Exam (elaborations) TEST BANK FOR An Introduction to Category Theory By Harold Simmons [Solution Manual]-Converted

1.1 Categories defined 1.1.1 Not needed?  1.1.2 These examples are dealt with in Section 1.5.  1.2 Categories of structured sets 1.2.1 (c) Consider the function f(r) = r(a) which sends each r 2 N to the rth iterate of  applied to a. A simple calculation shows this is a Pno-arrow. A proof by induction shows this is the only possible arrow.  1.2.2 Consider a pair (A;X) f - (B; Y ) g - (C;Z) of such morphisms. We show the function composite g  f is also a morphism, that is x 2 X =) g(f(x)) 2 Z for each element x of A. The morphism property of f and then g gives x 2 X =) y = f(x) 2 Y =) g(f(x)) = g(y) 2 Z as required. This doesn’t yet prove we have a category, but the other requirements – arrow composition is associative, and there are identity arrows – are easy.  1.2.3 The appropriate notion of arrow (A;R) f - (B; S) is a function between the carrying sets such that (x; y) 2 R =) (f(x); f(y)) 2 S for all x; y 2 A. This generalizes the idea used in Pre and Pos.  1.2.4 Consider a pair of continuous maps R - S  - T between topological spaces. A simple calculation gives (  ) =   which is the required property. 1  2 1. Categories 1.2.5 Let R = C[A;A]. We have a binary operation  on R, namely arrow composition. This operation is associative (by one of the axioms of being a category). We also have a distinguished element idA of R, the identity arrow on A. This is the required unit. (Strictly speaking, this do not show that R is a monoid, for we don’t know that R is a set. There are some categories for which C[A;A] is so large it is not a set. This is rather weired but it shouldn’t worry us.)  1.2.6 To show that Pfn is a category we must at least show that composition of arrows is associative. Consider three composible partial functions A f - B g - C h - D X [ 6 f - Y [ 6 g - Z [ 6 h - as indicated. We must describe h  (g  f) (h  g)  f and show that they are the same. We need A g  f - C h - D A f - B h  g - D U [ 6 g  fjU - Y [ 6 h - X [ 6 f - V [ 6 h  gjV - where a 2 U () a 2 X and f(a) 2 Y b 2 V () b 2 Y and g(b) 2 Z for a 2 A and b 2 B. We also need A h  (g  f) - D A (h  g)  f - D L [ 6 h  (g  fjU)jL - R [ 6 h  (g  fjV )jR - where a 2 L () 8>< >: a 2 U and (g  fjU)(a) 2 Z 9>= >; a 2 R () 8>< >: a 2 X and f(a) 2 V 9>= >; for a 2 A. We show L = R and the two function composites are equal. For a 2 L we have a 2 U, so that fjU(a) = f(a). Thus, remembering the definition of U we have a 2 L () a 2 X and f(a) 2 Y and (g  fjU)(a) 2 Z for a 2 A. Remembering the definition of V we have f(a) 2 V () f(a) 2 Y and g(f(a)) 2 Z and hence a 2 R () a 2 X and f(a) 2 Y and (g  fjU)(a) 2 Z for a 2 A. This shows that L = R. 1.2. Categories of structured sets 3