®F (s1) ®F (s2 )
? ?
 N (F (s2 ))
N (F (s1))
N (F (u))
which is the same as
F ¤ (M )u

¤
F ¤ (M )(s2 )
F (M )(s1 )
F ¤ (®)s2 F ¤ (N )u
? ?

¤ ¤
F (N )(s1 ) F (N )(s2 )
F ¤ (®)s2
which is naturality of F ¤ (M ).
that (id )¤ (M ) = M ± id = M and for ® : M
¡
!
3. We have for id :
¡ N , (id )¤ (®) = ® id = ®. Thus (id )¤ is the identity functor. If G : R
!
is another homomorphism of sketches, then (F ± G)¤ (M ) = M ± F ± G =
¡!
G¤ (M ± F ) = G¤ (F ¤ (M )) = (G¤ ± F ¤ )(M ) and similarly for ® : M ¡ N . Thus
!
(F ± G)¤ = G¤ ± F ¤ which shows that Mod (¡) is a contravariant functor.
88 Solutions for section 4.1
4. Suppose, say, that ¾ is a diagram. The other two possibilities are similar.
Suppose ¾ says that
s1 ± ¢ ¢ ¢ ± sn = t1 ± ¢ ¢ ¢ ± tm
Then ¾ is satis¯ed in F ¤ (M ) if and only if
F ¤ (M )(s1 ) ± ¢ ¢ ¢ ± F ¤ (M )(sn ) = F ¤ (M )(t1 ) ± ¢ ¢ ¢ ± F ¤ (M )(tm )
which is the same as
M (F (s1 )) ± ¢ ¢ ¢ ± M (F (sn )) = M (F (t1 )) ± ¢ ¢ ¢ ± M (F (tm ))
which is the same as the condition that M satisfy F (¾).
Solutions for Chapter 4
Section 4.1
1. Let C be the ¯ber over an object C. If X is an object of C , then P (X) = C
and P (idX ) = idC by de¯nition of functor It follows that idX is an arrow of C .
If f : X ¡ Y and g : Y ¡ Z are arrows of C , then P (f ) = P (g) = idC , so
! !
P (g ± f ) = P (g) ± P (f) = idC ± idC = idC , so g ± f is an arrow of C . Hence C is
a subcategory.
2. Let P : ¡ ! be a ¯bration with cleavage °. To see that F f is a functor,
note ¯rst that °(f; Y ) ± idY = idY ± °(f; Y ) so that F f(idY ) = idF fY . Suppose
u0 : Y 0 ¡ Y 00 . Then
!
°(f; Y 00 ) ± F f (u0 ) ± F f (u) = u0 ± °(f; Y 0) ± F f (u) = u0 ± u ± °(f; Y )
so by uniqueness F f (u0 ± u) = F f (u0) ± F f (u).
Now we show that F is a contravariant functor. If we show that F (idC ) is
the identity on arrows it will have to be the identity on objects because functors
preserve source and target. Let C be an object of and u an arrow of F (C). Then
F (idC )(u) is the unique arrow for which °(idC ; Y 0 ) ± F (idC )(u) = u ± °(idC ; Y ). By
SC{1, this requirement becomes idY 0 ± F (idC )(u) = u ± idY . Thus F (idC )(u) = u
as required.
Now suppose f : C ¡ D and g : D ¡ E in . Let u : Y ¡ Y 0 in F (C).
! ! !
Then F (g ± f )(u) must be the unique arrow for which
°(g ± f; Y 0) ± F (g ± f)(u) = u ± °(g ± f; Y )
But
u ± °(g ± f; Y ) = u ± °(g; Y ) ± °(f; F f (Y ))
= °(g; Y 0) ± F g(u) ± °(f; F f (Y ))
= °(g; Y 0) ± °(f; F f(Y 0)) ± F f(F g(u))
so by CA{2 and SC{2, F (g ± f ) = F (f ) ± F (g).
Solutions for section 4.1 89
3. a. We will also denote the functor by Á. It is a ¯bration if for every element
x of Z2 there is an element u of Z4 such that Á(u) = x (that is CA{1) and for
every v 2 Z4 and h 2 Z2 such that x + h = Á(v) (mod 2), there is a unique w 2 Z4
for which Á(w) = h and u + w = v (mod 4) (which is CA{2). This amounts to
saying that Á is surjective and every equation m + y = n (mod 4) can be solved
uniquely for y, which requires simple case checking (or knowing that Z2 and Z4
are groups). An analogous proof works for op¯bration.
b. A splitting would be a monoid homomorphism (this follows immediately
from SC{1 and SC{2) which would make Á a split epimorphism, which it is not
by Exercise 2 of Section 2.9.
4. a. P preserves source and target by de¯nition. Let (h; k) : f ¡ f 0 and (h0; k0 ) :
!
f 0 ¡ f 00 . Then
!
P ((h0; k0 ) ± (h; k)) = P (h0 ± h; k0 ± k) = k0 ± k = P (h0 ; k0 ) ± P (h; k)
Thus P preserves composition. The identity on an object (A; B) is (idA ; idB ), so
it follows immediately that P preserves identities.
b. Let f : C ¡ D in
! and let k : B ¡ D be an object of
! lying over D.
Let °(f; k) be the arrow (u; f ) of de¯ned in ES 4.1.5. P (°(f; k)) = f so CA{1
0
is satis¯ed. As for CA{2, let (v; v ) : z ¡ k in
! and let h be an arrow of such
0
that f ± h = v . Let w be the unique arrow given by the pullback property for
which u ± w = v and u0 ± w = h ± z. The last equation says that (w; h) is an arrow
of , and (u; f) ± (w; h) = (u ± w; f ± h) = (v; v 0 ) as required. The uniqueness
property of the pullback means that (w; h) is the only such arrow.
Section 4.2
1. The identity arrow for (x; C) is (x; idC ). Let (x; f ) : (x; C) ¡ (x0; C 0), (x0 ; f 0 ) :
!
0 0 00 00 00 00 00 00 000 000
(x ; C ) ¡ (x ; C ) and (x ; f ) : (x ; C ) ¡ (x ; C ) be arrows of G0 ( ; F ).
! !
00 00 ± 0 0 00 00 ±
f 0 ) ± (x; f ) = (x00; (f 00 ± f 0) ± f ) =
Then ((x ; f ) (x ; f )) ± (x; f ) = (x ; f
(x00 ; f 00 ± (f 0 ± f )) = (x00; f 00 ) ± ((x0 ; f 0 ) ± (x; f )) so composition is associative.
2. Let f : C ¡ C 0 in . Let (x; C) be an object of G0( ; F ) lying over C. De¯ne
!
·(f; (x; C)) to be (x; f ) : (x; C) ¡ (x0; C 0) where x0 = F f(x). Now suppose (x; g) :
!
(x; C) ¡ (y; C ) and suppose k : C 0 ¡ C 00 has the property that k ± f = g. The
00
! !
arrow required by OA{2 is (x ; k) : (x0 ; C 0) ¡ (y; C 00). This is well de¯ned since
0
!
F k(x ) = F k(F f (x)) = F (k ± f )(x) = F g(x) which must be y since (y; C 00) is
0
the given target of (x; g). Moreover it satis¯es OA{2 since (x0 ; k) ± (x; f) = (x; k ±
f) = (x; g). An arrow satisfying OA{2 must lie over k and have source (x0; C 0) and
target (y; C 00) so that it can compose with f to give g, so (x0; k) is the only possible
such arrow. To see that · is a splitting, suppose that (x0 ; f 0) : (x0 ; C 0 ) ¡ (x00 ; C 00 ).
!
0 0 0 0 0 0 0± 0± 0±
Then ·(f ; (x ; C )) = (x ; f ) and (x ; f ) (x; f) = (x; f f ) = ·(f f; (x; C))
as required. The veri¯cation for identities is even easier.
90 Solutions for section 4.2
3. The identity arrow is (idx ; idC ) : (x; C) ¡ (x; C). It is well de¯ned since
!
F (idC )(x) = x, so idx : F (idC )(x) ¡ x. Suppose that (u; f) is an arrow from (x; C)
!
to (x0 ; C 0), so that u : F f (x) ¡ x0. Similarly, let (u0; f 0 ) : (x0 ; C 0 ) ¡ (x00 ; C 00 ) and
! !
(u00 ; f 00 ) : (x00 ; C 00) ¡ (x000 ; C 000 ). Then
!
((u00; f 00 ) ± (u0 ; f 0 )) ± (u; f ) = (u00; F f 00(u0 ); f 00 ± f 0) ± (u; f )
= (u00 ± F f 00(u0 ) ± F (f 00 ± f 0)(u); (f 00 ± f 0 ) ± f)
and
(u00 ; f 00 ) ± ((u0 ; f 0) ± (u; f )) = (u00 ; f 00 ) ± (u0 ± F f (u); f 0 ± f )
= (u00 ± F f 00(u0 ± F f 0(u)); f 00 ± (f 0 ± f ))
The result follow from the facts that F and F f 00 are functors and composition in