ñòð. 32 |

4. Let f : C Â¡ C 0 and let (x; C) lie over C. DeÂ¯ne Â·(f; (x; C)) to be

!

(idF f (x); f ) : (x; C) Â¡ (F f (x); C 0 )

!

Let (u; g) : (x; C) Â¡ (y; C 00) so that u : F g(x) Â¡ y. Let k : C 0 Â¡ C 00 satisfy

! ! !

0 00

k Â± f = g. Then (u; k) : (F f (x); C ) Â¡ (y; C ) because the domain of u is F g(x) =

!

F (k Â± f )(x) = F (k)(F f (x)). Furthermore,

(u; k) Â± (idF f (x) ; f) = (u Â± F (idF f(x) ); k Â± f) = (u; g)

as required by OA{2. It follows as in the answer to the second problem that (u; k)

is the only possible arrow with this property.

5.

((t; m)(t0 ; m0 ))(t00 ; m00) = (tÂ®(m; t0); mm0)(t00; m00 )

= (tÂ®(m; t0)Â®(mm0; t00 ); mm0 m00 )

and

(t; m)((t0; m0)(t00; m00 )) = (t; m)(t; Â®(m0; t00 ); m0m00)

= (tÂ®(m; t0Â®(m0 ; t00 )); mm0m00)

However, by MA{4,

Â®(m; t0)Â®(mm0; t00 ) = Â®(m; t0)Â®(m; Â®(m0 ; t00 ))

and that is Â®(m; t0 Â®(m0 ; t00 )) by MA|2.

6. Let Â¯ take an object x of F (C) to (x; C) and an arrow u to (u; idC ). By GC{3,

if v : y Â¡ z then

!

(v; idC ) Â± (u; idC ) = (v Â± F (idC )(u); idC Â± idC ) = (v Â± u; idC )

so Â¯ preserves composition. It is clearly a bijection and preserves identities.

Solutions for section 4.3 91

Section 4.3

1. Suppose that Â® : F Â¡ G is a natural transformation. GÂ® preserves identities

!

because GÂ®(idX ; idC ) = Â®C(idX ; idC ) = (idÂ®C(x) ; idC ). Suppose that (u; f) : (x; C)

Â¡ (x0 ; C 0) and (u0 ; f 0 ) : (x0 ; C 0) Â¡ (x00; C 00). Then

! !

(u0 ; f 0 ) Â± (u; f ) = (u0 Â± F f 0(u); f 0 Â± f )

by GC{3 (Section ES 4.2). On the other hand,

GÂ®(u0; f 0) Â± GÂ®(u; f ) = (Â®C 00 u0; f 0) Â± (Â®C 0u; f )

(Â®C 00 u0 Â± Gf 0(Â®C 0 u); f 0 Â± f )

=

(Â®C 00 u0 Â± Â®C 00(F f 0 (u)); f 0 Â± f )

=

(Â®C 00 (u0 Â± F f 0 (u)); f 0 Â± f)

=

GÂ®(u0 Â± F f 0(u); f 0 Â± f )

=

where the third equality uses the naturality of Â® and the fourth uses the fact that

Â®C 00 is a functor. Thus GÂ® preserves composition.

2. GR{2 implies that G preserves identity natural transformations, since any

component of an identity transformation is an identity arrow. Let Â® : F Â¡ F 0

!

0 00 0 00

and Â¯ : F Â¡ F be natural transformations, where F , F and F are functors

!

to Cat. Then for (u; f ) : (x; C) Â¡ (x; C 0 ),

!

from

G(Â¯ Â± Â®)(u; f ) = ((Â¯ Â± Â®)C 0 u; f ) = (Â¯C 0 (Â®C 0 u); f )

and

(GÂ¯(GÂ®))(u; f ) = GÂ¯(Â®C 0 u; f ) = (Â¯C 0 (Â®C 0 u); f )

so G preserves composition.

Section 4.4

1. If and are monoids, then each has only one object. By WP{1, an object

G

and P : G(A) Â¡ !

of wr is a pair (A; P ) where A is the only object of

is a functor. But if has only one object and G(A) is discrete, there is only one

functor from G(A) to { it must take all the objects of G(A) to the only object

G

of . Hence wr has only one object, so is a monoid.

2. Let and be groups. Since they have only one object each, we can simplify

the notation in WP{1 through WP{3 and omit mention of the objects A, A0 and

A00. The value of G at the only object of is a category we will call . If f is

an element of the group , then Gf is an automorphism of the category . An

! . An arrow (f; Â¸) : P Â¡ P 0 consists

wrG is a functor P : Â¡ !

object of

and a natural transformation Â¸ : P Â¡ P 0 Â± Gf .

!

of an element f of the group

Each component of Â¸ is an element of the group so has an inverse; thus Â¸ is

92 Solutions for section 5.1

an invertible natural transformation. Since f is also a group element, it has an

inverse f Â¡1. Let Â¹ be the natural transformation whose component at an object

is (Â¸(Gf )Â¡1 (X))Â¡1 . Then the inverse of the arrow (f; Â¸) is (F Â¡1; Â¹). To

X of

verify this, we calculate

(F Â¡1 ; Â¹) Â± (f; Â¸) = (f Â¡1 Â± f; Â¹Gf Â± Â¸)

Now F Â¡1 Â± f is the identity of and for an object X of ,

(Â¹Gf Â± Â¸)X = Â¹(Gf (X )) Â± Â¸X

= Â¸((Gf )Â¡1 (Gf(X)))Â¡1 Â± Â¸X

= (Â¸X)Â¡1 Â± Â¸X = idX

so (f; Â¸) is invertible.

Solutions for Chapter 5

Section 5.1

1. Let the pullbacks be

g0 - g1 -

0 0

C0 C0 C1 C1

0 0

f0 f1

f0 f1

? ? ? ?

-C -C

C0 C0

g g

Let u : C0 Â¡ C1 and v : C1 Â¡ C0 be the inverse isomorphisms such that f0 Â± v =

! !

0

f1 and f1 Â± u = f0 . Then f1 Â± u Â± g0 = f0 Â± g0 = g Â± f0 so by the universal mapping

property of the right hand pullback, there is a unique u0 : C0 Â¡ C1 such that

0

!0

g1 Â± u0 = f1 Â± u and f1 Â± u0 = f0 . Similarly, there is a unique v 0 : C1 Â¡ C0 such

0 0 0

!0

that g0 Â± v 0 = f0 Â± v and f0 Â± v 0 = f1 . Thus f0 and f1 belong to the same subobject.

0 0 0 0

!

2. For f : C0 )Â¡ C, the square

f-

C0 C

idC0 idC

? ?

-C

C0

f

is a pullback. Thus the function induced by the identity of C assigns to each

subobject of C the subobject itself (or an equivalent one).

Solutions for section 5.1 93

3. If S and T are Â¯nite, so is the set of functions between them. In fact if we let

#S denote the number of elements, then #[T Â¡ S] = #(S)#(T ) . For if T = ;,

ñòð. 32 |