In this section, the notion of ℘-continuous map is introduced and some of their basic properties are discussed. Also, the category \(\mathbf {K}\acute {}\) of ℘-continuous maps, as objects and the morphisms of ℘-continuous maps as arrows, is introduced.
Definition 11
Let S and T be topological partial groups. Then, the map f:S→T is called ℘-continuous if fh:C→T is continuous, for each a ℘-test map h:C→S.
We note that every continuous map of topological partial group is ℘-continuous. So, the following maps are ℘-continuous:
-
(i)
The identity map I:S→S
-
(ii)
The partial identity map: eS:S→S,x↦ex
-
(iii)
The partial inverse map: γ:S→S,x↦x−1
-
(iv)
The maps ra and la.
Definition 12
Let f:S→T be a ℘-continuous map. Then, f is called a ℘−morphism if it is a partial group homomorphism.
We note that (i) and (iii) above are ℘-morphisms.
Theorem 4
If f:S→T and g:T→F are ℘-morphisms, then gf:S→F is also a ℘-morphism.
Proof
It is clear that gf is a partial group homomorphism. Let h:C→S be a ℘−test map. Since f is ℘-continuous, then fh:C→T is continuous. Now, (fh)−1(Te)=h−1(f−1(Te)), for each e∈E(T). Since f is a partial group homomorphism, then f−1(Te) is a maximal subgroup of S. So, (fh)−1(Te) is open in C, for each e∈E(T). That means that fh is a ℘-test map. Since g is ℘-continuous, then g(fh)=(gf)h:C→F is continuous. Then, gf is ℘-continuous. Hence, gf is a ℘-morphism. □
Definition 13
A subset V of the topological partial group S is called ℘-open if h−1[V] is open in C for each a ℘-test map h:C→S
From the above definition, we have that \(S_{e_{x}}\) is ℘-open in S.
Theorem 5
The family {℘−τS} of ℘-open sets form a topology on S.
Proof
It is clear that ϕ and S are ℘-open sets, since h−1[S]=C and h−1[ϕ]=ϕ. If U and V are ℘-open sets, then h−1[U] and h−1[V] are open sets in C. But \(h^{-1}[U\bigcap V]=h^{-1}[U]\bigcap h^{-1}[V]\) is open in C. So, \(U\bigcap V\) is a ℘-open set. Similarly, let (Uλ)λ∈L be a subfamily of ℘-open sets. Then, h−1[Uλ] are open in C, for each λ∈L. Since \(h^{-1}[\bigcup _{\lambda }U_{\lambda }]=\bigcup _{\lambda }h^{-1}[U_{\lambda }]\) is open in C. Hence, \(\bigcup U_{\lambda }\) is a ℘-open set. □
Definition 14
A subset A of the topological partial group S is called a ℘-neighbourhood of x∈S if there exists a ℘-open set U in S such that x∈U⊆A.
The family of all ℘-neighbourhoods of x∈S is called a ℘-neighbourhood system and is denoted by ℘−Nx
Proposition 1
A subset A⊆S of the topological partial group S is a ℘-open set if and only if it is a ℘-neighbourhood of each of its points.
Proof
Let A be a ℘-open set. Then, x∈A⊆A, for all x∈A. Hence, A is a ℘-neighbourhood of x. Conversely, for each x∈A, there exists a ℘-open set Ux such that x∈Ux⊆A. So, \(A=\bigcup _{x\in A}U_{x}\). Hence, A is a ℘-open set. □
Theorem 6
Let S be a topological partial group and x∈S. Then,
-
(i)
x∈N, for all N∈Nx
-
(ii)
If N∈Nx and N⊆M, then M∈Nx
-
(iii)
If N,M∈Nx, then \(N\bigcap M\in N_{x}\)
-
(iv)
If N∈Nx, then there exists M∈Nx such that N∈Ny, for each y∈M.
Proof
-
(i)
If N∈Nx, then there exists a ℘-open set U in S such that x∈U⊆N. Hence, x∈N.
-
(ii)
If N∈Nx, then there exists a ℘-open set U in S such that x∈U⊆N. Since, N⊆M, then x∈U⊆M. Hence, M∈Nx.
-
(iii)
If N,M∈Nx, then there exist two ℘-open sets U and V, respectively such that x∈U⊆N and x∈V⊆M. So, we have that \(x\in U\bigcap V\subseteq N\bigcap M\). Since \(N\bigcap M\) is a ℘-open set, then \(N\bigcap M\in N_{x}\).
-
(iv)
If N∈Nx, then there exists a ℘-open set M in S such that x∈M⊆N. Since M is a ℘-open set, then M∈Ny, for all y∈M. Since, N⊆M, then N∈Ny, for each y∈M.
□
Definition 15
Let S be a topological partial group and A⊆S. Then, x∈A is called a ℘-interior point of A if A is a ℘-neighbourhood of x.
The set of all ℘-interior points of A is called ℘-interior set and is denoted by ℘−A0.
Proposition 2
Let S be a topological partial group and A,B⊆S. Then,
-
(i)
℘−A0⊆A
-
(ii)
If A⊆B, then ℘−A0⊆℘−B0
-
(iii)
℘−A0 is a ℘-open set
-
(iv)
(℘−A0)0=℘−A0.
Proof
-
(i)
Let x∈℘−A0. Then, A∈Nx. So, x∈A.
-
(ii)
Let x∈℘−A0. Then, A∈Nx. Since, A⊆B, then B∈Nx and so x∈℘−B0. Hence, ℘−A0⊆℘−B0.
-
(iii)
Let x∈℘−A0. Then, A∈Nx. Thus, there exists N∈Nx such that A∈Ny, for all y∈N. That is, y∈℘−A0, for all y∈N. Hence, N⊆A. Thus, x∈N⊆℘−A0. So, A∈Nx. Therefore, ℘−A0 is a ℘-open set.
-
(iv)
Since ℘−A0⊆A, then from (ii) (℘−A0)0⊆℘−A0. It remains that ℘−A0⊆(℘−A0)0. This is given from x∈℘−A0. That is, ℘−A0∈Nx. Hence, x∈(℘−A0)0.
□
Corollary 2
A subset A of the topological partial group S is ℘-open if and only if ℘−A0=A.
Proof
It is obvious. □
Definition 16
A subset A of the topological partial group S is called ℘-closed if S−A is a ℘-open set.
Definition 17
Let S be a topological partial group and A⊆S. Then, x∈S∈ is called a ℘-closure point of A if \(A\bigcap N\neq \phi \), for each N∈℘−Nx.
The set of all ℘-closure points of A is called the ℘-closure of A and is written by \(\wp -\overline {A}\).
Proposition 3
Let A be a subset of the topological partial group S. Then, the family \(\tau _{A}= \{U\bigcap A: U\ {is} \ \wp -open\ in\ \textit {S}\}\) is a topology on A, which is called ℘-relative topology.
Proof
It is clear that ϕ, A∈τA since \(\phi =\phi \bigcap A\) and \(A=A\bigcap S\). Let M,N∈τA. Then, there exist two ℘-open sets U and V such that \(M=U\bigcap A\) and \(N=V\bigcap A\). So, \(M\bigcap N \in \tau _{A}\). Also, let V=(Vλ)λ∈L be a subfamily of τA. Then, for each λ, there are ℘-open sets Uλ such that \(V=U_{\lambda }\bigcap A\). Then, \(V=\bigcup _{\lambda \in L}V_{\lambda } =\bigcup _{\lambda \in L}(U_{\lambda } \bigcap A)= (\bigcup _{\lambda \in L}U_{\lambda })\bigcap A\). □
Theorem 7
Let f:S→T be ℘-continuous. Then, f∣A:A→T is ℘-continuous.
Proof
Let U⊆T be ℘-open. Now, \((f \mid A)^{-1}(U)=f^{-1}(U)\bigcap A\). Since f−1(U) is a ℘−open set in S, then f−1(U) is a ℘-open in A. □
Definition 18
Let S be a topological partial group and A be a subpartial group of S. Then, A with the ℘-relative topology is a topological partial group, called a topological subpartial group, denoted by A≤S.
Definition 19
Let S and T be topological partial groups and let (x,y)∈S×T. The set ℘−(S×T), where M∈Nx in S and N∈Ny in T is called a ℘-basic neighbourhood of (x,y).
Definition 20
A subset U of M×N is called a ℘-neighbourhood if there exists a ℘-basic neighbourhood M×N of (x,y) such that (x,y)∈M×N⊆U.
We note that if M and N are ℘-open sets in the topological partial groups S and T, respectively, then M×N is a ℘-basic neighbourhood of any (x,y)∈M×N.
Theorem 8
-
(i)
If A and B are ℘-open sets in S and T, respectively, then A×B is also ℘-open in S×T
-
(ii)
If C and D are ℘-closed sets in S and T, respectively, then C×D is also ℘-closed in S×T.
Proof
-
(i)
Let (x,y)∈U×V. Then, x∈U and y∈V. So, U∈℘−Nx in S and V∈℘−Ny in T. This implies U×V is a ℘-basic neighbourhood of (x,y). Since (x,y)∈U×V⊆A×B, then U×V∈N(x,y). Hence, A×B is also ℘-open in S×T.
-
(ii)
We have \((S\times T)-(C\times D)=(S-C)\times T\bigcup S\times (T-D)\). Since S−C and T−D are ℘-open sets in S and T, respectively, then (S−C)×T and S×(T−D) are ℘-open sets in S×T and so (S×T)−(C×D) is ℘-open set in S×T. That is, C×D is ℘-closed in S×T.
□
We note that the following maps are ℘-continuous, for each topological partial group S:
-
(i)
The projection maps P1:S×T→S and P2:S×T→T.
-
(ii)
The product map μ:S×S→S.
-
(iii)
The diagonal map ΔS={(x,x):x∈S}.
Theorem 9
If f:S→T and f:S→F are ℘-morphisms, then (f,g):S→T×F is also a ℘-morphism.
Proof
It is clear that (f,g) is a partial group homomorphism. Let h:C→S be a ℘-test map. Since f is ℘-continuous, then fh:C→T is continuous. Also, since g is ℘-continuous, then gh:C→T is continuous. So, (fh,gh)=(f,g)h:S→T×F is continuous. That is, (f,g) is ℘-continuous. Hence, (f,g) is a ℘-morphism. □
Theorem 10
If f1:S1→T1 and f2:S2→T2 are ℘-morphisms, then f1×f2:S1×S2→T1×T2 is also a ℘-morphism.
Proof
It is clear that f1×f2:S1×S2→T1×T2 is a partial group homomorphism. Since f1×f2=(f1 P1, f2 P2), then from the last theorem, we have that f1×f2 is ℘-continuous. Hence, f1×f2 is a ℘-morphism. □
Theorem 11
Let S and T be topological partial groups. Then, the following conditions are equivalent for any map f:S→T.
-
(i)
f is ℘-continuous
-
(ii)
f−1[U] is a ℘-open set in S for each ℘-open set U in T.
-
(iii)
f−1[U] is a ℘-closed set in S for each ℘-closed set U in T.
Proof
(i) → (ii) Let f be ℘-continuous and let U⊆T be ℘-open. So, h−1[f−1[U]]=(fh)−1[U] is open in C, for each ℘-test map h:C→T. Hence, f−1[U] is a ℘-open set in S.
(ii) → (iii) Let U be ℘-closed in T. So T−U is ℘-open in T. Therefore, f−1[T−U]=S−f−1[U] is ℘-open in S. Hence, f−1[U] is ℘-closed in S.
(iii) → (ii) Let U be ℘-open in T. So, T−U is ℘-closed in T. Therefore, f−1[T−U]=S−f−1[U] is ℘-closed in S. Hence, f−1[U] is ℘-open in S.
(iii) → (i) Let h:C→S be a ℘-test map and U⊆T be open. So, f−1[U] is ℘-open in S. Therefore, h−1[f−1[U]]=(fh)−1[U] is open in C. Hence, f is ℘-continuous. □
Definition 21
Let S and T be topological partial groups. Then, the map f:S→T is called ℘-open if f(U) is ℘-open in T for each ℘-open set U in S. Also, the map f:S→T is called ℘-closed if f(U) is ℘-closed in T for each ℘-closed set U in S.
Theorem 12
If f1:S1→T1 and f2:S2→T2 are ℘-open maps, then f1×f2:S1×S2→T1×T2 is also a ℘-open map.
Proof
Let U⊆S1×T1 be ℘-open and (x,y)∈U. Then, there exists a ℘-basic neighbourhood M×N of (x,y) such that (x,y)∈℘−(M×N)⊆U. So, (f1×f2)[M×N]⊆(f1×f2)[U]. Therefore, f1[M]×f2[N]⊆(f1×f2)[U]. Since f1 and f2 are ℘-open maps, then f1[M] and f2[N] are ℘-open sets in T1 and T2, respectively. Hence, f1×f2 is ℘-open. □
Theorem 13
The maps ra and la are ℘-open maps.
Proof
We only prove that ra is ℘-open as follows: Let U⊆S be ℘-open. Then, \(U\bigcap S_{e_{x}} \) is open in the maximal topological subgroup \(S_{e_{x}} \) and so is open in S. Now, we have two cases:
-
(i)
Let \(\phantom {\dot {i}\!}r_{a} \, |_{_{_{S_{e_{x}} }} } :S_{e_{x}} \to S_{e_{y} }\). So, \(\phantom {\dot {i}\!}r_{a} \, |_{_{_{S_{e_{x}} }} } (U \bigcap S_{e_{x} })=Ua \bigcap S_{e_{y} }\). We show that \(Ua \bigcap S_{e_{y} }\) is open in S as follows: Let h:C→S be a ℘-test map. Then, rah:C→S is a ℘-test map. Now, \((r_{a} h)^{-1}(Ua \bigcap S_{e_{y} })=h^{-1}((r_{a})^{-1}(Ua \bigcap S_{e_{y} }))=h^{-1}((r_{a})^{-1}(Ua) \bigcap (r_{a})^{-1}(S_{e_{y}}))=h^{-1}(U \bigcap S_{e_{x}})\phantom {\dot {i}\!}\). Since \(U \bigcap S_{e_{x}}\) is open in S, then \(\phantom {\dot {i}\!}h^{-1}(U \bigcap S_{e_{x}})\) is open in C. Hence, \(Ua \bigcap S_{e_{y} }\) is ℘-open in S.
-
(ii)
Let \(\phantom {\dot {i}\!}r_{a} \, |_{_{_{S_{e_{x}} }} } :S_{e_{x}} \to S_{e_{x}} \). Since, the right transformation \(\phantom {\dot {i}\!}r_{a} \, |_{_{_{S_{e_{x}} }} }\) is a homeomorphism of the topological maximal subgroups \(S_{e_{x} }\), then \(\phantom {\dot {i}\!}r_{a} \, |_{_{_{S_{e_{x}} }} } (U\bigcap S_{e_{x}})\) is open in \(S_{e_{x}} \). Since \(S_{e_{x}} \) is open in S, then \(\phantom {\dot {i}\!}r_{a} \, |_{_{_{S_{e_{x}} }} } (U\bigcap S_{e_{x}})=Ua\bigcap S_{e_{x}} \) is open in S. That means \(\phantom {\dot {i}\!}r_{a} (U)=\bigcup _{e_{x} \in E(S)}\, r_{a} |_{_{_{S_{e_{x}} }} } (U\bigcap S_{e_{x}})\) is ℘-open in S.
Similarly, we can prove that ℓa is ℘-open. □
Theorem 14
Let S be a topological partial group and A,B⊆S. Then, if A is ℘-open in S, then AB and BA are also ℘-open in S.
Proof
We only prove that AB is ℘-open in S as follows: Since \(AB=\bigcup _{b\in B}r_{b}(A)\), and rb(A) is ℘-open in S, then AB is ℘-open in S. Similarly, we can prove that BA is also ℘-open in S. □
Theorem 15
If S is a topological partial group, then every ℘-open topological subpartial group of S is ℘-closed.
Proof
Let A be a ℘-open topological subpartial group of S. Then, xA is ℘-open in S, for all x∈S. Since \(S-A=\bigcup _{x\neq A}xA\), then S−A is ℘-open. Therefore, A is ℘-closed. □
Theorem 16
The projection maps P1:S×T→S and P2:S×T→T are ℘-open maps.
Proof
we only prove that P1 is ℘-open, as follows: let W⊆S×T be ℘-open and x∈P1[W]. Then, there exists y∈T such that (x,y)∈W. Since W is ℘-open, then there exists a ℘-basic neighbourhood M×N of (x,y) such that (x,y)∈M×N⊆W. So, \(x\in M=P_{1}^{-1} \left [M\times N\right ]\subseteq P_{1} [W]\). Hence, P1[W]∈℘−Nx. Therefore, P1 is ℘-open. Similarly, we can prove that P2 is ℘-open. □
Let {Si:i=1,2,⋯,n} be a family of topological partial groups and \(S=\mathop {\otimes }\limits _{i=1}^{n} \, S_{i} \, \) be the cartesian product of topological partial groups. That is, S={x=〈xi〉:xi∈Si, ∀ i=1,2,…,n}.
Theorem 17
The partial group S with the cartesian product topology \(S=\mathop {\otimes }\limits _{i=1}^{n} \, S_{i} \,\) is a topological partial group.
Proof
The maps μ,γ and eS are ℘-continuous, since μ=〈μi (Pi×Pi)〉, γ=〈γiPi〉 and \(e_{S}=\left \langle e_{S_{i}} P_{i} \right \rangle \), respectively, where \(P_{i} :\mathop {\otimes }\limits _{i=1}^{n} \, (S_{i})\to S_{i} \), are the projection maps. □
Definition 22
Let S and T be topological partial groups. A topology ℘−τ∗ on T is called ℘−final with respect to the map f:S→T if, for any topological partial group F and all maps g:T→F, we have that g is ℘-continuous if gf:S→F is ℘-continuous.
Theorem 18
The ℘−τ∗ final topology on T with respect to the function f:S→T exists and is characterized by the following condition: If U⊆T, then U is ℘-open (℘-closed) in T if and only if f−1[U] is ℘-open (℘-closed) in S.
Proof
It is clear that ϕ and T are ℘-open sets in S. If U and V are ℘-open sets in T, then \(f^{-1}[U\bigcap V]=f^{-1}[U]\bigcap f^{-1}[V]\) is ℘-open in S. So, \(U\bigcap V\) is ℘-open in T. Similarly, let (Uλ)λ∈L be a subfamily of ℘-open sets in T. Then, \(f^{-1}[\bigcup (U_{\lambda })]\) are ℘-open sets in S. So, \(\bigcup U_{\lambda }\) is a ℘-open set in S. A similar proof applies with ℘-open replaced by ℘-closed. □
Definition 23
Let S and T be topological partial groups. Then, the map f:S→T is called ℘-identification if f is surjective and T has the ℘-final topology with respect to f.
Theorem 19
Let f:S→T be a ℘-continuous surjection. If f is a ℘-open (closed) map. Then, f is a ℘-identification map.
Proof
Let U⊆T be a ℘-open set. Then, f−1[U] is ℘-open in S. Since f is surjective, then f[f−1[U]]=U. Hence, f−1[U] is ℘-open in S if and only if U is ℘-open. A similar proof applies with open replaced by ℘-closed. □