By Mac Lane S. (Ed)

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7 holds as before. 8 N considers a N-allowable morphism h:[Q,M]~P~N between proper N-shapes, with [Q,M] h itself has the form q(

@ i). 1) not constant, form ~(<~> ® i); under s u p p l e m e n t a r y change, >M hypotheses, and with F h of the one is to show that The previous proof applies w i t h o u t since h cannot be of type string when its domain includes [Q,M]. 4 N. theorem, already formu- We are to show that two N - a l l o w a b l e - morphisms h, h':T >S 27 - of the same N-graph between proper equal.

26 ~:T T and S, there is an a l g o r i t h m ~S is N-allowable. By the results of §6, we need only test when it is N-constructible. This we do by the a l g o r i t h m of [I] plus the evident finite test that a graph is a string with no blocks. 4 N. and T proper, to N-shapes without change. is essential If h:T ~S The to is N - a l l o w a b l e in K, with S constant then T is constant. This is proved trivial because inductively as before, with one new case, the codomain of a string is never constant.

We omit the analogous case in w h i c h we should use the n a t u r a l i t y of~. ii) We use the n a t u r a l i t y of 6 in the c o n s t r u c t i o n given by - 57 - x ' ( y + z) © > x ' y + x'z a = x ( y + z) xy + xz We o m i t the a n a l o g o u s of 6 and cases in w h i c h we s h o u l d use the n a t u r a l i t y 7 S u p p o s e t h a t C is c o h e r e n t , there are r a p p e l s . k, p, is no o c c u r r e n c e that a ~ > b is a p a t h in w h o s e of n and t h a t a Then there exists a commutative s u c h t h a t a' ~, ~', > xy + xz 6#.

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