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.

Equalizer of idempotents

Exercise: If

f = m ∙ e

is an idempotent satisfying

f ∙ f = f

and

e ∙ m = 1_P

(where ‘∙’ denotes composition), then the section

m: P –> A

is an equalizer of the parallel pair of maps

f, 1_A: A –> A

(Conceptual Mathematics, p. 293).

Definition: An equalizer of a parallel pair of maps

f, g: A –> B

is a map

i: X –> A

satisfying

f ∙ i = g ∙ i

and every map

j: Y –> A

satisfying

f ∙ j = g ∙ j

is uniquely included in

i: X –> A

i.e. there is exactly one map

k: Y –> X

such that

j = i ∙ k

To show that

m: P –> A

is an equalizer of

f, 1_A: A –> A,

given

f = m ∙ e

f ∙ f = f

e ∙ m = 1_P,

we have to first show that

f ∙ m = 1_A ∙ m

i.e.

f ∙ m = m

f ∙ m = (m ∙ e) ∙ m = m ∙ (e ∙ m) = m ∙ 1_P = m

Next, we have to show that if any map

j: Y –> A

satisfies

f ∙ j = j

then there is exactly one map

k: Y –> P

satisfying

j = m ∙ k

First, we need a map from Y to P.  Given

j: Y –> A

and

e: A –> P

we can take the composite map

e ∙ j: Y –> A –> P

as the map k from Y to P i.e.

k = e ∙ j

and see if

m ∙ k = j

i.e. if

m ∙ (e ∙ j) = j

m ∙ (e ∙ j) = (m ∙ e) ∙ j = f ∙ j = j

(since we are given f ∙ j = j).

Finally, we have to show that

k: Y –> P

satisfying

m ∙ k = j

is unique i.e. if there is another map

k’: Y –> P

satisfying

m ∙ k’ = j

then

k’ = k

Given

m ∙ k’ = j = m ∙ k

post-composing with

e: A –> P

on both sides of

m ∙ k’ = m ∙ k

we find

e ∙ (m ∙ k’) = e ∙ (m ∙ k)

(e ∙ m) ∙ k’ = (e ∙ m) ∙ k

1_P ∙ k’ = 1_P ∙ k

k’ = k

Thus the section

m: P –> A

of the splitting of an idempotent

f = m ∙ e: A –> P –> A

is an equalizer of

f, 1_A: A –> A

Tailpiece: An equalizer of two idempotents

f, f’: A –> A

with a common section i.e.

f = m ∙ e

f’ = m ∙ e’

is the common section

m: P –> A

True or false? (For some reason, I’m not my usual verbose self :(  I’ll back soon to spice-up the story all the way to cohesion vs. quality of reflexive graphs vs. idempotents via points and pieces functors constructed in terms of equalizers and coequalizers.)