*A mapping f* : *X →* *Y from a space* *X* *onto a space* *Y* *is said to be* semi-quotient *provided a subset V of Y is open in Y if and only if f *^{←} (V) is semi-open in X.

Evidently, every semi-quotient mapping is semi-continuous and every quotient mapping is semi-quotient. The following simple examples show that semi-quotient mappings are different from semi-continuous mappings and quotient mappings.

*Let X* = *Y* = {1, 2, 3} *and let τ*_{X} = {*∅*, *X*, {1}, {2} {1, 2}, {1, 3}} *and τ*_{Y} = {*∅,* *X*, {1}, {2}, {1 2}} *be* *topologies on X and Y. Let f : X → Y be defined by f(x) = x, x ∈* *X. Since τ*_{Y} ⊂ *τχ, the mapping* *ƒ is* *continuous, hence semi-continuous. On the other hand, this mapping is not semi-quotient because f* ^{←} ({3}) *is semi-open in X although* {1, 3} *is not open in Y*.

*Let X =* {1, 2, 3, 4}, *Y =* {*a*, *b*}, *τχ* = {*∅, X*, {1}, {3}, {1, 3}}, *τγ* = {*∅, **Y*, {a}}. Define f : *X → Y by*; *f* (1) = *f*(3) = *f* (4) = a; *f* (2) = b. The *mapping f* *is not a* *quotient mapping because it is not continuous. On the other hand*, *f is semi-quotient: the only proper subset of Y whose preimage is semi-open in X is the set fag which is open in Y*.

The following proposition is obvious.

(a) *Every surjective semi-continuous mapping f : X → Y which is either s-open or s-closed is a semi-quotient mapping*.

(b) *If f : X → Y is a semi-quotient mapping and g : Y → Z* *a quotient mapping, then g ∘ f : X → Z* *is semi-quotient*.

We prove only (b). A subset *V ⊂* *Z* is open in *Z* if and only if *g*^{←}(*V*) is open in *Y* (because *g* is a quotient mapping), while the latter set is open in *Y* if and only if *f*(^{←}g^{←}(*V*)) is semi-open in *X* (because *f* is semi-quotient). So, *V* is open in *Z* if and only (*g ∘ f*)^{←}(*V*) is semi-open in *X*, i.e. *g ∘ f* is a semi-quotient mapping.

The restriction of a semi-quotient mapping to a subspace is not necessarily semi-quotient. Let *X* and *Y* be the spaces from Example 3.3, and *A =* {2, 4}. Then τ_{A} = {∅, *A*}. The restriction *f*_{A} : *A →* Y of *f* to *A* is not a semi-quotient mapping because *f*_{A}^{←}{(*a*)} *=* {4} is not semi-open in *A*.

To see when the restriction of a semi-quotient mapping is also semi-quotient we will need the following simple but useful lemmas.

([22, Theorem 1]). *Let X be a topological space, X*_{0} ∈ SO(X) *and A ⊂* *X*_{0}. Then A ∈ SO(X_{0} ) *if and only if A ∈* SO(*X*).

([23, Lemma 2.1]). *Let X be a topological space, X*_{0} a subspace of X. If A ∈ SO(*X*_{0}), *then A = B ⋂ X*_{0}, for some B ∈ SO(X).

([21]). *Let f : X → Y be* *a mapping, A a subspace of X saturated with respect to f, B a subset of X. If g* : *A → f(A) is the restriction of f to A, then*:

(1) *g*^{←}(C) = f^{←}(C) for any C ⊂ *f(A);*

(2) *f*(*A* *⋂* *B*) = *f*(*A*) *⋂* *f*(*B*).

Now we have this result.

*Let f : X → Y* be *a semi-quotient mapping and let A be a subspace of X saturated with respect to f, and let g : A →* *f*(*A*) *be the restriction of f to A. Then*:

(a) *If A is open in X, then g is a semi-quotient mapping;*

(b) *Iff is an s-open mapping, then g is semi-quotient*.

(a) Let *V* be an open subset of *f*(*A*). Then *V = W ⋂* *f*(*A*) for some open subset *W* of *X*, so that *g*^{←}(*V*) = *f*^{←} (*W* *⋂* *f*(*A*)) = *f*^{←}(*W*) *⋂* *A* is a semi-open set in *A*.

Let now *V* be a subset of *f*(*A*) such that *g*^{←}(*V*) is semi-open in *A*. We have to prove that *V* is open in *f*(*A*). Since *g*(*V*) is semi-open in *A* and A is open in *X* we have that *g*^{←}(*V*) is semi-open in *X*. By Lemma 3.7, *g*^{←}(*V*)= *f*^{←}(*V*); the set *f*^{←}(*V*) is semi-open in *X* since *f* is semi-quotient, hence *g*^{←}(*V*) is semi-open in *f*(*A*). This means that *V* is open in *Y* and thus in *f*(*A*). This completes the proof that *g* is a semi-quotient mapping.

(b) Let now *f* be *s*-open and *V* a subset of *f*(*A*) such that *g* ^{←}(*V*) is semi-open in *A*. Again we must prove that *V* is open in *f*(*A*). Since *g*^{←}(*V*)= *f*^{←}(*V*) and *g*^{←}(*V*) is semi-open in *A*, by Lemma 3.6 we have *f*^{←}(*V*) = *U* ∩ *A*, for some U semi-open in *X*. As *f* is surjective, it holds *f*(*f*^{←}(*V*) = *V*. By Lemma 3.7, then *V* = *f*(*f*^{←}(*V*)) = *f*(*U* ∩ *A*) = *f*(*U*) ∩ *f*(*A*). The set *f*(*U*) is open in *Y* because *f* is *s*-open, so that *V* is open in *f*(*A*). Other part is the same as in (a).

As a complement to Proposition 3.4 we have the following two theorems.

*Let X, Y and Z be topological spaces, f : X → Y a semi-quotient mapping, g : Y →* Z *a mapping. Then the mapping g o f : X → Z* *is semi-quotient if and only if g is a quotient mapping*.

If *g* is a quotient mapping, then *g ∘* f is semi-quotient as the composition of a semi-quotient and a quotient mapping (Proposition 3.4).

Conversely, let *g* o f be semi-quotient. We have to prove that a subset *V* of *Z* is open in *Z* if and only *ifg*^{←}(*V*) is open in *Y*. For, the set (*g ∘* *f*)^{←}(*V*) is semi-open in *X* and since *f* is a semi-quotient mapping we conclude that it will be if and only if *g*^{←}(*V*) is open in *Y*.

*Let f : X → Y* *be a mapping and g* : *X →* Z a *mapping which is constant on each set f*^{←}({*y*})*, y ∈* *Y. Then g induces a mapping h : Y →* Z such that g = *h ∘ f. Then*:

(1) *If f is pre-semi-open and irresolute, then h is a semi-continuous mapping if and only if g is semi-continuous;*

(2) *If f is semi-quotient, then h is continuous if and only if g is semi-continuous*.

Since *g* is constant on the set *f*^{←}({*y*}), *y* ∈ Y, then for each *y ∈* Y, the set *g*(*f*^{←}({*y*})) is a one-point set in *Z*, say *h*(*y*). Define now *h* : *Y* *→* *Z* by the rule

*h*(*y*)* = g*(*f*^{←}(*y*)) (*y ∈* *Y*).

Then for each *x ∈* X we have

*g*(*x*) = *g*(*f*^{←}(*f*(*x*))) = *h*(*f*(*x*)), i.e. *g = h º f*.

(1) Suppose g is a semi-continuous mapping. If *V* is an open set in *Z*, then *h*^{←}(*V*) = *f*(*g*^{←}(*V*)) ∈ SO(*Y*) because *f* is pre-semi-open and *g*^{←}(*V*) is semi-open in *X*. Thus *h* is a semi-continuous mapping.

Conversely, suppose *h* is a semi-continuous. Let *V* be an open set in *Z*. The set *g*^{←}(*V*) = *f*^{←}h^{←}(*V*)) is semi-open in *X* because *f* is irresolute and *h*^{←}(*V*) *∈* *SO*(Y). So, *g* is semi-continuous.

(2) If *g* is semi-continuous, then for any open set *V* in *Z* we have *g*^{←}(*V*) is a semi-open set in *X*. But, *g*^{←}(*V*) = *f*^{←}(*h*^{←}(*V*)). Since f is semi-quotient it follows that *h*^{←}(*V*) is open in Y. So, *h* is continuous.

Conversely, suppose *h* is continuous. For a given open set *V* in *Z*, *h*^{←}(*V*) is an open set in *Y*. We have then *g*^{←}(*V*) = *f*^{←}(*h*^{←}(*V*)) is semi-open in *X* because *f* is semi-quotient. Hence *g* is semi-continuous.

At the end of this section we describe now a typical construction which shows how the notion of semi-quotient mappings may be used to get a topology or a topology-like structure on a set.

Construction: Let *X* be a topological space and Y a set. Let* f : X→ Y* be a mapping. Define

*sτ*_{Q} := {*V ⊂* *Y* : *f*^{←}(*V*) ∈ *SO*(*X*)},

It is easy to see that the family *sτ*_{Q} ia a generalized topology on Y (i.e. ∅ ∈ *sτ*_{Q} and union of any collection of sets in *sτ*_{Q} is again in *sτ*_{Q}) generated by *f*; we call it the *semi-quotient generalized topology*. But *sτ*_{Q} need not be a topology on Y. It happens if *X* is an extremally disconnected space, because in this case the intersection of two semi-open sets in *X* is semi-open [24]. It is trivial fact that in the latter case *sτQ* is the finest topology *ς* on Y such that *f : X →* (Y, σ) is semi-continuous. In fact, *f : X →* (*Y, sτ*_{Q}) is a quotient mapping in this case.

In particular, let p be an equivalence relation on *X*. Let *p* : *X →* *X/ρ* be the natural (or canonical) projection from *X* onto the quotient set *X/ρ*: for each *x* in X, p sends *x* to the equivalence class *ρ*(x). Then the family *sτ*_{Q} generated by *p* is a generalized topology on the quotient set *Y/ρ*, and a topology when *X* is extremally disconnected. This topology will be called the *semi-quotient topology* on *X/ρ*. Observe, that we forced the mapping *p* to be semi-continuous, that is semi-quotient.

This kind of construction will be applied here to topologized groups: to *s*-topological groups and irresolute-topological groups.

The following example shows that a quotient topology on a set generated by a mapping and the semi-quotient (generalized) topology generated by the same mapping are different.

*Let the set X =* {1, 2, 3, 4} *be endowed with the topology*

*τ =* (∅,*X*, {1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}.

*Then the set SO*(*X*) *is*

(∅, *X*, {1},{2},{3},{1,2},{1,3},{2,3},{1,4},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4}}.

*Define the relation R on X by xRy if and only if x + y is even. Therefore*,

*R =* {(1,1),(1,3),(2,2),(2,4),(3,1),(3,3),(4,2),(4,4),}

*is an equivalence relation, and X/R =* {*R*(1),*R*(2)}* =* {{1,3},{2,4}}. *Let p : X →* *X/R be the canonical projection. Then, p*^{←}(*R*(1)) *=* {1, 3} ∈ *SO*(*X*), *and p*^{←}(*R*(2)) *=* f*2;* 4} ∈ *SO*(*X*) , *so that*

*sτ*_{Q} = {∅, *X/R*, {*R*(1)},{*R*(2)}}

*is the semi-quotient topology on X/R. On the other hand, p*^{←}({*R(*1*)}*) = {1,3} ∈ τ*, but p*^{←}({*R*(2)}) = {2,4} ∉* τ. Therefore, the quotient topology on X/R is*

*τ*_{Q} = {∅ *X/R*, {*R*(1)}}.

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