Every morphism f: X –> Y in a category C gives rise to a morphism in the category of Z-structures, by the associative law.
Let’s start with what we are given: associative law. This seemingly simple law from elementary school
(1 + 2) + 3 = 1 + (2 + 3)
is all that we need to do the exercise (all exercises, for that matter; see Conceptual Mathematics, p. 136 and p. 371).
The composite of three composable morphisms such as
A – p –> B – x –> X – f –> Y
can be calculated in two ways:
1. first calculate the composite of morphisms p and x, and then the composite of xp and f
2. first calculate the composite of morphisms x and f, and then the composite of p and fx
The associative law states that these two calculations give the same result, which is expressed as
f (xp) = (fx) p
All we have to do, in order to do the exercise, is turn the associate law
f (xp) = (fx) p
into the commutative diagram
xp –> fxp
^ ^
| |
x –> fx
Let’s start at the bottom-left corner
x: B –> X
which when pre-composed with
p: A –> B
takes us to the top-left corner
xp: A –> X
which when post-composed with
f: X –> Y
takes us to the top-right corner
fxp: A –> Y
Of course, we could have taken the other route i.e. start (as earlier) at the bottom-left corner
x: B –> X
which when post-composed with
f: X –> Y
takes us to the bottom-right corner
fx: B –> Y
which when pre-composed with
p: A –> B
takes us to the top-right corner
fxp: A –> Y
So, to sum up, reading the commutative diagram
xp –post–> fxp
^ ^
| |
pre pre
| |
x –post–> fx
satisfying
post o pre = pre o post
(where ‘o’ denotes composition) into the associative law
f (xp) = (fx) p
satisfied by
A – p –> B – x –> X – f –> Y
is all that’s needed to do the exercise.
YOU: Just do it!
OK. What we need to deliver is a category of Z-structures, and that’s starters. But, what on wordpress is Z-structure? First, let’s make explicit something that’s implicit. Z is a small family of objects and morphisms of the category C. The only condition on the small family Z is that if a morphism of C is in the small family, then both domain and codomain of the morphism are in the family. Let’s take
A – p –> B
as our small family Z i.e. our family Z consists of two objects (A, B) and one morphism (p). Every object X of the category C gives rise to a Z-structure, which has as many component sets as the number of objects in Z and as many structural functions (between these component sets) as the number of morphisms in Z. Since there are two objects (A, B) in our family Z, we have two component sets:
1. A-th component set A(X) is the set of all A-shaped figures in X (i.e. A –> X)
2. B-th component set B(X) is the set of all B-shaped figures in X (i.e. B –> X)
corresponding to the two objects (A, B). Since there is one morphism
p: A –> B
in Z, we have one structural function
p(X): A(X) <– B(X)
(note the opposite direction of p(X) compared to p) assigning to each element
x: B –> X
in the domain set B(X) an element in the codomain set A(X). But, which one?
p(X)(x) = xp
Summing up, what we have so far is the following diagram
A(X)
^
|
p(X)
|
B(X)
of two component sets and one structural function constituting a Z-structure denoted Z(X), which the object X (of the category C) gave rise to. Along similar lines, object Y (of C) gives rise to another Z-structure Z(Y) i.e.
A(Y)
^
|
p(Y)
|
B(Y)
Now we have to show that a morphism
f: X –> Y
in the category C gives rise to a morphism
Z(f): Z(X) –> Z(Y)
in the category of Z-structures i.e. a morphism from the object Z(X)
A(X)
^
|
p(X)
|
B(X)
to the object Z(Y)
A(Y)
^
|
p(Y)
|
B(Y)
Putting it all together, we need a pair of functions
f(A): A(X) –> A(Y)
f(B): B(X) –> B(Y)
making the diagram
A(X) –f(A)–> A(Y)
^ ^
| |
p(X) p(Y)
| |
B(X) –f(B)–> B(Y)
commute i.e. satisfy
p(Y) o f(B) = f(A) o p(X)
If that’s not enough headache we need to show that the morphism
f: A –> B
gives rise to these functions
f(A): A(X) –> A(Y)
f(B): B(X) –> B(Y)
This added headache turns out to be the solution in the following sense. Take the case of
f(A): A(X) –> A(Y)
which has to assign to each element
xp: A –> X
of the domain set A(X) an element of the codomain set A(Y), but which one? The solution is to post-compose with
f: A –> B
i.e. define the function
f(A): A(X) –> A(Y)
as
f(A)(xp) = f (xp)
Similarly, we define the function
f(B): B(X) –> B(Y)
as
f(B)(x) = fx
With
f(A): A(X) –> A(Y)
f(B): B(X) –> B(Y)
defined as above, the diagram
A(X) –f(A)–> A(Y)
^ ^
| |
p(X) p(Y)
| |
B(X) –f(B)–> B(Y)
does commute i.e. satisfies
p(Y) o f(B) = f(A) o p(X)
To be more explicit, let’s start at the bottom-left corner with a B-shaped figure in X
x: B –> X
i.e. an element of B(X).
LHS: p(Y) o f(B)
p(Y) o f(B) (x) = p(Y) (fx) = (fx) p
RHS: f(A) o p(X)
f(A) o p(X) (x) = f(A) (xp) = f (xp)
If
(fx) p = f (xp)
then the above diagram commutes. Fortunately the associative law
(fx) p = f (xp)
says just that. In other words the commutativity of the diagram (morphism of Z-structures)
A(X) –f(A)–> A(Y)
^ ^
| |
p(X) p(Y)
| |
B(X) –f(B)–> B(Y)
induced by the morphism
f: X –> Y
(of the category C) follows from the associativity of composition of morphisms
A – p –> B – x –> X – f –> Y
YOU: What if the family Z has more (or less) than the one morphism
p: A –> B
that we considered?
In general the Z-structure has a component set for each object and a structural function for each morphism in Z. These component sets and structural functions (that an object X of C gave rise to) together constitute an object of the category of Z-structures. A morphism in the category of Z-structures has as many component functions as the number of component sets (which is same as the number of objects in Z). All these component functions together are required to satisfy, in order to constitute a morphism, as many equations as the number of structural functions (which is same as the number of morphisms in Z). Morphisms (in so satisfying the commutativity equations) preserve, while transforming, all the structure of objects of the category of Z-structures.
Cutting to confession: we did not do the exercise at the level of abstraction that the exercise called for. However, I hope, working out the case of a particular small family of two objects and one morphism gave a feel for the calculations involved in general.
YOU: But, what does STRUCTURE and SMALL FAMILY have to do with cognition?
Where to begin…
abstraction: Particulars –> Generals
to be continued…
Science of Knowing 