In this section we apply the construction *of sτ*_{Q} described in Section 3 to topologized groups and establish some properties of their semi-quotients.

If G is a topological group and *H* a subgroup of G, we can look at the collection *G/H* of left cosets of *H* in G (or the collection *H\G* of right cosets of *H* in G), and endow *G/H* (or *G\H*) with the semi-quotient structure induced by the natural projection *p* : G → *G/H*. Recall that *G/H* is not a group under coset multiplication unless *H* is a normal subgroup of G.

The following simple lemmas may be quite useful in what follows.

**Lemma 5.1:** *([20]). **Let p* : *G* → G/H *be a canonicalprojection map. Then for any subset U of G, p*^{←}(*p*(*U*)) *= U * H*.

**Lemma 5.2:** *([25]). **Let* (*G*, *, *τ*)* be an s-topological group, K an s-compact subset of G, and F a semi-closed subset of G. Then F* * *K and K* * *F are semi-closed subsets of G*.

**Lemma 5.3:** *([4]). **Let* (*G*, *, *τ*)* be an s-topological group. Then each left (right) translation in G is an S-homeomorphism. Moreover they and symmetry mappings are actually semi-homeomorphisim (see [1, Remark 1]*).

**Lemma 5.4:** *([26]). **If* ƒ : *X* → *Y is a semi-continuous mapping and X*_{0}* is an open set in X, then the restriction $f{|}_{{X}_{0}}:{X}_{0}\to Y$ is semi-continuous*.

**Theorem 5.5:** *Let* (*G*, *, *τ*)* be an extremally disconnected irresolute-topological group and H its invariant subgroup. Then ρ* : $p:\left(G,\ast ,\tau \right)\to \left(G/H,\overline{\ast},s{\tau}_{Q}\right)$ * is pre-semi-open*.

**Proof:** *Let **V* ⊂ *G* be semi-open. By the definition of semi-quotient topology, *p*(*V*)⊂ *G/H* is open if and only if *p*^{←}(*p*(*V*))⊂ *G* is open. By Lemma 5.1 *p*^{←}(*p*(*V*))* = V * H*. Since *V* is semi-open, *V* * *H* is semi-open and so *p*(*V*)is semi-open. Hence *p* is pre-semi-open.

**Theorem 5.6:** *Let* (*G*, *, *τ*)* be an extremally disconnected irresolute-topological group, Η its invariant subgroup. Then $\left(G/H,\overline{\ast},s{\tau}_{Q}\right)$ is an irresolute-topological group*.

**Proof:** *First, we observe that **sτ*_{Q} is a topology on *G/H*. Let *x* * *H, y* * *H ∈ G/H* and let *W ⊂* *G/H* be a semi-open neighbourhood of $(x\ast !H)\overline{\ast}(y\ast {H}^{-1})$. By the definition of *sτ*_{Q} (induced by *p*), the set *p*^{←}(*W*)is a semi-open neighbourhooud of *x* * *y*^{-1}in G, and since *G* is an irresolute-topological group, there are semi-open sets *U* ⊂ SO(*G*, *x*)and *V* ⊂ SO(G, *y*)such that *U * V*^{-1}⊂ *p*^{←}(*W*). By Theorem 5.5, the sets *p*(*U*)* = U * H* and *p*(*V*)* = V * H* are semi-open in *G/H*, contain *x * H* and *y * H*, respectively, and satisfy
$$\left(U\ast H\right)\overline{\ast}{\left(V\ast H\right)}^{-1}=\left(U\ast {V}^{-1}\right)\ast H=p\left(U\ast {V}^{-1}\right)\subset p\left({p}^{\leftarrow}\left(w\right)\right)\subset W.$$This just means that $\left(G/H,\overline{\ast},s{\tau}_{Q}\right)$ is an irresolute-topological group.

**Theorem 5.7:** *Let* (*G, *, τ*)* be an s-topological group and H a subgroup of G. Then for every semi-open set U ⊂* *G*, *the set p*(*U*)* belongs to sτ*_{Q}. In particular, if G is extremally disconnected, then p is an s-open mapping from G to (*G/H, sτ*_{Q}).

**Proof:** *Let **V ⊂* *G* be semi-open. By definition of *sτ*_{Q}, p(*V*)* ∈ sτ*_{Q} if and only if *p*^{←}(*p*(*V*))* ⊂* *G* is semi-open, i.e. *V * H* is semi-open in G. But *V * H* is semi-open in *G* because *V ∈* SO(*G*) and *(G, *, τ*)is an s-topological group. Clearly, if *sτ*_{Q} is a topology, the last condition actually says that *p* is an s-open mapping.The following theorem is similar to Theorem 5.7.

**Theorem 5.8:** *If Η is an s-compact subgroup of an s-topological group* (*G*, *, *τ*), *then for every semi-closed set F ⊂* G, *the set p*(*G* \* F*)* belongs to sτ*_{Q}. If sτ_{Q} is a topology, then p is an s-perfect mapping.

**Proof:** *Let **F ⊂* *G* be semi-closed. By Lemma 5.2 the set *p*^{←}(*p*(*F*))* = F * H ⊂* *G* is semi-closed. By definition of *sτ*_{Q}, G/H \ (*F * H*) ∈* sτ*_{Q}.Let now *sτ*_{Q} be a topology on *G/H*. Take any semi-closed subset *F* of G. The set *F * H* is semi-closed in *G* and *F * H = p*^{←}(*p*(*F*)). This implies, *p*(*F*)is closed in the semi-quotient space *G/H*. Thus *p* is an *s*-closed mapping. On the other hand, if *z* * *Η ∈ G/H* and *p*(*x*)* = z * Η* for some *x ∈ G*, then *p*^{←}(*z * H*)* = p*^{←}(*p*(*x*))* = x * Η*, and by Lemmas 4.3 and 5.3 this set is s-compact in *G*. Therefore, *p* is *s*-perfect.

**Corollary 5.9:** *Let* (*G, *,τ*)* be an extremally disconnected s-topological group and Η its s-compact subgroup. If the semi-quotient space* (*G/H, sτ*_{Q})* is compact, then G is s-compact*.

**Proof:** *By Theorem 5.8, the projection **p : G → G/H* is *s*-perfect. Then by Theorem 4.4 we obtain that *p*^{←}(*p*(*G*))* = G * H = G* is *s*-compact.

**Theorem 5.10:** *Suppose that (G,*, τ) is an extremally disconnected s -topological group, H an invariant subgroup of G, p*:*G →* (*G/H, sτ*_{Q})* the canonical projection. Let U and V be semi-open neighbourhoods of e in G such that V*^{-}^{1}* *V ⊂* *U. Then* C1(*p*(*V*))* ⊂* *p*(*U*).

**Proof:** *Let **p*(*x*)* ∈* C1(*p*(*V*)). Since *V* * *x* is a semi-open neighbourhood of *x* ∈ *G* and, by Theorem 5.7, *p* is s-open, we have that *p*(*V* * *x*)is an open neighbourhood of *p*(*x*). Therefore, *p*(*V* * *x*)⋂ *p*(*V*)≠ø. It follows that for some *a*, *b* ∈* V* we have *p*(*a* * *x*)* = p*(*b*), that is *a* * *x* * *h*_{1}* = b* * *h*_{2}for some *h*_{1}*, h*_{2}∈ *Η*. Hence,
$$x={a}^{-1}\ast b\ast {h}_{2}\ast {h}_{1}^{-1}=\left({a}^{-1}\ast b\right)\ast \left({h}_{2}\ast {h}_{1}^{-1}\right)\in U\ast H$$since *a*^{-1} * *b* ∈* V*^{-1}* *V* ⊂ *U* and *H* is a subgroup of *G*. Therefore, *p* (*x*)∈* p*(*U* * *H*)* = U * H * H = U *H = p*(*U*).

**Theorem 5.11:** *Let (G,*, τ) be an extremally disconnected irresolute-topological group and H an invariant subgroup of G. Then the semi-quotient space* (*G/H, sτ*_{Q})* is regular*.

**Proof:** *Let **W* be an open neighbourhood of *p*(*e*_{G})* = Η* in *G/H*. By semi-continuity of *p*, we can find a semi-open neighbourhood *U* of *e*_{G} such that *p*(*U*)⊂ *W*. As *G* is extremally disconnected and irresolute-topological group, it follows from ${e}_{G}\ast {e}_{G}^{-1}={e}_{G}$ that there is a semi-open neighbourhood *V* of *e*_{G} such that *V* * *V*^{-1} ⊂ *U*. By Theorem 5.10 we have *C1*(*p*(*V*))⊂ *p*(*U*)⊂ *W*. By Theorem 5.7, *p*(*V*)is an open neighbourhood of *p*(*e*_{G}). This proves that (*G/H, sτ*_{Q}) is a regular space.If *(G*, *) is a group, *H* its subgroup, and *a* ∈* G*, then we define the mapping *λ*_{a} : G/H →* G/H* by *λ*_{a}(*x * Η*)* = a ** (*x * Η*). This mapping is called a *left translation of G/H by a* [20].

**Theorem 5.12:** *If* (G, *, *τ*) *is* *an* *extremally disconnected irresolute-topological group, Η a subgroup of G, and a ∈* *G*, *then the* *mapping λ*_{a} is a semi-homeomorphism and p ∘* ℓ*_{a} = λ_{a} ∘* p holds*.

**Proof:** *Since **G* is a group, it is easy to see that *λ*_{a} is a (well defined) bijection on *G/H*. We prove that *λ*_{a} ∘ *p =* *p* ∘ *ℓ*_{a}· Indeed, for each *x* ∈ *G* we have (*p* ∘ *ℓ*_{a})(*x*)* = p*(*a * x*)* =* (*a*x*)** Η = a** (*x * Η*)* = λ*_{a}(*p*(*x*))* =* (*λ*_{a} ∘ *p*)(*x*). This is required. It remains to prove that *λ*_{a} is irresolute and pre-semi-open.This follows from the following facts. Let *x* * *Η* ∈* G/H*. For any semi-open neighbourhood *U* of *e*_{G}*, p*(*x ** *U * H*)is a semi-open neighbourhood of *x* * *Η* in *G/H*. Similarly, the set *p* (*a * x * U * H*)is a semi-open neighbourhood of *a * x * Η* in *G/H*. Since*λ*_{a}(*p*(*x * U * H*)) = *p*(*l*_{a}(*x * U * H*)) = *p* (*a * x * U * H*),it follows that *λ*_{a} is a semi-homeomorphism.

**Definition 5.13:** *A mapping* ƒ : *X → Y is*:

*
***Theorem 5.14:** *Let* (*G,*,τ*_{g})* and* (*Η*, ·,*τ*_{η})* be extremally disconnected irresolute-topological groups and* ƒ : G → *H a semi-isomorphism. If G*_{0}* is an invariant subgroup of G and H*_{0}* = f* (*G*_{0})*, then the semi-quotient irresolute-topological groups* (*G/G*_{0}*,sτ*_{Q})* and (H/H*_{0}*,sτ*_{Q})* are semi-isomorphic*.

**Proof:** *Let **ρ : G → G*/*G*_{0}, *x ↦ x ** G_{0}, and *π : Η → Η/H*_{0}*, f*(*x*_{0})↦ *f*(*x*_{0}). *H*_{0}(*x*_{0}* ∈* G_{0}) be the canonical projections. Consider the mapping *φ* : G/Go *→ H/H*_{0}defined by*φ*(*x ** G_{0}) = *f*(*x*) · *f*(G_{0}), *x* ∈ *G*, *y = f*(*x*).Then for *x*_{1} * *G*_{0}, *x*_{2} * *G*_{0} ∈ *G*/*G*_{0} we have*φ*(*x*_{1} * *G*_{0}* *x*_{2} * *G*_{0}) = *φ*(*x*_{1} * *x*_{2} * *G*_{0}) = *f*(*x*_{1}* *x*_{2})* · f*(*G*_{0}) = *y*_{1}* · y*_{2}* · H*_{0}* = φ*(*x*_{1} * *G*_{0}) · *φ*(*x*_{2} * *G*_{0}),i.e. *φ* is a homomorphism. Let us prove that *φ* is one-to-one. Let *x* * *G*_{0}be an arbitrary element of *G/G*_{0}. Set *y = f*(*x*). If *φ*(x * G_{0}) = *H*_{0}, then π(y) = *H*_{0}, which implies x ∈ G_{0}, *y* ∈ H_{0}, and ker *φ* = Go. So, *φ* is one-to-one.Next, we have *φ*(*x* * G_{0}) = *y · H*_{0}, i.e. *φ*(*p*(*x*)) *= π*(*y*)* = π*(*f*(*x*)). This implies *φ ∘* *p = π ∘* f. Since f is a semi-homeomorphism, and p and π are s-open, semi-continuous homomorphisms, we conclude that *φ* is open and continuous. Hence *φ* is semi-homeomorphism and a semi-isomorphism.

**Theorem 5.15:** *Let* (G, *, τ) *be an extremally disconnected irresolute-topological group, H an invariant subgroup of G, M an open subgroup of G, and p G* → *G/H the canonical projection. Then the semi-quotient group M* * *H/H is semi-isomorphic to the subgroup p*(*M*)* of G/H*.

**Proof:** *It is clear that **M* * *H = p*^{←}(*p*(*M*)). As *p* is s-open and semi-continuous and *M* is open in *τ*, the restriction π of *p* to *M* * *H* is an s-open and semi-continuous mapping of *M* * *H* onto *p*(*M*)by Lemma 5.4. Since M is a subgroup of G and *p is a* homomorphism it follows that *p*(*M*)is a subgroup of *G/H, M * H* is a subgroup of G, and π: *M * H* →* p*(*M*)is a homomorphism. We have π^{←}(*π*(e_{G})) *= p*^{←}(p(e_{G})) *=* H, i.e. ker π = *H*. It is easy now to conclude that *M * H/H* and *p*(*M*)are semi-isomorphic.

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