y whenever x � y, which is equivalent to the order ≤ being closed
in (X, τ) × (X, τ) .

Now, we are in position to give the characterization of ordered topological
spaces whose T0−ordered reflections are T2−ordered.
Theorem 2.7. Let (X, τ, ≤) be an ordered topological space. Then the following
statements are equivalent:

(i) The T0 − ordered reflection X/≈ of X is T2 − ordered;
(ii) If x �(X,≤) y there exists an increasing neighborhood of x disjoint from

some decreasing neighborhood of y;
(iii) The graph G (�≤) of �≤ is closed in X × X.

Proof.

• (i) =⇒ (ii)
Let x, y be two points in X such that x �≤ y. Then qX (x) �

0

qX (y) . Since X/≈ is T2−ordered, there exists an increasing neighbor-
hood U of qX (x) disjoint from some decreasing neighborhood V of
qX (y). Now, we can see that q

−1
X (U) is an increasing neighborhood of

x disjoint from q−1X (V ), which is a decreasing neighborhood of y.
• (ii) =⇒ (iii)

Let x, y ∈ X such that (x, y) /∈ G (�≤) which means that x �≤ y.
Then, there exists an increasing neighborhood U of x disjoint from some
decreasing neighborhood V of y. Clearly, we can see that U × V is a
neighborhood of (x, y) and we have (U × V ) ∩ G (�≤) = ∅. Therefore,
G (�≤) is closed in X × X.

• (iii) =⇒ (i)
For this implication we can see that G (�≤) = q−1X × q

−1
X

(

G
(

≤0
))

.

Then , G
(

≤0
)

is closed and thus X/≈ is T2−ordered.
�

By the same way as in Corollary 2.3, the following result holds immediately.

Corollary 2.8 ([2, Theorem 3.12]). Let (X, τ) be a topological space. Then
the following statements are equivalent:

(i) X is a T(0,2)-space;

(ii) For each x, y ∈ X such that {x} 6= {y}, there are two disjoints open
sets U and V in X with x ∈ U and y ∈ V .

© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 47



S. Lazaar and A. Mhemdi

3. The class of morphisms in ORDTOP orthogonal to all
T0−ordered spaces

It is worth noting that reflective subcategories arise throughout mathemat-
ics, via several examples such as the free group and free ring functors in algebra,
various compactification functors in topology, and completion functors in anal-
ysis: cf. [14, p. 90]. Recall from [14, p. 89] that a subcategory D of a category
C is called reflective (in C) if the inclusion functor I : D −→ C has a left
adjoint functor F : C −→ D; i.e., if, for each object A of C, there exist an
object F(A) of D and a morphism µA : A −→ F(A) in C such that, for each
object X in D and each morphism f : A −→ X in C, there exists a unique
morphism f̃ : F(A) −→ X in D such that f̃ ◦ µA = f.

The concept of reflections in categories has been investigated by several
authors (see for example [3], [4], [5], [6],[9], [11], [15]). This concept serves the
purpose of unifying various constructions in mathematics.

Historically, the concept of reflections in categories seems to have its origin
in the universal extension property of the Stone-Čech compactification of a
Tychonoff space.

A morphism f : A −→ B and an object X in a category C are called
orthogonal [7], if the mapping homC(f; X) : homC(B; X) −→ homC(A; X)
that takes g to gf is bijective. For a class of morphisms Σ (resp., a class of
objects D), we denote by Σ⊥ the class of objects orthogonal to every f in Σ
(resp., by D⊥ the class of morphisms orthogonal to all X in D) [7].

The orthogonality class of morphisms D⊥ associated with a reflective subcat-
egory D of a category C satisfies the following identity D⊥⊥ = D [1, Proposition
2.6]. Thus, it is of interest to give explicitly the class D⊥. Note also that, if
I : D −→ C is the inclusion functor and F : C −→ D is a left adjoint functor of
I, then the class D⊥ is the collection of all morphisms of C rendered invertible
by the functor F ( i.e. D⊥ = {f ∈ homC : F (f) is an isomorphism of D})[1,
Proposition 2.3].

This section is devoted to the study of the orthogonal class ORDTOP⊥
0
;

hence we will give a characterization of morphisms rendered invertible by the
functor of the T0−ordered reflection.

Recall that a continuous map q : Y → Z is said to be a quasihomeomorphism
if U → q−1(U) defines a bijection O(Z) → O(Y ) [8], where O(Y ) is the set of
all open subsets of the space Y.

Then the following definition is more natural.

Definition 3.1. Let f : (X, τ, ≤) −→ (Y, γ, ⊑) be an increasing continuous
map between two ordered topological spaces. f is said to be an ordered −
quasihomeomorphism if U 7−→ f−1 (U) defines a bijection between the set of
saturated open (resp. closed) sets of Y and the set of saturated open (resp.
closed) sets of X.

© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 48



On some properties of T0−ordered reflection

Examples 3.2.

(1) qX : X −→ X/≈ is an ordered-quasihomeomorphism.
(2) Let q : (X, τ, ≤) −→ (Y, γ, ⊑) be an increasing continuous map between

two ordered topological spaces.

If
q̃ : (X, τ) −→ (Y, γ)

x 7−→ q (x) is a quasihomeomorphism then q is an
ordered-quasihomeomorphism.

The converse does not hold as shown in the following example:
(3) Let X = [0, 3] with the topology induced by the usual topology of R.

Define on X the order � by G (�) = {(a, b) : a, b ∈ Q ∩ X and a ≤ b}∪
{(√

5, x
)

: x ∈ (Q ∩ X) ∪
{√

2,
√
5
}}

∪
{(

x,
√
5
)

: x ∈ (Q ∩ X) ∪
{√

2,
√
5
}}

∪
△X.

qX is an ordered-quasihomeomorphism which is not a quasihomeo-
morphism: ]0, 2[ is an open set, since it is not saturate then there is no
an open subset V of X/≈ such that ]0, 2[ = q−1

X
(V ).

Let us give an important property of ordered-quasihomeomorphisms.

Proposition 3.3. If f : X −→ Y , g : Y −→ Z are continuous increasing
maps between ordered topological spaces and two of the three maps f, g , g ◦ f
are ordered − quasihomeomorphisms, then so is the third one.

Proof.

• Suppose that f and g are two ordered-quasihomeomorphisms. For any
saturated closed subset U of X, let V be the unique saturated closed
subset of Y such that U = f−1 (V ) and let W the unique saturated
closed subset in Z such that V = g−1 (W) . It is clear that W is the

unique saturated closed subset of Z such that U = (g ◦ f)−1 (W) . We
conclude that g ◦ f is an ordered-quasihomeomorphism.

• Suppose that g and g◦f are ordered-quasihomeomorphisms. Let U be a
saturated closed subset in X. Since g◦f is an ordered-quasihomeomorphism,
there exists a unique saturated closed subset W in Z such that U =
(g ◦ f)−1 (W) = f−1

(

g−1 (W)
)

. Now, V = g−1 (W) is a saturated

closed subset of Y satisfying U = f−1 (V ) . Let us show that V is
the unique saturated closed subset of Y such that U = f−1 (V ) . For
this, let V ′ be a saturated closed subset in Y such that U = f−1 (V ′) .
There exists a unique saturated closed subset W ′ in Z such that V ′ =
g−1 (W ′). So

(g ◦ f)−1 (W) = U = f−1 (V ′) = f−1
(

g−1 (W ′)
)

= (g ◦ f)−1 (W ′) .

Finally W = W ′ and consequently V = g−1 (W) = g−1 (W ′) = V ′.
• Suppose that f and g ◦ f are ordered-quasihomeomorphisms. If V is
a saturated closed set in Y,f−1 (V ) is a saturated closed set in X.
Then there exits a unique saturated closed set W in Z such that
(g ◦ f)−1 (W) = f−1 (V ). It is easy to show that W is the unique

© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 49



S. Lazaar and A. Mhemdi

saturated closed set in Z such that V = g−1 (W) . We conclude that f
is an ordered-quasihomeomorphism.

�

Now, let’us introduce the following definition:

Definition 3.4. Let f : (X, τ, ≤) −→ (Y, γ, ⊑) be an increasing continuous
map between two ordered topological spaces. We say that f is strongly −
increasing (for short s − increasing) if it satisfies : x �≤ y if and only if
f (x) �(Y,⊑) f (y) for all x, y ∈ X.
Examples 3.5.

(1) Let (X, τ, ≤) be an ordered topological space. Then qX is a s-increasing
map.

(2) An increasing map need not to be s-increasing map. Indeed, take
(X, τ, ≤) of the example in 3.2 (3) and f the following map.

f : [0, 3] −→ [0, 3]
x 7−→ 0

Clearly for any α ∈ [0, 3] \
(

Q ∪
{√

2,
√
5
})

we have f (α) �≤ f (0) but
α �≤ 0.

In order to give the main result of this section, we introduce the following
definitions.

Definitions 3.6. Let f : (X, τ, ≤) −→ (Y, γ, ⊑) be an increasing continuous
map.

(1) f is said to be T − injective (or T − one − to − one) if, for each x, y in
X : if there exists a monotone open subset of X which contains one of
this points but not the other, then, the points f (x) , f(y) of Y , can be
separated by a monotone open subset of Y.

(2) f is said to be T − surjective (or T − onto) if, for each point y ∈ Y,
there exists x ∈ X such that we can not separate y and f (x) by a
monotone open subset of Y.

(3) f is said to be T −bijective if it is both T −injective and T −surjective.
Examples 3.7.

(1) Every onto continuous increasing map is T-onto.
(2) A T-onto map need not be onto as shown the following example : Let

X = {0, 1, 2} with the topology τX = {∅, X, {0, 2} , {1}} and the order
≤X defined by his graph G (≤X) = {(0, 0) , (0, 1) , (0, 2) , (1, 1) , (1, 2) , (2, 2)} .
The map f : (X, τX, ≤X) −→ (X, τX, ≤X) such that f (X) = {0} is
T-onto but not onto.

(3) A T-one-to-one map need not be one-to-one : qX : (X, τX, ≤X) −→
(

X/≈, τX/≈, ≤0X
)

is T-one-to-one but not one-to-one.
(4) A one-to-one map need not be T-one-to-one : Let τd the discrete topol-

ogy on X. Then the map f : (X, τd, ≤X) −→ (X, τX, ≤X) defined by
f (x) = x for all x ∈ X is a one-to-one map but not T-one-to-one.

© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 50



On some properties of T0−ordered reflection

Before giving the main result of this section we need a lemma:

Lemma 3.8. Let f : (X, τ, ≤) −→ (Y, γ, ⊑) be an increasing continuous map.
Then the following properties hold:

(1) f is T-injective if and only if
≈

f is injective.

(2) f is T-surjective if and only if
≈

f is surjective.

(3) f is T-bijective if and only if
≈

f is bijective.

Proof.

(1) • Suppose that
≈

f is injective : Let x, y ∈ X. If we can separate x
and y by a monotone open subset of X then qX (x) 6= qX (y) .
Since

≈

f is injective then
≈

f (qX (x)) 6=
≈

f (qX (y)) which means
qY (f (x)) 6= qY (f (y)) . Therefore, we can separate f (x) and f (y)
by a monotone open subset of Y .

• Conversely, suppose that f is T-injective : Let x, y ∈ X be such
that qX (x) 6= qX (y) which means that we can separate x and
y by one monotone open subset of X. Since f is T-injective we
can separate f (x) and f (y) by one monotone open subset of Y

which means that qY (f (x)) 6= qY (f (y)) and then
≈

f (qX (x)) 6=
≈

f (qX (y)) .

(2) • Suppose that
≈

f is surjective : If y ∈ Y , since
≈

f is a surjective

map there exists x ∈ X such that
≈

f (qX (x)) = qY (y). Thus, we
have qY (f (x)) = qY (y) and we can not separate f (x) and y by
a monotone open subset of Y.

• Conversely, suppose that f is T-onto. If we can’t separate f (x)
and y (x ∈ X, y ∈ Y ) then we have

≈

f (qX (x)) = qY (y) , and we

conclude that
≈

f is an onto map.
(3) An immediate consequence of (1) and (2).

�

Now, we are in a position to give the main result of this section.

Theorem 3.9. Let f : (X, τ, ≤) −→ (Y, γ, ⊑) be an increasing continuous
map between two ordered topological spaces. Then the following statements are

equivalent:

(1)
≈

f is an isomorphism;
(2) f satisfies the following properties.

(i) f is s-increasing.
(ii) f is T-onto.
(iii) f is ordered-quasihomeomorphism.

© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 51



S. Lazaar and A. Mhemdi

(X, τ, �)
f

//

qX

��

	

(Y, γ, ⊑)

qY

��
(

X/≈, τ/≈, ≤0
)

≈

f

//
(

Y/≈, γ/≈, ⊑0
)

Proof.

(1) ⇒ (2)

• By Lemma 3.8, f is T-onto .
• f is s-increasing.

Since
≈

f is an isomorphism, then qX (x) ≤0 qX (y) if and only if
≈

f (qX (x)) ⊑0
≈

f (qX (y)) which means that f (qY (x)) ⊑0 f (qY (y)) .
Now, by Remarks 2.6, we can see that x �≤ y if and only if f (x) �⊑
f (y) .

• By Proposition 3.3 and Example 3.2 it’s clear that f is an ordered-
quasihomeomorphism.

(2) ⇒ (1)

• According to Lemma 3.8,
≈

f is a surjective map.

•
≈

f is injective.
By Lemma 3.8, it is sufficient to show that f is T-one-to-one. To

do this result, let x, y ∈ X and U an open monotone neighborhood of
x such that y /∈ U. Since f is an ordered-quasihomemorphism, there
exists a saturated open subset V of Y such that f−1 (V ) = U. Let us
show that V is monotone.

Without loss of generality we can suppose U increasing.
Let a, b ∈ Y such that a ∈ V and a ⊑ b.
Since f is T-onto, there exists α ∈ U and β ∈ X such that T (f (α)) =

T (a) and T (f (β)) = T (b) .
Now, we can see that f (α) �⊑ f (β) and thus α �≤ β. As U is

increasing we have β ∈ U. Therefore f (β) ∈ V and b ∈ V.

•
≈

f
−1

is increasing.
Let y, y′ ∈ Y such that qY (y) ⊑0 qY (y′) . Since f is T-onto there

exist x, x′ ∈ X such that T (f (x)) = T (y) and T (f (x′)) = T (y′) .
Then, we have qY (f (x)) = qY (y) , we have also qY (y) = qY (f (x)) =
≈

f (qX (x)) so that
≈

f
−1

(qY (y)) = qX (x) . By the same way, we have
≈

f
−1

(qY (y
′)) = qX (x

′). Since qY (y) ⊑0 qY (y′) we have qY (f (x)) ⊑0
qY (f (x

′)) which means that f (x) �⊑ f (x′) . Now, according the fact

© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 52



On some properties of T0−ordered reflection

that f is s-increasing, we have x �≤ x′ which is equivalent to qX (x) ≤0

qX (x
′) and finally

≈

f
−1

(qY (y)) ≤0
≈

f
−1

(qY (y
′)) .

• Now, let we show that
≈

f is an open map.
Let U be an open set of X/≈. Since q−1

X
(U) is an open saturated

subset of X and f is an ordered-quasihomeomorphism then there ex-
ist a saturated open subset V of Y such that f−1 (V ) = q−1X (U) =

f−1
(

q−1Y

(

≈

f (U)

))

. Let us show that q−1Y

(

≈

f (U)

)

= V.

Let y ∈ V, since f is a T-onto map there exists x ∈ X such that
T (f (x)) = T (y) . By saturation of V, f (x) ∈ V and consequently x ∈

f−1 (V ) = q−1X (U) = q
−1
X

(

≈

f
−1 (≈

f (U)

))

= f−1
(

q−1Y

(

≈

f (U)

))

,

and thus, f (x) ∈ q−1
Y

(

≈

f (U)

)

. Now the saturation of q−1
Y

(

≈

f (U)

)

shows that y ∈ q−1Y
(

≈

f (U)

)

. We conclude that V ⊆ q−1Y
(

≈

f (U)

)

.

The second inclusion is proved similarly.

Therefore
≈

f (U) is an open subset of Y/≈.

Finally,
≈

f is a bijective open morphism with its inverse
≈

f
−1

is increasing; so

that
≈

f is an isomorphism. �

We close this paper by giving an immediate consequence of this theorem.

Corollary 3.10 ([2, Theorem 2.4]). Let q : (X, τ) −→ (Y, γ) be a continuous
map. Then the following statements are equivalent:

(i) q is a topologically onto quasihomeomorphism;
(ii) T0(q) is a homeomorphism.

Acknowledgements. The authors gratefully acknowledge helpful corrections,

comments, and suggestions of the anonymous referee which reinforce and clarify

the presentation of our paper.

© AGT, UPV, 2014 Appl. Gen. Topol. 15, no. 1 53



S. Lazaar and A. Mhemdi

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