CONTENTS

J. Duhok Univ., Vol.14, No.1 (Pure and Eng. Sciences), Pp 9-16, 2011
is a preopen subset of A. Hence f \A is cpcontinuous.
(2). Let V be a cp-open set of Y. Then
� 1
�1
f (V)=� { f � (V) : � in A} and since
each f � is cp-continuous, it follows that
�1
each f (V) is a preopen set in U � and by
�
Lemma 1.4,
�1
�
� 1
(V) is a preopen set on X.
f (V) is a preopen set in X. So
f
Theorem 3.12. If f : X �Y is preirresolute
function and g :Y � Z is cp-continuous, then
g � f : X � Z is cp-continuous.
Proof. Let V be any cp-open set in Z. Since g is
cp-continuous, so g 1 � (V) is preopen set in Y.
Since f is preirresolute, then we obtain
f
� 1 � 1
� 1
( g (V))= ( g � f ) (V) is a preopen set
in X. Hence g � f is cp-continuous.
Theorem 3.13. Let f : X � Y be M-preopen
and g : Y � Z be a cp-open function, then
g � f is cp-open function.
Proof. Let U be any preopen set in X. Since f
is M-preopen, then f (U) is preopen set in Y.
since g is cp-open function, so g ( f (U))=
( g � f )(U) is cp-open set in Z. By Theorem
3.7(ii), g � f is a cp-open function.
Theorem 3.14. Let f : X � Y and g : Y � Z
be any two function, then.
1. If g � f is cp-open function and f is
preirresolute, then g is a cp-open function,
2. If g � f is preirresolute and f is Mpreopen
surjective, then g is cp-continuous
function,
3. If g � f is cp-open function and g is
cp-continuous injective, then f is a M-preopen
function,
4. If g � f is cp-continuous function
and g is cp-open injective, then f is
preirresolute function.
Proof. (1). Let V be any preopen set in Y. Since
� 1
(V) is preopen set
f is preirresolute, then f
in X. Since g � f is cp-open function, so
( g � f )( f
� 1
(V))= g ( f ( f
� 1
(V)))= g (V)
is cp-open set in Z. By Theorem 3.7(ii), g is a
cp-open function.
(2). Let V be any cp-open set in Z, thus it is also
preopen set. Since g � f is preirresolute, then
( g � f ) 1 � (V) is preopen set in X. Since f is
M-preopen surjective, implies that
f (( g � f ) 1 � (V))= f ( f
� 1 � 1
( g (V)))
= g 1 � (V) is preopen set in Y. By Theorem
3.5(ii), g is cp-continuous function,
(3). Let U be any preopen set in X, g � f is cpopen
function, then by Theorem 3.7(ii),
( g � f )(U) is a cp-open set in Z. Since g is
cp-continuous injective, so g 1 � (( g � f )(U)))=
g 1 � ( g ( f (U)))= f (U) is a preopen set in Y.
Hence f is M-preopen function.
(4). Let V be any preopen set in Y. Since g is
cp-open function, so g (V) is a cp-open set in Z.
Since g � f is cp-continuous and g is
injective function, then ( g � f ) 1 � ( g (V))=
f
� 1 � 1
( g ( g (V)))= f
� 1
(V) is a preopen set in
X. Hence f is a preirresolute function.
Theorem 3.15. Let f : X �Y be a cpcontinuous,
M-preclosed function from a
prenormal space X on to a space Y. If either X
or Y is pre-T 1 , then Y is pre-T 2 .
Proof (i). Y is pre-T 1 . Let y 1 , y 2 �Y and
y 1 � y 2 . So {y 1 }, {y 2 } are preclosed strongly
compact subset of Y, by Theorem 3.5(iii), we
� 1
have f ( y 1 ) and f
� 1
( y 2 ) are preclosed
subset of a prenormal space X, then there exist
two disjoint preopen sets U 1 and U 2 in X
containing them. Since a function f is M-
preclosed, the set V 1 =Y\ f (X\U 1 ) and
V 2 =Y\ f (X\U 2 ) are preopen set in Y . Also are
disjoint and containing y 1 and y 2 respectively,
so Y is pre-T 2 .
(ii). X is pre-T 1 , Let f (x) be a point of Y. {x}
is preclosed in X. Since f is M-preclosed, then
{ f (x)} is a preclosed set of Y. Hence Y is pre-
T 1 and the proof is complete by part(i).
Theorem 3.16. Let f : X � Y be any function
with P1 -closed graph, X has the property P and
Y is strongly compact, then f is cpcontinuous.
Proof. For each x�X and each cp-open set V of
Y containing f (x), then V is preopen set
containing f (x). Hence Y\V is preclosed and
Y\V is a subset of strongly compact Y, so Y\V is
strongly compact. Since f has a P1 -closed
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