This is a story about how ‘a’ gets to be ‘the’. Mathematics is, in a sense, about understanding something known until all traces of surprise are extinguished (yes, I don’t like surprises ;)
We all know that
and if we are asked ‘what, if any, are the numbers which when multiplied with 2 give 2?’ we readily answer: 1. In a similar vein, if we are given a map
f: A –> B
and asked ‘what, if any, are the maps which when pre-composed with f give f?’ we, going by the (first or left) identity law of composition of maps
f · 1A = f
(where ‘·’ denotes composition; Conceptual Mathematics, p. 17), answer that the identity map
1A: A –> A
is one such map. Let us now consider a pair of maps
pA: P –> A, pB: P –> B
(with a common domain P). What, if any, are the maps to the common domain P which upon pre-composition with the pair of maps (pA, pB) give the pair (pA, pB)? Once again, we find that the identity map
1P: P –> P
is one such map satisfying both
pA · 1P = pA and pB · 1P = pB
Now, if
P, pA: P –> A, pB: P –> B
is the product of A and B, then, going by the definition of product (Conceptual Mathematics, p. 217), there is exactly one map
q: Q –> P
from the common domain Q of any pair of maps to the factors A, B
qA: Q –> A, qB: Q –> B
to the common domain P of the product projections
pA: P –> A, pB: P –> B
which satisfies both
pA · q = qA and pB · q = qB
Since the pair of product projections
pA: P –> A, pB: P –> B
qualifies as ‘any pair of maps to the factors A, B’ in the above definition of product, there is exactly one map
p: P –> P
which satisfies both
pA · p = pA and pB · p = pB
But we already know, from the identity law we talked about a min ago, of a map
1P: P –> P
which satisfies both
pA · 1P = pA and pB · 1P = pB
So, if there’s going to be only one map
p: P –> P
satisfying both
pA · p = pA and pB · p = pB
then that one map must be
1P: P –> P
This is the key to the following exercise in which we are asked to show that products are unique up to unique isomorphism.
Exercise 12 (Conceptual Mathematics, p. 217):
If (P, pA: P –> A, pB: P –> B) and (Q, qA: Q –> A, qB: Q –> B) are both products of A and B, then the unique map f: P –> Q which satisfies both qA · f = pA and qB · f = pB is an isomorphism.
If
(Q, qA: Q –> A, qB: Q –> B)
is a product of A and B, then, given any pair of maps to the factors
pA: P –> A, pB: P –> B
with a common domain P, there is exactly one map
f: P –> Q
which satisfies both
qA · f = pA and qB · f = pB
Now, if
(P, pA: P –> A, pB: P –> B)
is a product of A and B, then, given any pair of maps to the factors
qA: Q –> A, qB: Q –> B
with a common domain Q, there is exactly one map
g: Q –> P
which satisfies both
pA · g = qA and pB · g = qB
Next, composing these two maps we get two endomaps
g · f: P –> Q –> P
and
f · g: Q –> P –> Q
Let’s look at the endomap on P
g · f: P –> P
which we can compose with the pair of projection maps
pA: P –> A, pB: P –> B
to get a pair of maps
pA · (g · f): P –> A, pB · (g · f): P –> B
Let’s see what these maps are:
pA · (g · f) = (pA · g) · f = qA · f = pA
pB · (g · f) = (pB · g) · f = qB · f = pB
Since there is only one map
1P: P –> P
which satisfies both
pA · 1P = pA
pB · 1P = pB
the above endomap (g · f) must be the identity 1P. Similar reasoning reveals that the other composite (f · g) is the identity 1Q. Thus the product (P, pA: P –> A, pB: P –> B) of A and B is unique up to unique isomorphism.
Science of Knowing 