GAZETA DE MATEMÁTICA 19

Topological

Î by John B. Pan, S. J. Fu Jan CalvsrsSty, Taipei, Formol a

Introduction. Let S be a , R be a homomor- phic equivalence relation on S, and let S/R A topological semigroup is a system con- be the quotient . We define an operation sisting of a set S , an operation •, (we omit on S/R in the following manner. Suppose this dot and write this operation by juxtapo- that A and B are two arbitrary elements in sition), and a topology T, satisfying the S/R, then AB = C if for any aeA and following conditions: be B we have abeC. This operation is well-defined because R is a homomorphic 1) for any x,yeS, aryeS; equivalence relation. Also it is associative, because the semigroup S is associative. The- 2) for x,y,ze S, (xy)z = x{yz); refore the quotient set S/R with the opera- 3) the operation • is continuous in the tion just defined is a semigroup. We call it topology T. the quotient semigroup. We say a semigroup S satisfies the con- A topological subsemigroup H of a semi- dition A if for every open set U of S, the S is a topological subspace of S and subset (»J (U)) is also open, where n is also a subsemigroup of S. the natural mapping from S onto S/R. An equivalence relation R defined on a In this paper *ve shall prove the following semigroup S is called homomorphic if for theorems: any a ,b ,c ,

The family U of all subsets U* of S/R relation RK on S. such that n_I(U*) is open in S is a topology for S/R and is called the quotient topology THEOREM 3, Let S and T be two topolo- for S/R. gical semigroups and let g be an open homo- We use the term homomorphism to mean morphism from iS onto T. Then continuous homomorphism. In general, we use the terms mapping, function to mean a) S/Rg is a topological semigroup with continuous mapping, . the quotient topology; \

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b) the natural mapping n from S on to Now let U* be an open set in S/R. By S/BK is an open homomorphism; the definition of the quotient topology for

c) the mapping b from S/Re onto T S/R (U*) is open. Hence n is continuous. defined by b (A) = g (a) for any a e A Let U be an open set in S. Since S

as a subset of S and AeS/Rg tJ a satisfies the condition A, «-'[«(U)] is open. topological isomorphism. Then by the definition of the quotient topo- logy, n(U) is open. THEOREM 4. (The First Isomorphism Now we wish to show that the semigroup Theorem). Let S and T be two topological operation in S/R is continuous. Let A and semigroups both satisfying the condition A . B be two arbitrary elements in S/R such Let g be an open homomorphism from S onto that AB = C. Suppose that W* is an open T and let R* be a homomorphic equivalence neighborhood of C. Then W = w~i(W*) is relation defined on T. Then there is a homo- an open neighborhood of 0, considered as mcrphic equivalence relation R on S and a subset of S . Since the semigroup operation there is a mapping h from S/R onto T/R* in S is continuous, for every a e A and which is a topological isomorphism. every b 6 B such that ab = c , there is an open neighborhood U„ of a and an open At the end of the paper, we give an exam- neighborhood V» of b such that UJcW. ple to illustrate the theorems. Choose such a neighborhood U„ for every aeA and such a neighborhood V& for every 5 e B. Then THEOREMS

acA a«A THEOREM 1, If the semigroup S satisfies l«B the condition A , then the quotient set S/R is a topological semigroup icith the quotient topo- Now (J Un is an open neighborhood of A logy, and the natural mapping N from S onto aeA S/R is an open topological homomorphism. in S, and n is an open mapping. It follows

that »j~ [J UaJ is an open neighborhood of PROOF. We have shown that S/R is an aeA abstract semigroup. Now we wish to show the element A in S/R. Similarly that the natural mapping n from S to tad S/R is an abstract homomorphism. Let X is an open neighborhood of the element B in and Y be two equivalence classes mod R, S/R. Since [ (J U„] [ (J V»]e W, we have and let XY=Z. Then by definition of the aeA beB operation in S/R, for any xeX and yeY, xyeZ. Since the natural mapping n assigns *[UU»>[U V»]="[UU«UV>] each element to the class it belongs, we am A 6eB aeA t o B have c«(W)=W n(X) = X, n(Y) = Y, and n(xy) = n(z) ~ Z. Hence we have found an open neighborhood These equations together with the equation » [ U U„] of A and an open neighborhood XY = Z imply that n (x y) = n (x) n (y) . aeA

This shows that the natural mapping n is an n i [J V4J of B such that abstract homomorphism from S onto S/R. bt D GAZETA DE MATEMÁTICA 21

u n Vi c * [ U °] [ U ] S/Rg be the quotient set. Then 8(Rt is a asA AeB semigroup. Let n be the natural mapping from S onto S/R . We show that the This shows that the semigroup operation in g semigroup S satisfies the condition A. S/R is continuous. With this, the proof of Let XI be an open subset in S. Since g the theorem is complete. is an open map, g{U) is open in T. Also g is continuous. Hence the subset iT1^^)] THEOREM 2. If S and T are two semi- is open in S. But g~l [g ( U)] = e S \g (t) groups and g is a homomorpkism from S •=g{y) for some y e U\ and «-'[«(U)] onto T, then g induces a homomorphic equi- -*» fare S \ g(x) — g{y) for some y G U j hence valence relation R on S. K n-1 [n ( £/)] = g~l [r/ ( U)] and «-'[»(J7)] is open. This shows that S satisfies the con- PROOF. We define a relation Rj, on S in dition A. the following manner. Suppose that a and Since S satisfies the condition A, the a* are two elements of S, then parts a) and 6) follow from theorem 1. Before proving part c), we wish to show a = a* mod Rg if and only if g (a) = g (a*), that the mapping h defined in the theorem is well-defined. Evidently, Rg is an equivalence relation. Let A be any element of S/Rg and let We show that Rg is homomorphic, i. e., if a, a*, b, 6* e S such that a a* mod R;, and a* and a** be any two elements of A as a subset of S. Then 6 = 6* mod Rg , then a 6 = a* 6* mod Rg . Now a = a* mod-ffj implies g (a) g (a*), a* — a** mod Rg. and 6 = 6* mod Rg implies ^ (6) = <;(&*), These two equations imply that g (a)g(b) This implies = g (a*) g (6*). Since g is a homomorphism, we have g(a)g(6) = g(a6} and g(a*)g(b*) ?(«*) = ff («**)• = gr(a*6"). Hence g(a b) = g (a* 6*). This means that a b = a* 6* mod Rg . This com- Ilence pletes the proof.

THEOREM 3. Let S and T be two topo- logical semigroups and let g be an open This shows that h is well-defined. homomorphism from S o>ifo T. Then Also A is a one to one mapping. For each A e S/Rg there corresponds a unique value a) S/RK is a topological semigroup with the quotient topology ; b) the natural mapping n from S onto

S/Rs is an open homomorphism; in T as shown above. Now since g is a c) the mapping h from S/RG onto T mapping from S onto T, for each teT defined by b (A) = g(a) for any a e A there is an element ae S such that t = g(a),

as a subset of S and A e S/Rg is a by definition of Rg,a = b mod Rg if and only topological isomorphism. if g(a)=*g(b). It follows that for each g(a) t, there is one and only one equiva- PROOF. By theorem 2, g induces a homo- lence class AmodRg such that h(A)*^g(a) morphic equivalence relation Rg on S. Let = t. Hence h is a one to one mapping. 22 GAZETA DE MATEMÁTICA 22

We further show that ft is an algebraic If the semigroup S satisfies the condition homomorphism. Let A and B be any two A, then the quotient set S/R is a topolo- elements in S/Rg. Then gical semigroup with the quotient topology, and the natural mapping n from S onto h (A B) = g (a b) « g (a) g(t>) = h (A) h(B), S/R is an open topological homomorphism. Conversely, if g is an open homomorphism where a and b are arbitrary elements of A from S onto a semigroup T, then T is and B respectively. This shows that ft is topologically isomorphic to the quotient se-' an algebraic homomorphism. migroup S/Rg, where Rg if a homomorphic We show also that A is continuous. Let equivalence relation defined by

A be an element in S/Rg such that h(A) = t, and let W be an open neighborhood of f. a Rgb if and only if (a) = ^ (5); a,beS Since h(A)=>g(a) for every aeA, and since g is continuous, for every aeA, there THEOREM 4. (The First Isomorphism Theo- is an open neighborhood Ua of a such that rem). Let S and T be two topological semi- g(Ua)a W. Choose such an open neighbor- groups both satisfying the condition A . Let hood Ua for every aeA. Then [J {Ua) is g be an open homomorphism from S onto T tie A and let R* be a homomorphic equivalence a neighborhood of A in S and relation defined on T , Then there is a homo- aeA morphic. equivalence relation K on S and is an open neighborhood of the element A there is a mapping h from S/R onto T/R* which is a topological isomorphism. in S/Rg. But ff[[J(^)]=A{«[U(^)]} aeA aeA a W. So for any neighborhood IV of h{A), PROOF. Since R* is a homomorphic we have found a neighborhood equivalence relation on T, by theorem 1, aeA T/R* is a topological semigroup and the of A such that A|«[ (J(t/n)]}

EXAMI'LE. TO illustrate some of the fore- The equivalence classes modi? are of the going theorems we give the following form: ja:) x (0 , 00) . We denote the set of example. all equivalence classes modii by S/R. Wo Let (0,oo) be the semigroup of positive define an operation in S/R in the following 00 real numbers with addition as its operation manner. Let [«jxiO, ) and j^ix(0,°o) and with the usual topology as its topology. be any two elements in S/R. Then Let HxfO.^+lylXiO.c^jar+yjXtO,«).

Since for any two positive real numbers x and let the vector addition be defined in and y the number x + y is unique, the S; i.e., operation defined on S is well-defined. This (*,$) + {x*, y*) = (x + x*,y + y*). operation is associative, because the operation of addition in the set of positive real numbers The set S with the vector addition is a is associative. Hence the set of equivalence semigroup. classes modii with the operation of addition We topologize the semigroup S with the is a semigroup. usual product topology P\ i.e., the family We define the natural mapping n from S of subsets onto S/R by assigning each element (x,y) to the equivalence class |ftj x(0,<»). We B = [(Ux V)\U,V are open in CO,»)] show that the mapping n is an algebraic homomorphism. Let (x,y) and (x*,y*) be is the base for the topology P in S. two arbitrary elements in S. Then n(x,y) We define a relation R on S as follows: = Mx(0,™) and «(a^y) = j«*|x(0,oo) for (x,y),(x*,y')eS, (x, y) R (**, y*) if and 71 [(a?, y) + (ar*, y*)] = \x + x* j X (0, «). and only if x — x*. It is easy to see that But this relation R is an equivalence relation, because the equation x = x* is reflexive, »(* ,l/)+n(x*, y*) = |a?|x(0 , «) + [a:*j symmetric, and transitive, We show that the equivalence relation R is also homomorphic. X(0,co)= ja; + | X (0 , 00) . Suppose that (a?! , y,) , (arf , yj) , (a>a , ya), Hence > l/v) 6 & such that n (x , y) + n (x*, y*) = n(x ,y) + (xy*). (KJ , y{) R (x2 , y2) and («J , y\) R («3 , y$). This shows that the mapping 11 is an abstract Then x = x and x\ = . From these t 2 homomorphism. equations we have Now we topologize the semigroup S/R with the quotient topology with respect to X) + x* = x3 + X2 . the mapping n. That is, a subset J7x(0,») Hence is open in S/R if and only if ii"1[i7x(0,oo)] is open in S. We observe that (a:, + x* , yx R (a-2 -f x\ , j'a + y'2), a-1 [*7X(0,=o)] = D"x(0,»). This means that the relation is a homomor- phic equivalence relation. Hence a subset i7x(0,«=) of S/R is open 24 GAZETA DE MATEMÁTICA 24 if and only if the U is open in the usual and jyj x (0, «>} be any two elements in topology of (0 , oo) . S/R such that If a subset UxV is open in S, then MX(0,<»)+|ifl><(0,«>)-|a: + y|X(0,oo). the subset Let Wx (0, oo) be an open neighborhood n-i[Z7x(0,<»)J = *7x(0,») of jj? + y\ x(0 , uu) . Then since the addition is continuous in the semigroup of positive is also open in S, Hence S satisfies the real numbers, for an open neighborhood W condition A. of x + y, there are open neighborhoods U We show that the natural mapping n from of x and V of y such that U+V^W. S onto S/R is continuous and open. Let Choose £7x(0,oo) as an open neighborhood Ux(0, oo) be an open set in S/R. Then of |a?jx(0,oo) and 7x(0,«) as an open [U~X(0,t»)] which equals i/x(0,co) is neighborhood of jyjxCOft»). open in S. Hence n is continuous. Now let Then UxV be an open subset of S. Then n{Ux.V)= i7x(0,co) is open in S/R Ux(0,oo) + Fx (0, oo) _ (U+ V) according to the observation of the last X{0,o°)cWX(0t