TEST BANK FOR An Introduction to Category Theory By Harold Simmons

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