Characterization of Quotient Maps in Lo- cales

3.2 Characterization of Quotient Maps in Lo-

map of topological spaces to conclude that S is open in Q. Since h⁻¹(S) is open in X+·pl(Q), the set f⁻¹(S) is open in X and

i⁻¹£

Q−f¡

X−f⁻¹(S)¢¤

≥i⁻¹(S)wpl(S), hence

Q−f¡

X−f⁻¹(S)¢

≥pl(S).

Since Q−f(X−f⁻¹(S)) is an open sublocale of Q, then it is spatial. It remains to show thatS ≥pt(S)≥Q−f(X−f⁻¹(S)) to complete the proof because this would imply that S ≥ pt(S) ≥ Q−f(X−f⁻¹(S)) ≥ pl(S) and hence that S is spatial since any locale is the join of its largest pointless sublocale and its points provided that the largest pointless sublocale exists.

But this follows from the fact that

S∨f[X−f⁻¹(S)] = f[f⁻¹(S)]∨f[X−f⁻¹(S)]

=f[f⁻¹(S)∨(X−f⁻¹(S))]

=f(X)

=Q. 2

Definition 3.2.3 (See e.g. Siweya [16])

A topological space X is said to be a sober space if every irreducible closed subset ofXis the closure of exactly one singleton ofX. An irreducible closed subset ofX is a non-empty closed subset ofX which is not the union of two proper closed subsets of itself.

For the proof of the next result, recall that the category Sob of sober spaces into Loc is a spatial subcategory of the category Loc of locales and locale maps. See Johnstone [9].

Example 3.2.4

A surjection f : X → Q of spatial metrizable locales which is a regular epimorphism, but which does not satisfy (QO).

Take any quotient map f : X =`

q∈QQ_q→Q induced by the natural maps f_q : Q_q → Q in the category Top of topological spaces from a completely metrizable space X =`

q∈QQ_q onto the irrationals Q where Q_q (for q ∈Q) has the same underlying set asQ, but all the points exceptqare isolated and qhas the same neighborhoods inQandQ_q.Since the embeddingSob ,→Loc of the category Sob of sober spaces into Loc as spatial locales has a right adjoint, it preserves all regular epimorphisms, so f : X → Q is a regular epimorphism in Loc. All pointless sublocales of Q pullback to the empty sublocales of X because of the following:

(i) pl(X), the largest pointless sublocale is an O_δ inQ (a countable intersection of open sublocales).

(ii) pulling back alongf preserves all meets.

(iii) the O_δ’s in complete spaces are spatial (See Isbell [7]), hence pointless O_δ’s are empty.

So, if S is any non-zero pointless sublocale of Q, thenf⁻¹(S) = 0 is open in

X, but S is not open in Q, which means thatf does not satisfy (QO). 2

Definition 3.2.5 (Plewe [12])

Let f : X → Y be a surjection of locales. Then f is said to satisfy the condition

(WQ) if, for all complemented sublocales S ∈ Sub(Y), whenever f⁻¹(S) is open in X then S is open inY;

(EQC) if, for all sublocales S = U ∨(V −W), where U, V and W are open sublocales of Y, whenever f⁻¹(S) is open in X then S is open inY. Proposition 3.2.6

A surjectionf :X →Y inLocsatisfies (EQC) if and only if it satisfies (W Q).

Proof:

Suppose that f : X → Y satisfies (W Q). Then by definition it follows that f satisfies (EQC).

Conversely, suppose that f : X → Y satisfies (EQC) and let C be a complemented sublocale of Y such that f⁻¹(C) is open in X. We want to show that C is open inY. Let U =int(C),H =cl(C−U) and P =intH(C−U).

Then U∨P is open inY and P is locally closed in Y. The setP is dense in H because every dense and complemented sublocale has dense interior (the complement D of C − U is disjoint from D(H) the smallest dense sublocale of H ), and hence so is clH(D). Now f⁻¹(U ∨P) is open in f⁻¹(C), hence open in X. So applying (EQC) we find that U ∨P ⊆ C is open in

Y. Because U is the interior of C we can conclude that U ∨P = U, and then that P = 0 since U ∧P = 0. Since P is dense in H, it follows that H = 0, so thatC−U = 0. HenceC =U is open inY, i.e.,f satisfies (W Q).2

Definition 3.2.7

A locale L is called a subfit locale if each open sublocale of L is the join of closed sublocales ofL or equivalently if each closed sublocale ofLis the join of open sublocales of L.

Remark 3.2.8

(a) In [14], Simmons called these locales conjunctive whereas Herrlich and Pultr inform [5] us that in spaces the corresponding term used was weakly regular

(b) Subfitness is equivalent to the following condition:

(∗) whenever ab in L there exists a c∈L such that a∨c= 16=b∨c In view of the definition of open and closed sublocales in Remark 1.2.7, we know that ˇb v ˆa iff a∨b = 1 which implies that the statement “each open sublocale can be written as a join of closed sublocales” can now be rewritten thus:

(∗∗) : (x∧a6=y∧a)⇒(∃c∈L such that a∨c= 1 and x∨c6=y∨c).

(**) ⇒ (*) : Therefore, take ab and setx=a and y=b in (**). Then a∨c (=x∨c) = 16=y∨c(=b∨c).

(*) ⇒ (**) : Given x∧a 6= y∧a, suppose that x∧a x∧y∧a in (*). Then there exists a c∈L such that (x∧a)∨c= 16= (x∧y∧a) from which a∨c= 1 andx∨c6=y∨c follow.

Proposition 3.2.9

Iff :Y →X and g :Z →X are closed surjections whereY andZ are subfit domains, then the induced map Y ×_X Z →X is again a surjection.

Proof:

Suppose that f : Y → X and g : Z → X are closed surjections with subfit domains and consider the map h : Y ×_X Z → X induced by f and g. We must show that h is a surjection. Note that it is sufficient to show that Y ×_X Z = 0 implies that X = 0, because of the following:

(a) closed surjections are stable under pullback along complemented sublocales.

(b) subfitness is inherited by complemented sublocales.

(c) the image h(Y ×_X Z) in X is the intersection of complemented sublocales C of X for which the restriction of f and g to X−C have zero pullback.

Now the topology of the pullback Y ×X Z is presented as follows:

T(Y)⊗_T_(X₎T(Z) = (T(Y)×T(Z)|cov), where cov(y, z) is the union of the following covers:

(i) {{(y_i, z)|i∈I} |W

y_i =y}

(ii) {{(y, z_i)|i∈I} |W

z_i =z}

(iii) {{((y∧f^∗(x)), z)} |g^∗(x)≥z}

(iv) {{(y,(g^∗(x)∧z))} |f^∗(x)≥y}

Clearly these covers are stable. Now let the sieve

S ={(y, z)| ∀closed sublocales F ≤Y, G≤Z :F ≤Uy and G≤U_z =⇒f(F)∧g(G) = 0}

We claim that the sieveS is closed:

To this end, under the covering of the first type, let {(y_i, z)|i∈I} ⊆ S be arbitrary, and let F ≤ U^W_y_i and G ≤ U_z be closed sublocales of Y and Z, respectively. Then if U_y_i =W

F_i,j where each F_i,j is closed, then f(F)∧g(G) =f

h F ∧W

i,jF_i,j i

∧g(G)

=fhW

i,j(F ∧F_i,j) i

∧g(G)

=hW

i,jf(F ∧F_i,j) i

∧g(G)

≤hW

i,jf(F_i,j) i

∧g(G)

i,j[f(F_i,j)∧g(G)]

= 0

because each F_i,j ≤U_y_i and G≤U_z.

For covers of the second type, let {(y, z_i) | i ∈ I} ⊆ S be arbitrary and let F ≤ U_y and G ≤ U_z_i be closed sublocales of Y and Z, respectively. If U_z_i =W

G_i,j where each G_i,j is closed, then f(F)∧g(G) =f(F)∧g[G∧W

i,jG_i,j]

=f(F)∧g[W

i,j(G∧G_i,j)]

=f(F)∧W

i,j[g(G∧G_i,j)]

≤f(F)∧W

i,jg(G_i,j)

i,j[f(F)∧g(G_i,j)]

= 0.

because each F ≤U_y and G≤U_z_i.

Now for the covers of the third type, let [(y∧f^∗(x), z)]∈ Sand assume that g^∗(x) ≥ z where g^∗ = g : Z → X. Let F ≤ U_y and G ≤ U_z be any closed sublocales of Y and Z, respectively. We want to show thatf(F)∧g(G) = 0.

Since X is subfit (subfitness is preserved under closed surjections), there

exists a set{H_i |i∈I}of closed sublocales ofX such thatU_x = W

i∈I

H_i. Now [f(F)∧Hi]∧g(G) = f£

F ∧f⁻¹(Hi)¤

∧g(G) = 0

because F ∧f⁻¹(H_i) ≤ U_y∧f^∗_(x) for all i∈ I. Since g^∗(x) ≥ z implies that g⁻¹(Ux)≥Uz and hence Ux ≥g(Uz)≥g(G), from which it follows that

f(F)∧g(G) =f(F)∧g(G)∧U_x

=f(F)∧g(G)∧ W

i∈I

H_i

= W

i∈I

[f(F)∧H_i∧g(G)]

= 0.

Finally, for the covers of the fourth type let [y,(g^∗(x)∧z)]∈ S and assume that f^∗(x) ≥ y where f^∗ = f : Y → X. Let F ≤ Uy and G ≤ Uz be any closed sublocales of Y and Z, respectively. We want to show that f(F)∧ g(G) = 0.SinceX is subfit (subfitness is preserved under closed surjections), there exists a set{K_i |i∈I}of closed sublocales of X such thatU_x = W

i∈I

K_i. Now

[f(F)∧K_i]∧g(G) = f£

F ∧f⁻¹(K_i)¤

∧g(G) = 0

because F ∧f⁻¹(K_i) ≤ U_g^∗_(x)∧z for all i ∈ I. But f^∗(x) ≥ y gives rise to f⁻¹(U_x)≥U_y and hence U_x ≥f(U_y)≥f(F), so that

f(F)∧g(G) =f(F)∧g(G)∧Ux

=f(F)∧g(G)∧ W

i∈I

= W

i∈I

[f(F)∧K_i∧g(G)]

= 0.

Now S is maximal if and only if (1,1) ∈ S, i.e., f(Y)∧g(Z) = 0. Since f and g are surjections, we havef(Y)∧g(Z) = 0 if and only if X = 0. So we conclude that Y ×_X Z = 0 if and only if X = 0. 2

Theorem 3.2.10

Closed surjections f : X → Y with subfit domain X are regular epimorphisms.

Proof:

Letf :X →Y be a closed surjection with subfit domainX. Thenf :X →Y is a quotient map with the kernel pair X ×_Y X ⇒^π¹

π2

X. To show that f is a regular epimorphism, we need only show that f ≈ Coeq(π₁, π₂). So let K ⊆X be any closed sublocale such thatπ₁⁻¹(K) = π⁻¹₂ (K). Now restricting the map f :X →Y to a map f⁻¹◦f(K)→f(K) if necessary (subfitness is inherited by closed subspaces), we may assume that f(K) =Y. We want to show thatK =X. But this will follow if we can show thatK∧H = 0 implies that H = 0 for any closed sublocale H ⊆ X, because then the subfitness of

X implies that

X−K =_

{H closed in X |K∧H = 0}= 0.

So, letHbe any closed sublocale ofXsuch thatK∧H= 0. Then we restrict the domain of f to K∨H. The resulting map g =f ¹_K∨H: K∨H → Y is still a closed surjection onto Y and the domain of g is the disjoint union of K and H. Furthermore, both K and H have isomorphic pullbacks along π₁ and π2 because

K×_Y H ≤K×_Y (X−K)

=K×(W

{H⊆X |K ∧H = 0})

=F ×_Y 0

= 0.

Finally, we restrict the codomain Y to W = g(H) to arrive at the closed surjection

g⁻¹◦g(H) = Z ^{h=g g}⁻¹^◦g(G) ^//W.

Then both T = K ∧ Z and H are mapped by h onto W and because T×_WH = 0 and bothT and H are equalized by the kernel pair (π₁⁻¹, π⁻¹₂ ) of h. But since the restrictions of htoT are also closed surjections, Proposition (3.2.9) implies that T ×_WH maps onto W, and so W = 0 so that H = 0. 2

Proposition 3.2.11

Closed maps are stable under pullback along complemented inclusions.

Proof:

Suppose that C is complemented inA and let g :B →C be the pullback of f :B →A along the inclusion i:C ,→A. Now from the following figure

B ^id^B ^//

²²

²²C _i ^//A

we find that if f : B → A is a closed map and K is a closed sublocale of f⁻¹(C), then (applying Motivating Example 1.2.8)

g[id⁻¹_B (K)] =g(K)

=f(F)

=f[cl_B(K)∧f⁻¹(C)]

=f[cl_B(K)]∧C

which is closed in C because f[clB(K)] is closed in A. So g is also closed.2

Proposition 3.2.12

A simple coveringB →^f Ais a regular epimorphism if and only if it satisfies (W Q).

Proof: We refer to condition (RE) in Proposition 2.2.9.

Recall that in the category Loc of locales regular epimorphisms are always extremal epimorphisms (Proposition 3.1.2), so we need only show that (W Q) implies (RE). To this end, we proceed as follows: if K is any closed sublocale ofA then (A−K)∨

H₀−H−

is a complemented sublocale ofA with complement H_c∨H−

m. Now since f⁻¹[(A−K)∨(H₀−H−

m)] = f⁻¹(A−K)∨f⁻¹

H₀−H−

= [f⁻¹(A)−f⁻¹(K)]∨[f⁻¹(H₀)−f⁻¹

³ H−

´ ]

= [B −f⁻¹(K)]∨[[f⁻¹(K)−f⁻¹

³ H−

´ ]∧U]

= [(B −f⁻¹(H)]∨[(B−f⁻¹

³ H−

´ )∧U]

is open in B, condition (W Q) implies that (A−K)∨

H0−H_m⁻

is open in A and therefore H₀−H−

m is open in K. 2

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Cat. Str. 9 (2001), 245-258.

[3] Chen, X.: Stably closed frame homomorphisms, G´eom. Diff. Cat, 2 (1996), 123-144.

[4] Dowker, C. H and Strauss, D. P.: Paracompact frames and closed maps, Symp. Math. 16 (1975), 93-116.

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Dalam dokumen ON CLOSED AND QUOTIENT MAPS OF LOCALES - ULSpace (Halaman 60-74)