Unique up to Unique Isomorphism

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

2 x 1 = 2

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 · 1_A = f

(where ‘·’ denotes composition; Conceptual Mathematics, p. 17), answer that the identity map

1_A: A –> A

is one such map.  Let us now consider a pair of maps

p_A: P –> A, p_B: 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 (p_A, p_B) give the pair (p_A, p_B)?  Once again, we find that the identity map

1_P: P –> P

is one such map satisfying both

p_A · 1_P = p_A and p_B · 1_P = p_B

Now, if

P, p_A: P –> A, p_B: 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

q_A: Q –> A, q_B: Q –> B

to the common domain P of the product projections

p_A: P –> A, p_B: P –> B

which satisfies both

p_A · q = q_A and p_B · q = q_B

Since the pair of product projections

p_A: P –> A, p_B: 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

p_A · p = p_A and p_B · p = p_B

But we already know, from the identity law we talked about a min ago, of a map

1_P: P –> P

which satisfies both

p_A · 1_P = p_A and p_B · 1_P = p_B

So, if there’s going to be only one map

p: P –> P

satisfying both

p_A · p = p_A and p_B · p = p_B

then that one map must be

1_P: 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, p_A: P –> A, p_B: P –> B) and (Q, q_A: Q –> A, q_B: Q –> B) are both products of A and B, then the unique map f: P –> Q which satisfies both q_A · f = p_A and q_B · f = p_B is an isomorphism.

If

(Q, q_A: Q –> A, q_B: Q –> B)

is a product of A and B, then, given any pair of maps to the factors

p_A: P –> A, p_B: P –> B

with a common domain P, there is exactly one map

f: P –> Q

which satisfies both

q_A · f = p_A and q_B · f = p_B

Now, if

(P, p_A: P –> A, p_B: P –> B)

is a product of A and B, then, given any pair of maps to the factors

q_A: Q –> A, q_B: Q –> B

with a common domain Q, there is exactly one map

g: Q –> P

which satisfies both

p_A · g = q_A and p_B · g = q_B

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

p_A: P –> A, p_B: P –> B

to get a pair of maps

p_A · (g · f): P –> A, p_B · (g · f): P –> B

Let’s see what these maps are:

p_A · (g · f) = (p_A · g) · f = q_A · f = p_A

p_B · (g · f) = (p_B · g) · f = q_B · f = p_B

Since there is only one map

1_P: P –> P

which satisfies both

p_A · 1_P = p_A

p_B · 1_P = p_B

the above endomap (g · f) must be the identity 1_P.  Similar reasoning reveals that the other composite (f · g) is the identity 1_Q.  Thus the product (P, p_A: P –> A, p_B: P –> B) of A and B is unique up to unique isomorphism.

Why does 1 x 2 = 2?

Reason being the method of mathematics, we can safely substitute ‘how’ for ‘why’ and look at how we calculate products.  Calculating the product of, say, two numbers

1 and 2

is, in essence, selecting a number

2

from all the numbers, which immediately raises the question: how do we select (Conceptual Mathematics, p. 292)?  If I were to select a shirt to buy, then I’d be looking at sizes; and if I were to select a wine to drink, then I’d be tasting to select.  The ‘how’ of selecting, thus, depends on the ‘what’ that’s being selected.  So, what is product (Conceptual Mathematics, pp. 216-7)?

A product of two objects

A and B

(in a category C) is an object

P

structured by two projection maps

p_A: P –> A, p_B: P –> B

(to the two factors A, B) such that given any pair of maps

f_A: F –> A, f_B: F –> B

(to the factors A, B from a common domain object F), there is exactly one map

f: F –> P

(to the product object P) satisfying both

p_A · f = f_A and p_B · f = f_B

(where ‘·’ denotes composition).  Now, using the definition of product, let us calculate the product of two objects in, say, the category of sets.  The product of two sets

A and B

is a set

P

structured by two projection maps

p_A: P –> A and p_B: P –> B

such that for every pair of functions

f_A: F –> A and f_B: F –> B

there is exactly one function

f: F –> P

(to the product set P) satisfying both

p_A · f = f_A and p_B · f = f_B

The key to calculating products is the 1-1 correspondence in the definition of product:

for each pair of functions

f_A: F –> A and f_B: F –> B

there is exactly one function

f: F –> P

which we summarize as

F –> P

———————————

F –> A, F –> B

Zooming out for a moment, all that there is to a set is its elements; and elements of a set A are functions from 1 (= {•}) to the set A (Conceptual Mathematics, pp. 230-1 and 256-8).  So if we know all the functions

1 –> P

from 1 to the product set P, then we know P.  But, how are we going to find all the functions

1 –> P

especially when we do not know what P is.  We can use the 1-1 correspondence

F –> P

———————————

F –> A, F –> B

to find all the functions from 1 to P, and, in turn, the product set P along with the projection maps p_A and p_B as follows.

Consider two sets

A = {a} and B = {b₁, b₂}

The product of the given two factor sets A and B is a set

A x B

structured by two projection maps

p_A: A x B –> A and p_B: A x B –> B

Now we have to find out the product set A x B, along with its two projection maps p_A and p_B.  Since any set is completely determined by its elements, we need to know the elements of the product set

A x B

What are the elements of the product set A x B?  Elements of a set are functions to the set from a one-element set 1 = {•}.  So we have a complete specification of the product set A x B once we have all the functions to A x B from 1 i.e.

1 –> A x B

Since A x B is a product set, functions to A x B are in 1-1 correspondence with pairs of functions to the factors A, B.  For each ordered pair of functions to the factors from a common domain F i.e. for each pair of functions

f_A: F –> A, f_B: F –> B

there is exactly one function

f: F –> A x B

from F to the product set A x B, which when composed with the projection functions

p_A: A x B –> A, p_B: A x B –> B

gives the pair of functions we started with i.e.

p_A · f = f_A, p_B · f = f_B

Taking F = 1, we have the 1-1 correspondence

1 –> A x B

——————————————

1 –> A, 1 –> B

between ordered pairs of elements of the factors A, B and the elements of the product set A x B.  Thus, corresponding to the ordered pair of elements of the factors A, B

(element a of the set A, element b₁ of the set B)

we have the element

(a, b₁)

of the product set

A x B

Similarly, corresponding to the other ordered pair of elements in the factors A, B

(a: 1 –> A, b₂: 1 –> B)

we have the element

(a, b₂): 1 –> A x B

in the product set A x B.  Since

(a: 1 –> A, b₁: 1 –> B)

and

(a: 1 –> A, b₂: 1 –> B)

are all the ordered pairs of elements of the factors A, B, we have two elements

(a, b₁): 1 –> A x B

and

(a, b₂): 1 –> A x B

in the product set A x B i.e.

A x B = {(a, b₁), (a, b₂)}

Next, we need the projection maps

p_A: A x B –> A, p_B: A x B –> B

which can also be calculated using the definition of product:

for each ordered pair of functions to the factors A, B from a common domain F i.e. for each pair of functions

f_A: F –> A, f_B: F –> B

there is exactly one function

f: F –> A x B

from F to the product set A x B, which when composed with the projection functions p_A, p_B gives the pair of functions we started with i.e.

p_A · f = f_A

p_B · f = f_B

With F = 1, we have

p_A · (a, b₁) = a and p_A · (a, b₂) = a

p_B · (a, b₁) = b₁ and p_B · (a, b₂) = b₂

(where ‘·’ denotes composition of functions).  Since

(a, b₁) and (a, b₂)

are all the elements of the product set

A x B

and since the value of the projection maps

p_A: A x B –> A, p_B: A x B –> B

at each one of the elements

1 –> A x B

of the product set A x B is given by

A x B –> A · 1 –> A x B = 1 –> A

A x B –> B · 1 –> A x B = 1 –> B

(where ‘·’ denotes composition of functions), we have complete specification of the two projection maps p_A, p_B structuring the product set

A x B = {(a, b₁), (a, b₂)}

as

p_A (a, b₁) = a and p_A (a, b₂) = a

p_B (a, b₁) = b₁ and p_B (a, b₂) = b₂

Summing up, using the definition of product, we can calculate the product (of the given objects) along with its projection maps, provided we know the basic shapes of the objects (cf. elements or 1-shaped figures in the case of sets).  However, there is an additional step in the case of more structured objects such as graphs, wherein we need to figure out the incidence relations between basic shapes (dot, arrow) of the product graph, which can also be done using the definition of product (Conceptual Mathematics, pp. 273-4).