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: e_{S}:S→S,x↦e_{x}

(iii)
The partial inverse map: γ:S→S,x↦x^{−1}

(iv)
The maps r_{a} and l_{a}.
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}(T_{e})=h^{−1}(f^{−1}(T_{e})), for each e∈E(T). Since f is a partial group homomorphism, then f^{−1}(T_{e}) is a maximal subgroup of S. So, (fh)^{−1}(T_{e}) 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 {℘−τ_{S}} 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 (U_{λ})_{λ∈L} be a subfamily of ℘open sets. Then, h^{−1}[U_{λ}] 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 ℘−N_{x}
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 U_{x} such that x∈U_{x}⊆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∈N_{x}

(ii)
If N∈N_{x} and N⊆M, then M∈N_{x}

(iii)
If N,M∈N_{x}, then \(N\bigcap M\in N_{x}\)

(iv)
If N∈N_{x}, then there exists M∈N_{x} such that N∈N_{y}, for each y∈M.
Proof

(i)
If N∈N_{x}, then there exists a ℘open set U in S such that x∈U⊆N. Hence, x∈N.

(ii)
If N∈N_{x}, then there exists a ℘open set U in S such that x∈U⊆N. Since, N⊆M, then x∈U⊆M. Hence, M∈N_{x}.

(iii)
If N,M∈N_{x}, 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∈N_{x}, then there exists a ℘open set M in S such that x∈M⊆N. Since M is a ℘open set, then M∈N_{y}, for all y∈M. Since, N⊆M, then N∈N_{y}, 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 ℘−A^{0}.
Proposition 2
Let S be a topological partial group and A,B⊆S. Then,

(i)
℘−A^{0}⊆A

(ii)
If A⊆B, then ℘−A^{0}⊆℘−B^{0}

(iii)
℘−A^{0} is a ℘open set

(iv)
(℘−A^{0})^{0}=℘−A^{0}.
Proof

(i)
Let x∈℘−A^{0}. Then, A∈N_{x}. So, x∈A.

(ii)
Let x∈℘−A^{0}. Then, A∈N_{x}. Since, A⊆B, then B∈N_{x} and so x∈℘−B^{0}. Hence, ℘−A^{0}⊆℘−B^{0}.

(iii)
Let x∈℘−A^{0}. Then, A∈N_{x}. Thus, there exists N∈N_{x} such that A∈N_{y}, for all y∈N. That is, y∈℘−A^{0}, for all y∈N. Hence, N⊆A. Thus, x∈N⊆℘−A^{0}. So, A∈N_{x}. Therefore, ℘−A^{0} is a ℘open set.

(iv)
Since ℘−A^{0}⊆A, then from (ii) (℘−A^{0})^{0}⊆℘−A^{0}. It remains that ℘−A^{0}⊆(℘−A^{0})^{0}. This is given from x∈℘−A^{0}. That is, ℘−A^{0}∈N_{x}. Hence, x∈(℘−A^{0})^{0}.
□
Corollary 2
A subset A of the topological partial group S is ℘open if and only if ℘−A^{0}=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∈℘−N_{x}.
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∈τ_{A} since \(\phi =\phi \bigcap A\) and \(A=A\bigcap S\). Let M,N∈τ_{A}. 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=(V_{λ})_{λ∈L} be a subfamily of τ_{A}. Then, for each λ, there are ℘open sets U_{λ} 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∈N_{x} in S and N∈N_{y} 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∈℘−N_{x} in S and V∈℘−N_{y} 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 \((S\times T)(C\times D)=(SC)\times T\bigcup S\times (TD)\). 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 P_{1}:S×T→S and P_{2}:S×T→T.

(ii)
The product map μ:S×S→S.

(iii)
The diagonal map Δ_{S}={(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 f_{1}:S_{1}→T_{1} and f_{2}:S_{2}→T_{2} are ℘morphisms, then f_{1}×f_{2}:S_{1}×S_{2}→T_{1}×T_{2} is also a ℘morphism.
Proof
It is clear that f_{1}×f_{2}:S_{1}×S_{2}→T_{1}×T_{2} is a partial group homomorphism. Since f_{1}×f_{2}=(f_{1} P_{1}, f_{2} P_{2}), then from the last theorem, we have that f_{1}×f_{2} is ℘continuous. Hence, f_{1}×f_{2} 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 f_{1}:S_{1}→T_{1} and f_{2}:S_{2}→T_{2} are ℘open maps, then f_{1}×f_{2}:S_{1}×S_{2}→T_{1}×T_{2} is also a ℘open map.
Proof
Let U⊆S_{1}×T_{1} 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, (f_{1}×f_{2})[M×N]⊆(f_{1}×f_{2})[U]. Therefore, f_{1}[M]×f_{2}[N]⊆(f_{1}×f_{2})[U]. Since f_{1} and f_{2} are ℘open maps, then f_{1}[M] and f_{2}[N] are ℘open sets in T_{1} and T_{2}, respectively. Hence, f_{1}×f_{2} is ℘open. □
Theorem 13
The maps r_{a} and l_{a} are ℘open maps.
Proof
We only prove that r_{a} 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, r_{a}h: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 ℓ_{a} 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 r_{b}(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 \(SA=\bigcup _{x\neq A}xA\), then S−A is ℘open. Therefore, A is ℘closed. □
Theorem 16
The projection maps P_{1}:S×T→S and P_{2}:S×T→T are ℘open maps.
Proof
we only prove that P_{1} is ℘open, as follows: let W⊆S×T be ℘open and x∈P_{1}[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, P_{1}[W]∈℘−N_{x}. Therefore, P_{1} is ℘open. Similarly, we can prove that P_{2} is ℘open. □
Let {S_{i}: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=〈x_{i}〉:x_{i}∈S_{i}, ∀ 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 e_{S} are ℘continuous, since μ=〈μ_{i} (P_{i}×P_{i})〉, γ=〈γ_{i}P_{i}〉 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 (U_{λ})_{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. □