ñòð. 30 
?
f
?
? ? ?
 
M (s) Â£ M (t) N0 (s) Â£ N0 (t) N (s) Â£ N (t)
Solutions for section 1.3 85
certain functions have been labeled as being surjective or injective (although oth
ers are and some are bijective). These use the facts that the product of two injec
tive functions is injective and the product of two surjective functions is surjective.
That f is surjective follows from the fact that the composite of two surjective
functions is surjective and if the composite of two functions is surjective, so is the
second one. Dually, the fact that f is injective uses the facts that the composite
of two injective functions is injective and that if the composite of two functions
is injective, the Â¯rst one is.
The dual argument, using the fact that a sum of injectives is injective and a
sum of surjectives is surjective, gives the corresponding result for discrete cocones.
It is worth noting that these arguments fail if either the cones or cocones
fail to be discrete or if the category in which the models are taken is other than
the category of sets. The reason is that even in the category of sets, the arrow
induced between equalizers of epis is not necessarily epic and between coequalizers
of monos will not be monic. (Equalizers and coequalizers are deÂ¯ned in chapter 8.)
In categories other than sets, even the sum of epics will not generally be epic, nor
the sums of monics monic.
b. There are two ways of doing this. One is to show that the intersection of
submodels is a submodel. Then the intersection of all submodels is clearly the
smallest submodel. Another is to take the image of the initial term model in the
component of that model. Let M be the initial term model, N is the given model
and N0 this submodel. If N1 Âµ N is any other submodel, there is an initial term
model M 0 in the component of N1 that has an arrow M 0 Â¡ N1. But N1 Âµ N and
!
0
so is in the same component as N . Thus M = M and the map M Â¡ N1 Âµ N is !
the original map M Â¡ N . Since that factors through N1, it follows that N0 Âµ N1 .
!
Solutions for Chapter 2
Section 2.1
1. Add a new unary operation i : c1 Â¡ c1 and the following diagrams:
!
c2 c2
Â¡ @ Â¡ @
id Â¡ hid; ii @ iÂ¡ hi; idi @
i id
Â¡ @ Â¡ @
Â¡
Âª @
R Â¡
Âª @
R
? ?
c1 Â¾ p  c1 c1 Â¾ p  c1
c2 c2
p p
1 2 1 2
86 Solutions for section 2.3
hid; ii hi; idi
 
c1 c2 c1 c2
@ @
id@ id@
c c
@ @
R
@ ? R
@ ?
c1 c1
2. Let (C0; C1 ; s; t; u; c) and (D0 ; D1 ; s; t; u; c) be two models of the sketch for
categories. Note that we have used the common convention of using the same
letter to stand for the arrow in the sketch and in the model (in every model, in
fact). A homomorphism consists of an arrow F0 : C0 Â¡ D0 and an arrow F1 : C1
!
Â¡ D1 that satisfy some conditions forced by the fact that a homomorphism is
!
a natural transformation. First oÂ®, we have s Â± F1 = F0 Â± s, t Â± F1 = F0 Â± t and
u Â± F0 = F1 Â± u. These mean that the homomorphism preserves source, target and
identity arrows. These identities induce a unique arrow F2 : C2 Â¡ D2 such that
!
p1 Â± F2 = F1 Â± p1 and p2 Â± F2 = F1 Â± p2 . We further suppose that c Â± F2 = F1 Â± c.
These conditions mean that the homomorphism preserves composition; thus the
homomorphism is a functor.
Section 2.2
1. Referring to Diagram (ES 2.1), we see that incl is the arrow opposite 6
Â± true : 1
Â¡ b in a cone, which means that in a model it is the arrow opposite an arrow
!
from 1. But every arrow from 1 is monic (see Section 2.8, Exercise 7). Hence incl
is monic.
Section 2.3
1. It is the set of all diagrams (not necessarily commutative) of the form
f
A B
@
g
@
h
R
@?
C
2. It is the set of all commutative diagrams of the form
f
A B
@
g
@ h
R
@?
C
3. For each object C of , let M 0(c) = f(x; c) j x 2 M (c)g. If t : c Â¡ d is an
!
arrow in , let t(x; c) = (t(x); d). Then the Â¯rst projection is an isomorphism
M 0 Â¡ M and it is clear that M (c) \ M (d) = ; for c 6d.
! =
Solutions for section 3.1 87
Solutions for Chapter 3
Section 3.1
1. A model of is a sketch homomorphisms Â¡ ! . This assigns to the single
node of an object E and that is all. If we denote the single object of by
w, then the functor M 7 M (e) is evidently an isomorphism between the objects
!
and those of Mod( ; ). If Â® : M Â¡ N is a natural transformation, the
!
of
Â®e : M (e) Â¡ N (e) is an arrow of and is subject to no conditions. Thus means
!
that Nat(M; N ) = Hom(M (e); N (e)) and so the functor is an isomorphism.
Section 3.2
1. Nat has one such node, denoted 1, as does Stack. The disjoint union has two
such nodes, but when the coequalizer is formed they are identiÂ¯ed to a single
node in . This node, and only this, must become a singleton in any model.
Section 3.3
1. U Â£ U is induced by the sketch homomorphism from to the sketch for
monoids (given in 7.2.1 and 7.3.2) that takes e to s Â£ s.
2. For any u : s1 Â¡ s2 in
! , we have F (u) : F (s1) Â¡ F (s2) which gives a
!
commutative square

M (F (u))
M (F (s1 )) M (F (s2 ))
ñòð. 30 