Experiencing math (of course)

Consider a set

A = {-1, 1}

The absolute value of both -1 and 1 is 1, which we can think of as an endomap

α: A -> A

with

α(-1) = 1 and α(1) = 1

Calculating the absolute value of the absolute value of -1, we notice that it is equal to the absolute value of -1:

α(α(-1)) = α(-1)

We also know that

-1 x -1 = 1 and 1 x 1 = 1

which we can think of as a function (squaring)

f: A -> B

with A = {-1, 1} as above and B = {1}

and

f(-1) = 1 and f(1) = 1

Of course (this is not the ‘of course’ of the title), we can equip the set B with an endomap

β: B -> B

β(1) = 1 (we can think of the endomap β as taking absolute value)

Calculating the absolute value of the absolute value of 1 (in B), we notice that it is equal to the absolute value of 1:

β(β(1)) = β(1)

which looks like the equation

α(α(-1)) = α(-1)

So we say that the element 1 (in B) to which the element -1 (in A) is mapped to by the function (squaring)

f: (A, α: A -> A) -> (B, β: B -> B)

satisfies an equation

β(β(1)) = β(1)

which is ‘same’ as the one satisfied by -1 (in A):

α(α(-1)) = α(-1)

all of which reads like an instance of:

Now, of course, any map f: (A, α: A -> A) -> (B, β: B -> B) sends ‘a’ in A to elements f(a) = b in B satisfying the ‘same equations.’ —Conceptual Mathematics, page 184

And I have yet to feel the ‘of course’ in the above sentence… long ways to go!

I thought of the title thanks to Vicky Nanjappa’s Experiential Journalism; Thanks Vicky!

(Note that the result of squaring after taking absolute value [of elements -1, 1 in the set A] equals that of taking absolute value after squaring [the elements in A] i.e.

square after absolute = absolute after square

fα(-1) = βf(-1)

and

fα(1) = βf(1)

in view of which we state the equality of the two composite functions (fα, βf):

fα = βf

which means that our function (squaring)

f: (A, α: A -> A) -> (B, β: B -> B)

is indeed a map in the category of endomaps.)

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