6 !T# where T# is an epimorphism and <1l, 3>y = j'T' where r'
is an epimorphism. But since yX = (j2xj2xj2)6, it follows that
<'1f,3>yX = (2xj2)6 so that (j2xj2)0#T# = j'T'A. Hence the fol-
lowing diagram commutes.
(RxXR) C (XRxR) -> RoR
(RxX) n (XxR) XRxXR
T' j2Xj2
RoR -----------------'> XxX
1
Since T# is an epimorphism and j' is an extremal monomorphism, by
the diagonalizing property (0.19) there exists a unique morphism & such
that j'S = (j2xji2)# and CT# = T'X. But this says that
(RoR,(j2xj2)i#) < (RoR,j'). Since (R,j) is transitive (RoR,j') < (R,j)
hence (RoR,(j2xj 2)#) < (R,j) =: (R,(j2xj2)i). Hence there exists a mor-
phism o from RoR to R such that (j2xj2)io = (j2xj2)y1#. Again, j2xj2 is a
monomorphism so that ipo = p# which says that (RoR,4#I) < (R,y) hence
(R,y) is transitive.
3.7. Theorem. If (R,j) is a relation from X to Y and 7rlj is an epi-
morphism then (RoR-l,j#) is reflexive on X.
Proof. It will first be shown that the following diagram commutes.
RxX -------- ---> (XxY) xX
<1RKlj> 01
R -- -- > (RxX) r (XxR-l) 1 ;- XXYxX
< 2
XxR -------------------- Xx(YxX)
Consider the following products: (XxY,7Tl,r2), (XxYxX,T~1,7r2, 73),
(RxX,p1,o2), (XxR-l,pl*,p2*), ((XxY)xX,l*,'2*), (Xx(YxX), 1i 2.),
(YxX, 1, 12), and (XxX,-T1, 2).
Now,
1 (jxlX)= il1*(jxlx) r ljPl
2O1(jxl) = T27Tl*'(jx = 2" -
730 (jxl)1R, 1R ,3IJ> = T2*(jxlx)<1R,'1>J = ; 2< R iP> = Tr13
102(iXxj) = iI(Ixxj*)> = P1*Tj,T> = Tlj.
72 2(1XxjT)<1]jT> = (T 2(IXXj*) = ,IJ*j * =
^Tfj*' = 11j = r2J
B302(1Xxj*)<3)j.T> = 722(iXxj*)<' .,T> = 22j*T = 2<2,lI>j = 1J"
Thus by the definition of pr-'uct the diagram commutes. Hence
there exists a morphism E so that X1E = <1R,'Ij> and X2E = .
From the above it is easy to see that
1l< ,1T3>YE = TlY = 1Tlj = T = 3Y 2YE. Recall that (Ax,iX) is
the equalizer of T1 and T2 hence there exists a morphism such that
ix = <11l,73>YE.
Let (RoR ,j#) be the indicated composition of relations and let
T# denote that epimorphism for which j#T#/ = y. Thus, combining
the above results, <7Tij,>Tj> = YE -= j#T#E = iX.
Since (Ax,ix) and (X,) are isomorphic as extremal subobjects
of XxX (1.21), there exists an isomorphism X such that <1,1x>X = iX.
Consequently, Xt = iX = Y = .
Now 7ITX = l XX = X} = Xrl<7T J,7 rj> = 7rlj and by hypothesis
Trlj is an epimorphism; thus, since is an isomorphism,it follows that
( must be an epimorphism.
Thus has ((,iX) as its epi-extremal mono factcrization.
But this means that (Ax,ix) is the intersection of all extremal subob-
j.cts of XxX through which factors (0.21). Recall that
= j#t#/E, thus (A ,iv) < (RoR- ,ji#) which was to be proved.
3.8. Corollary. If (R,j) is symmetric on X then (RoR,ipV), the composi-
tion of (R,i) -,ith (R,.) on XR, is reflexive on XR.
Proof. Since (R,j) is symmetric on X then (Ri) is symmetric on XR (3.5)
hence (O,,) 1 (R-1. :*) (1.13). Referring to the diagram in (3.41) fol-
lowing the c0 'inition of the canonical embedding it is immediate that
pl is an e; -rphism since plI = Bkt- and each of B, k, and T1 is an
-1 ,.
epimorphisr. Thus (RoR ,1) (RoR,/#) is reflexive on XR (3.7).
3.9. Corollary. If (R,j) is a quasi-equivalence on X then (R,j) is an
equivalence relation if and only if lrlj is an epimorphism (respectively
if and only if r2j is an epimorphism).
Proof. If (R,j) is an equivalence relation then (R,j) is reflexive and a
quasi-equivalence. Thus by Proposition 1.24, 7rj and T2j are retractions
hence epimorphisms.
Conversely, if 1lj is an epimorphism then applying the theorem
-1
and Proposition 1.30, (A ,iX) < (RoR ,j#) < (RoR,j') < (R,j) so that
(R,j) is reflexive and hence is an equivalence relation. (If T2j is an
-1 -1 -1 -1
epimorphism then (Ax,i ) < (R oR,j#*) < (R oR ,j'*) < (R ,j*) and
-1
(R ,j*) E (R,j).)
3.10. Corollary. If (R,j) is-a quasi-equivalence on X then (Rd:) is an
equivalence relation on XR.
Proof. (R,i) is a quasi-equivalence on XR (3.6) and (RoR,ijI#) is reflexive
on XR (3.8). Thus (AXR, XR) < (RoR,y#) < (R,w) whence (R,',) is reflexive
and thus is an equivalence relation on XR.
3.11. Proposition. If (R,j) is a quasi-equivalence on X then (R,j) and
(RoR,j') are isomorphic relations on X.
Proof. By Corollary 3.10 (R,i) is an equivalence relation on XR whence
(RoR,ir#) and (R,,) are isomorphic relations on XR (1.32). Recall that
there exists a mcrphism E such that the following diagram commutes (3.6).
(RxXR)ri (XRxR) --------------;> RoR
T'X ( 2Xj 2)
RoR -;- XxX
j'
Thus (RoR,(j2xj2)i#) < (RoR,j'). But as mentioned above
(RoR,i#) (R,t) hence there exists an isomorphism A# such that
I^,#jt = So by the definition of the canonical embedding (3.4),
j'gX# = (j2xj2) #lx = (j2xj2)y = j. But this implies that
(R,j) < (RoR,j'). Thus since (R,j) is transitive, (R,j) (RoR,j')
which was to be proved.
3.12. Proposition. Let (R,j) be a relation on X. Then (R,j) < (AX,iX)
if and only if R is symmetric on X and (R,i) < (AXR,iXP).
Proof. If (R,j) < (AX,iX) then there exists a morphism a such that
j = iXa. Thus iTl = rliXa = T2ixa = 72j whence
n <,f2,Tl>j = T2j = IrFj = Tr2<'r2,Ttl>j. Thus by the definition of product
j = j. Consequently the epi-extremal mono factorization of
j is (1R,j) and so (R,j) (R-1,j*); i.e., (R,j) is symmetric.
Recall that j = (j2xj2)d (3.4). Thus
lj FTl(j2Xj2)' = j2p14 and 72j = 2(j2xj2)) = j2P2P'. But "lj = '2j
hence j2P = j2P2'. Since j2 is a monomorphism it follows that
pl -= p21,'. Recall that (AXR,iXR) is the equalizer of pI and p2. Hence
there exists a morphism; B such that iXR5 = 4j'. This implies that
(R,,) < (AXRiXR))
Conversely, if R is symmetric and (R,') < ('XR,iXR) then there
exists a morphism 8 such that d = ixRP; hence
P1 = P liXR = P2iXRB = P2. Since (j2xj2) = j, we have
1 j r "Q(j2xj2) = j 2P1 = J2P2 7T 2'(j 2xj 2) = F2j. Thus r1j = Tr2j
so that there exists a morphism a such that j = iX&. This means that
(R,j) <. (Ax, ix).
3.13. Definition. Let (R,j) be a relation on X. Then R is said to be
a circular relation if and only if RoR < R-1
This notion is due to MacLane and Birkhoff [20] exercisee 3,
page 14).
3.14. Proposition. Let (R,j) be a relation on X. Then R is a circular
relation if and only if R1 is a circular relation.
Proof. If R is circular then RoR < R-1. Thus
R-loR-1 = (RoR)-1 < (R-1)-1 R (1.38, 1.;2 and 1.11). Hence R-1 is
circular.
Conversely, if R1 is circular then by the above, (R-1)'1 =R is
circular.
3.15. Theorem. Let (R,j) be a relation on X. Then R is an equivalence
relation on X if and only if R is reflexive and circular.
Proof. If R is an equivalence relation then R is reflexive. Since R is
transitive and symmetric, RoRR R-' hence R is circular.
Conversely, if R is reflexive and circular then R-1 is reflexive
(1.17) and R-1 is circular (3.14). Hence
R R'1oA.. < '-loR1 < R (1.31 and 1,12) whence R is symmetric. Thus
R R-1 (1.11).
Now RoR R-1oR-1 < R hence R is transitive. Thus R is an equiva-
lence relation.
SECTION 4. IMAGES
4.1. Definition. Let (R,j) be a relation from X to Y and let (A,a) and
(B,b) be extremal subobjects of X and Y respectively. Consider
(Rir(AxY),y) and (Rfi(XxB),6). Let (T ,a) and (? .2) be the epi-extremil
1 2
mono factorizations of i y and r 6 respectively. Denote the domain of a
2 1
by AR and the domain of B by RB. Thus the following diagrams commute.
R
^ Y 2
axly
Rt(Axy) ----> Xxy c Y
2 AXY ARIrw,
R
R 1 '(XxB) >-.------ -- XxY -----> X
XWX ^ RB
4.2. Remark. Since (X,1 ) and (Y,I ) are extremal subobjects of X and Y
X Y
respectively, then (R '(XxY),y) (R,j) and (RA (XxY),6) 5 (R, j) whence
(XR,&) is precisely the extremal subobject (XR.,j) used in the canonical
nQboddiin (3.-'). Since X = Y in 3.4 then also (RY,3) is precisely
(RX,j ) used inl 3.4.
4.3. Example. In the category Set, for (A,a) < (X,1 ), (B,b) < (Y,1 )
and (R,j) < (XxY,1 ),
XxY
AR j{yY: tL exists acA such that (a,y)ER}
RB {xeX: there exists bcB such that (x,b)eR}.
This is easily seen since R(\(AxY) L {(a,y): acA, (a,y)eR} and
R(I(XxB) {(x,b): beB, (x,b)eR}, and AR is the set of all second terms
of elements of R((AxY) and RB is the set of all first terms of elements
of Rf(XxB).
In the category Top AR and RB have precisely the same underlying
sets as above. They are endowed with the subspace topology determined by
the topology of XxY.
In the category Top2, AR and RB have" precisely the same underlying
sets as in Top for it is easy to verify that AR and RB are closed sub-
sets of X and Y respectively. Recall that the image of a morphism in Top
is the closure of the set theoretic image (0.15).
4.4. Theorem. If (R,j) is a relation from X to Y and (A,a) is an extremal
subobject of X then (AR,a) and (R-1A,B) are isonorphic extremal subobjects
of Y.
Proof. Consider the following commutative diagrams.
R
A
RAxY) -------------- XxY --- -------------Y
AxY _axl,,
R1
6 iT
R- 1 (YxA) ------- -- YxX -'--Y--
1 xa
Y
YxA RA
j <7f iT>
j
R ------- Xx ------ YxX
R-1
It can be shown in a straightforward manner that
(ax] ) = (1 xa)
2 1 Y Y 21
where pl and p2 are the projections of AxY. Hence
(1 xa)
= (axl )X = y = j1 = j* .
Y 12 1 2 2 1 Y2 2 1 2 1 1 1
Thus by the definition of intersection there exists a morphism 5 such
that SS = <2 ,Tr >Y = j*TX (1 xa)X Hence
21 1 Y 2'1 2
f 6 = f Y = y = ai But fi 6 = Bi C. Thus, since (AR,a) is
1 1 21 2 1 1 2
the intersection of all extremal subobjects through which T.2y factors
(0.21), it follows that (AR,o) < (R-1A,B).
Similarly, it follows that
< >2'1 *X = -> 16 = j (axl,])2,-X whence -h4rc
<2' 1 2 1 3 (ax] )
exists a moiphisnm such that y'* = -16. Then
2 < ,T7 >-16 = f 6 = ST2 = y* = aT ;*. Again, since (R1A, 8) is the
ntrscton f all exremal subobjecs 2
intersection of all extremal suhobjects through which h ,1 6 factors,
1
(R-IA,B) < (AR,a). Consequently (R-IA,B) E (AR,a).
4.5. Corollary. If (R,j) is a relation from X to Y and (B,b) is an extre-
mal subobject of Y then (RB,B) and (BR-1,a) are isomorphic as extremal
subobjects of X.
Proof. Recall ((R-1)-1,j#) E (R,j) (1.11). Letting (R-l,j*) play the role
of (R,j) and (E,b) the role of (A,a) in the theorem, the following is
obtained: (BR-1,Q) ((R-)-1B,B/#) E (RB,B).
4.6. Corollary. If (R,j) is a symmetric relation on X and (A,a) is an
extremal subobject of X then (AR,a) and (RA,B) are isomorphic as extremal
subobjects of X. (In particular, (XR,j ) and (RX,j ) are isomorphic as
extremal subobjects of X as was shown directly in 3.4.)
Proof. Recall that (R-1,j*) E (R,j) (1.13). Hence by the theorem
(AR,a) (R-1A,B) = (RA, ).
4.7. Proposition. Let (Al,al) and (A2,a2) be extremal subobjects of X and
(R,j) be a relation from X to Y. If (Al,al) < (A2,a2) then
(A1R,al) < (A2R,a2).
Proof. By hypothesis there exists a morphism v so that a2P = a1. Thus,
there exists a morphism ( such that the following diagram commutes.
R('(A1xY) Y xy
A AlXYA axl xy
1 -Y 2
R (A,2xy) ------. .. .-- ..-X XxY
Thus w2Y1 = i2Y25 whence, because (A1R,al) is the intersection of
all extremal subobjects through which T2rY1 factors and r2y2C factors
through (A2R,a2), (AIR,al) < (A2R,a2) which was to be proved.
4.8. Proposition. Let (Bl,bl) and (B2,b2) be extremal subobjects of Y and
(R,j) be a relation from X to Y. If (Bl,bl) < (B2,b2) then
(RBi,B1) < (RB2,B2).
Proof. (RB1 ) 1*)E (B1R(-1,12) < (B212) (RB22) (4.5 and 4.7).
4.9. Proposition. Let (R,j) and (S,k) be relations from X to Y and (A,a)
be an extremal subobject of X. If (R,j) < (S,k) then (AR,a) < (AS,a).
Proof. In a manner similar to that in the proof of 4.7 one can establish
the existence of a morphism $ such that the following diagram commutes.
R ---------- S
j t k
RTA(AxY) S A(axy) --- Xxy
axi
Axy
Hence the following diagram commutes.
R /(Axy)
S 1 (AxY) ---------- XxY
Thus, since (AR,a) is the intersection of all extremal subobjects
through which yT, factors, and 2 factors through (AS,a), it follows that
(AR,cL) < (AS,5) which was to be proved.
4.10. Proposition. Let (R,j) and (S,k) be relations from X to Y and (B,b)
be an extremal subobject of Y. If (R,j) < (Sk) then
(RB,B) < (SBj).
Proof. (RB,8) (BR-1,B*) < (BS-1,~*) E (SB,8) (4.5, 1.12, and 4.9).
4.11. Proposition. Let (R,j) be a relation from X to Y and let (Al,al)
and (A2,a2) be extremal subobjects of X. Then
((AlfA^R:A2 ) <_ (AIRAA2R,5).
Proof. Since (AljnA2,a) < (A1,al) and (A1CfA2,a) < (A2,a2) it follows
that ((A,! A2)R,a) < (A1R,a1) and ((A(fA2)R,a) < (A2R,a2) (4.7). Thus
((AiCA2)R,a) < (A1RflA2R,Q) (1.19).
4.12. Proposition. Let (R,j) be a relation from X to Y and let (B.,bl)
and (B2,b2) be extremal subobjects of Y. Then
(P.(B1^ B2),B) < (RB1 nRB2,B).
Proof. (R(B nB2),B) ((B B2 )R-1,B*) < (BiR-lnB R-1,*)
(RB1nRB2,B) (4.5 and 4.11).
4.13. Proposition. Let (RI,jl) and (R2,32) be relations from X to Y and
let (A,a) be an extremal subobject of X. Then
(A(R1AR2),a) < (AR,( AR2, ).
Proof. It is clear that there exist morphisms 1, and C2 such that the
following diagram commutes.
R1 N(AxY) "Y
(r.,fl R2) (AxY) )x ----> xxY
'R2-A (Axy)
69
Thus rr2y = 1r2Yl1 = T2Y2 2. Again since (A(R1(\R2),a) is the inter-
section of all extreral subobjects through which ir2y factors it follows
that (A(R(.R2) ,) < (AR1,51) and (A(R1 FR2),a) (AR2,a2). Hence
(A(RI AR2),a) < (AR1 AAR2, ) (1.19).
4.14. Proposition. Let (R1,jl) and (R2,j2) be relations from X to Y and
let (B,b) be an extremal subobject of Y. Then
((R AR )B, ) < (RB 2R B,B).
% 1 2 1 2
Proof. ((RflR2)B,6) E (B(RI.R2))-1, *) < (BR3-~I BR-1 ,1) =
1 2 1 2 i 2
(R1 BR 2B,6) (4.5 and 4.13).
4.15. Proposition. Let (R,j) be a relation from X to Y then (R,j) and
(R1 (RYxY),y) are isomorphic as extremal subobjects of Xxy.
Proof. Consider the following commutative diagrams.
R 1) > Xxy
RY ------ XxY
T 1 T 1
RY ---
ji
RYxY ,
-
R fA(RY7Y) :>--------------------------->^ XxY
R
Since (jlxl ) = = j, there exists a morphism E
such that yv =- j
R
RYxy
.^^.. yxY YXiy
R n---( R ((RYxY) XXY
P, R
Thus (R,j) < (R n(RYxy),y). Clearly the reverse inequality holds
so that (R,j) E (R(T(RYxY),y).
4.16. Proposition. Let (R,j) be a relation from X to Y. Then (R,j) and
(Rr%(XxXR),6) are isomorphic relations from X to Y.
Proof. Analogous to the proof of 4.15.
4.17. Corollary. Let (R,j) be a relation from X to Y. Then (R,j) and
(Rt'(RY^-XR),6) are isomorphic relations from X to Y.
Proof. (R.j) : (RO(RYxY),v) E (R(l(XxXR),'6) (4.15 and 4.16). But since
(RY,jl) and (XR,j2) are extremal subobjects of X and Y respectively it
folJows that ((RYxY)A(Xx'R),a) E (RYxXR,). Thus
(R,j) E ((Rr;(RYxY))n(Rn(X\-XR)),B) E (RA(RYxXR),).
4.18. Proposition. Let (R,j) be a relation from X to Y and let (A,a) be
an extremal subobject of X. Then (AR,a) ((RYAlA)R,&).
Proof. It follows from Proposition 1.5 that RI(((RYAA)xY) and
RPf((RYxY)A(AxY)) are isomorphic relations from X to Y. By Proposition
4.15. (R,j) and (Ro(RYxY,y) are isomorphic relations from X to Y. Thus
RI~(AxY) and R.i((RYAA)>:Y) are isomorphic relations from X to Y. Conse-
quencly by the definition of image (4.1), (AR,a) and ((RYftA)R,&) are
isomorphic as extremal subobjects of Y.
4.19. Corollary. Let (R,j) be a relation from X to Y. Then ((RY)R,a) and
(XR,j,) are isomorphic ns extremal subobjects of Y.
Proof. Let (X,1,) play the role of (A,a) in 4.18.
4.20. Corollary. Let (R,j) be a relation from X to Y and let (B,b) be an
extremal subobject of Y. Then BR-1 and (Br.XR)R-1 are isomorphic as ex-
tremal subobjects of X.
Proof. Immediate.
4.21. Proposition. Let (R,j) be a relation from X to Y. Then (RoR-1,j#)
and (RoR-71,(RYxX),y) are isomorphic relations on X.
Proof. Consider the following diagram.
RYxX
CTix1
xljxl
X1 01
y
(RxX) ( (XxR-1) ---- ---.-, XxyxX ---> XxX
02
XxR-1 XX(YXX)
T xj*
RoR-1
To see the diagram is commutative it need only be observed that
(jxl)(Qrxl) = < lT3>01(jY1). To show this note that
(jxl1)(Qlxl) = (jl'-xl) = (Tljxl) and
aT Cl(jxl) = 1 1(jx) !jpl Ir r(jxl),
7r0 1(jxl) = 3C1(jxl) = p2 = T2(rrjxl).
Thus, since (RoR"1 ,j#) is the intersection of all extremal sub-
objects through which <'7"l,3>y factors,it follows that
(RoR-l,j#) < (RYxX,j xl). Whence (RoR-],j#) < (RoR-1fI (RYxX),?).
4.22. Theorem. Let (R,j) be a relation from X to Y. Then (RY,jl) and
((RoR-)X,B) arc isomorphic as extremal subobjects of X.
Proof. Consider the following products: (XxYxX, 1,2, 3),
(Xx(YxX),p1,'2), ((XxY)xX,Pl *,P*), (RxX,PlP2), and (XxR-1, 1, 2).
Referring to the diagram in the proof of 4.21 it is easy to see
that: irlY = 1( = (jxl)Aj = il*(xl)X = r1jp1X1. Thus
<1 'l~' 3 = "> jll = jlT 1 1l
RoR
(RoR- )x >>
o;'XxX
1
Since <7- ,I3>' Y = j#i//' and rTlj# 5, the following diagram
cc;Lmu:es.
-Cl p
iT#
(R (RoR-1)X
XI I
RY .--
But since has the diagonal property (0.19) and Tr# is an epi-
mcrphism and j is an extremal monomorphism then there exists a morphism
: such that ji = B and pl1?1 = T,. Thus ((RoR-1)X,B) < (RY,j).
Next it will be shown that the following diagram is commutative.
. .----.----- ----
jxl
RxX ---------- (XxY)xX
<1R'i l
R --- --X)R (RxX)l(XXR-1) .- -- xxyxX
2 02
XxR-1 Xx(YxX)
ixj*
= lP1*(jx1)<1Rlj> = UiP1
2P1 (jx 1)1 lj> = 2j 1~1R j> = 2j.
= 2P ':(jxl) = T2JPl = 2""
= P2*(jx)<1Rlj> = P2 = !j.
= \ 2(!xj*)< ,lj, > = IT J"2 = --j =j
= T2p2(]xj*) = 7T2j*2 = I2j*T =
T2< 2',1i j = 1j .
Consequently there exists a morphism * such that the above diagram
commutes and such that yS* = .
Thus <';1,lr,>Y$: = and hence the following diagram is
commutative.
(RxX),(R (KRx)--- -------- XxyX
.T I F l1
.______ (' ' X
S(RR 1)X R_----- --. X
R ----i;"R -~----'--"" '
1
2e01(jxl)<1Rnlj>
30 ] (jxl)
"1 2(lxj*)< l 3," >
,2 '02xj*)
i3O2 (lxj-)
74
Since (RY,j1) is the intersection of all extremal subobjects
through which irj factors, it follows' that (RY,j1) ((RoR-1)X,B). Thus
(RY,j1) ((RoR-1)X,B) which was to be proved.
4.23. Corollary. Let (R,j) be a relation from X to Y. Then (XR,j2) and
((R-loR)Y,8) are isomorphic as extremal subobjects of Y.
Proof. (XR,j2) (R-lX,j3) E (((R-1)o(R-1)-1)Y, ) = ((R-loR)Y,P)
(4.4, 1.11 and 4.22).
SECTION 5. UNIONS
5.1. Definition. If {(Ri,ji): icI} is a family of relations from X to Y
then let ((FRi,j) be the intersection of all relations (i.e., extremal
icI
subobjects of XxY) "containing" each (Ri,ji) (where containment is in the
sense of "factoring through" as noted in Remark 0.12). (' Ri,j) shall
ieI
be called the relation theoretic union of the family {(Ri,ji): iIl}.
5.2. Examples. In the category Set the relation theoretic union is the
usual set theoretic union together with the inclusion map.
In the category Top the relation theoretic union is the usual set
theoretic union endowed with the subspace topology determined by the top-
ology of XxY together with the inclusion map.
In the category Top2 the relation theoretic union is the closure
of the set theoretic union together with the inclusion map.
In the categories Grp and Ab the relation theoretic union is the
subgroup generated by the set theoretic union of the relations.
5.3. Proposition. Let {(Ri:ji): isl} te a family of relations from X to
Y, let (LRi,k) denote the usual categorical union of subobjects, let
iel
(o.j) be the epi-extremal mono factorization of k and let the codomain
of a (domain of j) be denoted R. Then r and ( JRi are isomorphic rela--
iel
tions from X to Y.
Proof. Since (~Ri,j) is the intersection of all extremal subobjects
igI
containing each (R.iji) and each (Rj,ji) < ( iRi,k) and
iel
(CjRlk) (Rj) and since j is an extiemal monomorphism then
icI
(UtRi,j) < (R,j).
icd
Since (U R ,k) is the intersection of all subobjects which "con-
isI
tain" each (Ri,ji) then (UR,k) < (\jR ,j). Since j is an extremal
isl ieI
monomorphism and (R,j) is the intersection of all extremal subobjects
which "contain" (U R,k) then (R,j) < (Q)R ,j). Thus
ieI iEI
(R,j) E (U-R ,j)..
ieI
5.4. Remark. Notice that by the definition of relation theoretic union,
if (R ,j1), (R2,j2), and (S,k) are relations from X to Y and if
(R!,Jl) (S,k) and (R2,j) < (S,k), then (P R2j) < (S,k) (cf. 1.19).
.5.5 Proposition. Let (R ,j ), (R ,j ), (Sk ) and (S ,k ) be relations
= 1 1 2' 2 1 1 2 2
from X to Y. If (R ,ji) < (R2 j2) and (Slkl) < (S2,k2) then
(R 1- S1,) < (R2JS2k).
Proof. (R ,j ) (Rj ) < (R2' S,k) and (S,k ) < (S2,k) < (R2 S2,k)
_of I ] 2,<2 -2 1' 1 < 2' 2) 2 2
whence (r ..s ,j) (R2 L S2 k) (5.4).
5.6. .ma.'.rk. The following proposition can be strengthened with the ad--
ditional hypothesis that the category has finite coproducts (5.34);
however. J i s included here because it is of interest in its own right.
5.7. Proposition. Let (R,j) be a relation from X to Y and let (S,k) and
(T,m) be relations from Y to Z. Then ((RoS) U (RoT),B) (Ro(S')T), ).
Proof. Consider the following commutative diagrams.
s ----__ k
^s '^^ 7; '~^"~----------
S
S T -----
S- ''"-"."-. Z
rPp
RxZ ---------------> (XxY)xZ
S7A jxZz
x 3 z
In ,\ - ~
(RxZ)[1Xx(SQ}T) >>
Xxyxz
-0
'4 \
Xx(S ) T) ----------- Xx(YxZ)
1XXa
RxZ )'>- (XxY)xZ
jxj1
x5
RxZ f XxS
A6
XxS *I -'- V XX(YxZ)
lxxk
RxZ >, ---- -(Xxy)xZ
7" jx1Z 31-
Y3
R'Z rlXxT
02
XxT >- Xx(YxZ)
1Xxm
Yl <1, 7T3>
RxZ rXx (S T) )---- --- --- XxyxZ -------- XxZ
I Ro(S T)
Y2
T2 RoS '' -
1401
>- IXyxYZ
r'-02
XxyxZ
r.
41-------------------- -----------------
~-------------------- ----------,
--
-.
R.xZfIXxT ->--- XxyxZ XxZ
Y3 < 1'f 3 3>
SRoT '
RoS _
(RoS) ) (RoT) ------ XxZ
X
RoT B3
Since (S,k) < (S( UT,a) and (T,m) < (S kT,a) it readily fol-
lows that ((RxZ)I}(XxS),y2) < ((RxZ)A(Xx(S J T)),y1) and that
((RxZ) n(XxT),Y3) < ((RxZ) n(Xx(S (QT)),yl). Thus there exist imorphis.ns
.! ard 2 such that Yl1, = Y2 and Y12 = Y3. Hence
<~3,'3>' 1 iI Y2 and YiS2 = < ,3>Y3