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).