An easy way to get rid of racism, casteism, etc. is to make it hard to practice it.  You see I can discriminate against you only after separating you, which is impossible if everybody belongs to the same race.  Let’s see what this means.  We have a bunch of people, say,

P = {you, me, her, him}

Race is a property that people have i.e. a function

race: P –> R

on P with values in the set of races R.  I can discriminate against you based on your race only if

race (you)

is different from that of one or another in P.  But, for that to happen there must be more than one race in the set of races R.  If the set of races were a singleton set

1 = {human}

then the property

race: P –> 1

takes the same value

human

at every person in P i.e.

race (you) = human

race (me) = human

race (her) = human

race (him) = human

That’s enough saving-the-world for today ;)

What we need to take back from this excursion is functions such as

cP: P –> 1

with singleton set 1 = {•} as codomain.  These functions when viewed as properties on P cannot be used to tell apart elements of the domain set P since all the elements of P take the same value

which is the only element in the codomain set 1.

As you know there is a universe called the category of functions F where functions such as the above

cP: P –> 1

reside along with other functions such as

f: X –> Y

i.e. functions whose codomain is not necessarily a singleton set.  These functions are the objects of the category of functions F.  Then there are morphisms from one object

f: X –> Y

to another object

f’: X’ –> Y’

A morphism from

Y

^

|

f

|

X

to

Y’

^

|

f’

|

X’

is a pair of functions

p: X –> X’

q: Y –> Y’

making the diagram

Y         — q –>             Y’

^                                  ^

|                                   |

f                                   f’

|                                   |

X         — p –>             X’

commute i.e. satisfying

q o f = f’ o p

where o denotes composition.

Then there is the category of sets S whose objects are sets such as

X

and whose morphisms are functions such as

f: X –> Y

Now imagine a process which assigns to each object (set)

X

in the category of sets S the object (terminal set 1-valued function)

1

^

|

cX

|

X

in the category of functions F.

Since there are morphisms (functions)

f: X –> Y

in addition to objects in the category of sets, we have to do something about them.  Let’s send each morphism

f: X –> Y

in S to the commutative square

1          — 11 –>            1

^                                  ^

|                                   |

cX                                 cY

|                                   |

X         — f –>              Y

with

11 o cX = cY o f

Is this a functor

C: S –> F

from the category of sets S to the category of functions F?

Let’s check.

C: S –> F

is a pair of functions

COb: SOb –> FOb

CMp: SMp –> FMp

assigning objects (functions in FOb) to objects (sets in SOb) and morphisms (commutative squares in FMp) to morphisms (functions in SMp) in a way respectful of domain, codomain, identity, and composition as follows:

1. Respecting domain:

SMp      — CMp –>         FMp

|                                   |

domS                           domF

|                                   |

v                                  v

SOb       — COb –>         FOb

with

COb o domS = domF o CMp

where

domS: SMp –> SOb

assigns to each morphism (function)

f: X –> Y

in the category of sets S its domain object (set)

X

and

domF: FMp –> FOb

assigns to each morphism (commutative square)

Y         — q –>             Y’

^                                  ^

|                                   |

f                                   f’

|                                   |

X         — p –>             X’

in the category of functions F its domain object (function)

Y

^

|

f

|

X

2. Respecting codomain:

SMp      — CMp –>         FMp

|                                   |

codS                             codF

|                                   |

v                                  v

SOb       — COb –>         FOb

with

COb o codS = codF o CMp

where

codS: SMp –> SOb

assigns to each morphism (function)

f: X –> Y

in the category of sets S its codomain object (set)

Y

and

codF: FMp –> FOb

assigns to each morphism (commutative square)

Y         — q –>             Y’

^                                  ^

|                                   |

f                                   f’

|                                   |

X         — p –>             X’

in the category of functions F its codomain object (function)

Y’

^

|

f’

|

X’

3. Respecting identity:

SMp      — CMp –>         FMp

^                                  ^

|                                   |

idS                               idF

|                                   |

SOb       — COb –>         FOb

with

CMp o idS = idF o COb

where

idS: SOb –> SMp

assigns to each object (set)

X

in the category of set S its identity morphism (function)

1X: X –> X

and

idF: FOb –> FMp

assigns to each object (function)

Y

^

|

f

|

X

in the category of functions F its identity morphism (commutative square)

Y         — 1Y –>           Y

^                                  ^

|                                   |

f                                   f

|                                   |

X         — 1X –>           X

4. Respecting composition:

CMp (g o f) = CMp (g) * CMp (f)

which says that the value of the functor at the composite of two morphisms

CMp (g o f)

(note that o denotes composition in the category of sets S) is equal to the composite of the functor values at the two morphisms

CMp (g) * CMp (f)

(note that * denotes composition in the category of functions F).

So if

C: S –> F

with

COb (X) =

1

^

|

cX

|

X

and

CMp (f: X –> Y) = (11 o cX = cY o f) =

1          — 11 –>            1

^                                  ^

|                                   |

cX                                 cY

|                                   |

X         — f –>              Y

wants to be a functor, then it must satisfy

COb o domS = domF o CMp

COb o codS = codF o CMp

CMp o idS = idF o COb

CMp (g o f) = CMp (g) * CMp (f)

COb o domS = domF o CMp

COb o domS (f: X –> Y) = COb (X) = cX: X –> 1

domF o CMp (f: X –> Y) = domF (11 o cX = cY o f) = cX: X –> 1

Next we check ‘respecting codomain’:

COb o codS = codF o CMp

COb o codS (f: X –> Y) = COb (Y) = cY: Y –> 1

codF o CMp (f: X –> Y) = codF (11 o cX = cY o f) = cY: Y –> 1

Next we check ‘respecting identity’:

CMp o idS = idF o COb

CMp o idS (X) = CMp (1X: X –> X) = (11 o cX = cX o 1X) =

1          — 11 –>            1

^                                  ^

|                                   |

cX                                 cX

|                                   |

X         — 1X –>           X

idF o COb (X) = idF (cX: X –> 1) = (11 o cX = cX o 1X) =

1          — 11 –>            1

^                                  ^

|                                   |

cX                                 cX

|                                   |

X         — 1X –>           X

Finally we check ‘respecting composition’:

CMp (g o f) = CMp (g) * CMp (f)

CMp (f: X –> Y) = (11 o cX = cY o f) =

1          — 11 –>            1

^                                  ^

|                                   |

cX                                 cY

|                                   |

X         — f –>              Y

CMp (g: Y –> Z) = (11 o cY = cZ o g) =

1          — 11 –>            1

^                                  ^

|                                   |

cY                                 cZ

|                                   |

Y         — g –>             Z

CMp (g) * CMp (f) = (11 o cX = cZ o g o f) =

1          — 11 –>            1

^                                  ^

|                                   |

cX                                 cZ

|                                   |

X         — gf –>            Z

CMp (g o f) = CMp (gf: X –> Z) = (11 o cX = cZ o g o f) =

1          — 11 –>            1

^                                  ^

|                                   |

cX                                 cZ

|                                   |

X         — gf –>            Z

Thank god, we can now say we have a functor

C: S –> F

sending each set

X

in S to the function

cX: X –> 1

in F and each function

f: X –> Y

in S to the commutative square (11 o cX = cY o f) =

1          — 11 –>            1

^                                  ^

|                                   |

cX                                 cY

|                                   |

X         — f –>              Y

NO!

I didn’t even get to say what it is that we are trying to do here…

Fine!!  What is it?

The functor

C: S –> F

is right adjoint to the functor

P: F –> S

sending each function

f: X –> Y

in the category of functions F to the set

X

in the category of sets S and each commutative square (q o f = f’ o p) =

Y         — q –>             Y’

^                                  ^

|                                   |

f                                   f’

|                                   |

X         — p –>             X’

in F to the function

p: X –> X’

in S.

The functor

p: F –> S

is (just in case you haven’t noticed) none other than the functor

points: F –> S

we looked at in the context of

discrete –| points

where the functor

discrete: S –> F

sends each set

X

in S to its identity function

1X: X –> X

in F.  The identity function

1X: X –> X

which when thought of as a property on X, is such that no two elements of X are alike (going by the values of the property 1X) somewhat like the unique selves each one of us are.  At the other extreme is the property

cX: X –> 1

based on which (i.e. on the values of cX all of which are one and the only one in 1) no two elements of X can be distinguished (cf. the one and only human race to which we all belong).  Let us call the functor

C: S –> F

sending each set

X

in S to the function

cX: X –> 1

in F, to contrast it with the discrete functor, codiscrete.  With terminology in place, our task now is show that

codiscrete: S –> F

is right adjoint to

points: F –> S

To say

points –| codiscrete

means, in terms of figures and inclusions (Conceptual Mathematics, pp. 372 – 7), given an object

X

in domain category of sets

S

of the right adjoint

codiscrete: S –> F,

every figure

points (f) –> X

in X (whose shape

points (f)

is given as a value of the left adjoint

points: F –> S

at an object

f: A –> B

in the category of functions

F)

is uniquely included in

points (codiscrete (X)) –> X

More explicitly, there is a unique

f –> codiscrete (X)

such that

points (f) –> X = points (f –> codiscrete (X)) o points (codiscrete (X)) –> X

Let’s start: given

f: A –> B

we find

points (f) = A

next, given

X

we find

codiscrete (X) = cX: X –> 1

With these we have to show that

A –> X = points (f –> cX) o points (cX) –> X

Since

points (cX: X –> 1) = X

we have to show

A –> X = points (f –> cX) o X –> X

But what is

points (f –> cX)

The functor

points: F –> S

sends the commutative square (f –> cX)

B         –>       1

^                      ^

|                       |

f                       cX

|                       |

A         –>       X

in F to the function

A –> X

in S, i.e.

points (f –> cX) = A –> X

Now what we have to show boils down to

A –> X = A –> X o X –> X

Taking the endomap on X as identity

1X: X –> X

we have

A –> X = A –> X o X – 1X –> X

= A –> X

So we get to say

codiscrete: S –> F

is right adjoint to

points: F –> S

We have already seen (in an earlier post) that

points: F –> S

is right adjoint to

discrete: S –> F

Putting all three together

discrete –| points –| codiscrete

we arrive at Cantor’s genius abstracting (from Menge) sets of lauter Einsen that are

distinct but indistinguishable

which have been confusing mathematicians until Lawvere clarified it as the mathematical structure of

OK!  But what about the functor

pieces: F –> S

Yes, we have the adjoint string

pieces –| discrete –| points –| codiscrete

relating the category of sets S to the category of functions F.  When we look at the functor

discrete: S –> F

in the middle of the adjoint string

pieces –| discrete –| points

we notice that

discrete: S –> F

is the common section of the functors

pieces: F –> S

and

points: F –> S

This is in contrast to the UIAO of

discrete –| points –| codiscrete

where the middle functor

points: F –> S

is the common retract of the functors

discrete: S –> F

and

codiscrete: S –> F

I don’t know about you but I’m dead tired… the original idea:

study science in general and mathematics in particular to know how we know

is beginning to feel lot like a dream aeons away to walk into and yet too real to walk away ;) or is it :(

Ramayana, I remember reading somewhere, is one of those stories nobody in India can remember when they heard it for the first time.  In case you never heard, here it is:

A = {Ravana, Rama}

B = {Sita}

There is one onto function

love: A → B

from A to B with

love (Ravana) = Sita and love (Rama) = Sita

which is our way of saying that both Ravana and Rama are in love with the one and only Sita.  As is often the case with triangular love stories, Sita does not like Ravana but for some reason likes Rama i.e. there are two 1-1 functions

like: B → A

with

like (Sita) = Rama

and

hate (Sita) = Ravana

We can, with these two sections (1-1 functions) along with their common retract (onto function), sum up Ramayana in three arrows

A

^          |             ^

|           love         |

like          |        hate

|           v             |

B

which is what the mathematical structure of unity and identity of adjoint opposites (UIAO) is all about (or that’s UIAO according to Granny ;)

Of course, there’s more: beginning at

B

we could go to

A

via either

like: B → A

or

hate: B → A

and then from A return to B via

love: A → B

Either way we get the identity on B:

love º like = 1B: B → B

love º hate = 1B: B → B

where ‘º’ (read ‘after’) denotes composition.

Yes, of course, we could have started at

A

and gone to

B

via

love: A → B

and then returned to

A

via either

like: B → A

or

hate: B → A

Either way we get idempotents (two different ones) on A:

like º love = e0: A → A

hate º love = e1: A → A

I don’t know about you but I’m getting distracted with all this talk about like n love, so let’s loose them.  Composing the idempotents

e0: A → A

e1: A → A

we find

e0 º e0 = e0

e0 º e1 = e0

e1 º e0 = e1

e1 º e1 = e1

The first and the last equations (of the above four) are the familiar

e º e = e

from the definition of idempotent.  However the summation of all four as

ej º ei = ej

is something to sleep with.  One not so thoughtful question is if the above equation (reproduced below)

ej º ei = ej

holds true for all pairs of idempotents.

I don’t know about all but I can think of a structure in which the opposite i.e.

ej º ei = ei

is true (Exercise 15, page 145 of Conceptual Mathematics, see also pp. 192 – 4).  That structure is in a sense opposite to our UIAO

A

^          |           ^

|           r           |

s1         |           s2

|           v          |

B

where we have a pair of sections

s1: B → A

s2: B → A

with a common retract

r: A → B

That structure (i.e. the structure opposite to the above UAIO) is

A

|           ^          |

r1         |           r2

|           s           |

v          |           v

B

which is the structure of reflexive graphs (Conceptual Mathematics, page 145) consisting of a pair of retracts

r1: A → B

r2: A → B

with a common section

s: B → A

Consider, as an illustration of the reflexive graph structure, a set of arrows

A = {l, a, l’}

and a set of dots

D = {d, d’}

which we imagine as follows:

arrow l is a loop starting and ending at dot d

arrow a starts at d and ends at d

arrow l’ is another loop at d

Thus imagined graph can be modeled in the category of sets as a pair of retracts

s: A → D

t: A → D

with a common section

p: D → A

The function

s: A → D

with

s (l) = d, s (a) = d, and s (l’) = d’

assigns to each arrow in the set of arrows A its source dot in the set of dots D.

The function

t: A → D

with

t (l) = d, t (a) = d’, and t (l’) = d’

assigns to each arrow in A its target dot in D.

The function

p: D → A

with

p (d) = l and p (d’) = l’

assigns to each dot in D its preferred loop (an arrow) in A.

Now of course we can [pre-]compose the common preferred loop function

p: D → A

with the source, target functions

s: A → D

t: A → D

to get identity on the set of dots D i.e.

s º p = 1D: D → D

t º p = 1D: D → D

We might as well check to see if we indeed have identity:

s º p (d) = s (l) = d                               t º p (d) = t (l) = d

s º p (d’) = s (l’) = d’                           t º p (d’) = t (l’) = d’

With this we have everything we need to say what a reflexive graph is:

A

|           ^          |

s           |           t

|           p          |

v          |           v

D

satisfying

s º p = 1D

t º p = 1D

If we [post-]compose the common preferred loop function

p: D → A

with the source, target functions

s: A → D

t: A → D

we get two idempotents on the set of arrows A i.e.

p º s = e0: A → A

p º t = e1: A → A

The above two idempotents, as you would expect, satisfy:

e0 º e0 = e0

e1 º e1 = e1

We might as well check:

e0 º e0 = p º s º p º s

= p º 1D º s

= p º s

= e0

Next:

e1 º e1 = p º t º p º t

= p º 1D º t

= p º t

= e1

e0 º e1 = ?

e1 º e0 = ?

Let’s see

e0 º e1 = p º s º p º t

= p º 1D º t

= p º t

= e1

Next we calculate

e1 º e0 = p º t º p º s

= p º 1D º s

= p º s

= e0

Collecting all four

e0 º e0 = e0

e0 º e1 = e1

e1 º e0 = e0

e1 º e1 = e1

and summing up as

ej º ei = ei

we find, if nothing else, how idempotents formed of common section (above reflexive graphs) differ from those formed of common retract (UIAO with which we started), which satisfy

ej º ei = ej

In any case, now that we have a reflexive graph

A

|           ^          |

s           |           t

|           p          |

v          |           v

D

satisfying

s º p = 1D

and

t º p = 1D

we might as well think of the category of reflexive graphs (which has reflexive graphs as objects).  The moment we hear category, we think of morphisms.  What is a morphism in the category of reflexive graphs?

First, note that the structure (Conceptual Mathematics, pp. 149 – 51) of reflexive graphs consists of two component sets

A, D

and three component maps

s: A → D

p: D → A

t: A → D

which means that a morphism of reflexive graphs involves two maps of sets satisfying three equations.  More clearly, a morphism

f: G → G’

from a reflexive graph G

GA

|           ^          |

sG         |           tG

|           pG        |

v          |           v

GD

to a reflexive graph G’

G’A

|           ^          |

sG’        |           tG’

|           pG’        |

v          |           v

G’D

is a pair of set maps

fA: GA → G’A

fD: GD → G’D

satisfying three equations

fD º sG = sG’ º fA

fA º pG = pG’ º fD

fD º tG = tG’ º fA

Before we go any further, how does this category of reflexive graphs differ from the not too distant category of irreflexive graphs?  Looking at irreflexive graphs

A

|           |

s           t

|           |

v          v

D

and comparing them with reflexive graphs

A

|           ^          |

s           |           t

|           p          |

v          |           v

D

satisfying

s º p = 1D and t º p = 1D

we notice three differences:

2 structural component maps vs. 3 structural component maps

parallel [pair of] structural maps vs. opposed [pairs of] structural maps

no equation to constrain structural maps vs. 2 equations constraining structural maps

Fine!  Are these visible differences between irreflexive graphs and reflexive graphs telling of any differences in their behavioral repertoire???

Let me go find out…

Happy Ambedkar Jayanti!

All that there is is way too big to wrap around in warm embrace… wait, what am I thinking ;)  I meant to say that even though reality is seemingly much too much to grasp and go, there seem to be within the vastness of out-there, few small things—things reflective of the cosmic scheme of things—suitable for domesticating the galactic wilderness.  Pictures (in visual depictions) and words (in verbal descriptions) exemplify the resourcefulness of small things.  Thanks to the goodness of small things, we get to know.

Now, if only this working method of knowing were to bootstrap into a model of knowing reflecting

1. how we think about the things that are

and

1. how we make things that we think of

All that’s needed i.e.

and

is there in scientific knowledge.  If only the procedural knowledge about knowing implicit in science were to become explicit—declarative understanding of knowing—science of knowing.  Then I wouldn’t be yelling at scientists: be little more conscious and little less motor in running between things and thoughts ;)  Seriously, what got me into small is my recent encounter with discrete objects in the context of self-foundations.  If you are still a non-believer, then start with

and then make friends with

1. Small family of objects (Conceptual Mathematics, pp. 150 – 1, 369 – 71; Sets for Mathematics, pp. 154 – 5, 248 – 50)

along with

1. Separating (Conceptual Mathematics, page: 215, 245)

and not to be forgotten are

1. Exercises (Conceptual Mathematics, page: 250, 272, 340)

which reminds me that it’s time to exercise:

In the category F, whose objects are functions, there are infinitely many non-isomorphic objects with one piece (and also infinitely many with exactly one point; Exercise 2, Conceptual Mathematics, page 362).

Let’s consider few sets:

Furniture = {table, chair}

Animals = {cat, dog}

Vehicles = {bus, bike}

How many points are there in Furniture?

First, what’s a point?

A point is a morphism from a terminal object.  In the category of sets S that we are in now, a singleton set

1 = {•}

is the terminal object.  Here’s a point

table: 1 → Furniture

of Furniture.  There’s another point

chair: 1 → Furniture

All told, there are two points in Furniture.  So is the case with Animals and Vehicles.  These are all sets with two points.  More importantly, they are isomorphic.  For example, there is an isomorphism

f: Animals → Furniture

with

f (cat) = chair and f (dog) = table

satisfying

f f -1 = 1Furniture and f -1 f = 1Animals

where

-1: Furniture → Animals

with

-1 (chair) = cat and -1 (table) = dog

In general, in the category of sets, all sets with 1 point are isomorphic, and so is the case with 2-point sets, 3-point sets…  But this is not the case in the category of functions F.  An object in the category of functions is a function

f: A → B

What is a point of a function?

Once again, it’s a morphism from the terminal object of the category of functions.

Now, what is terminal object in the category of functions?

The identity function

11: 11

is the terminal object of the category of functions.

I almost forgot, what is a morphism in the category of functions?

A morphism from a function

f: A → B

to a function

g: C → D

is a pair of functions

p: A → C

q: B → D

making the diagram

A –p–> C

|           |

f           g

|           |

v           v

B –q–> D

commute i.e. satisfy

q f = g p

Now we are equipped to look at the points of a function, say,

f: A → B

with A = {you, me}, B = {happy}, and

f (you) = happy

f (me) = happy

A point of

f: A → B

is a morphism

1p–> A

|           |

11         f

|           |

v           v

1q–> B

with p (•) = you and q (•)= happy.  The other point of f corresponds to f (me) = happy.  In general, the number of points in a function is equal to the size (|A| = 2) of the domain (A) of the given function (f).

Let us now consider a slightly different function

g: A → C

with C = {happy, ecstatic} and

g (you) = happy

g (me) = ecstatic

Once again there are two points in g.  However the two objects

f: A → B

and

g: A → C

with the same number (2) of points are not isomorphic (unlike the case of the category of sets).  In general, by keeping the domain (size) fixed and by varying the codomain (size) we can get as many non-isomorphic objects as we like with same number (|domain|) of points.

How about pieces?  The number of pieces in a function

f: A → B

is equal to the size of its codomain (|B|).  Once again we can find as many non-isomorphic objects as we like with same number (|codomain|) of pieces simply by varying the domain (size) while keeping the codomain (size) fixed.

Fine, but what does this exercise have to do with small objects or discrete objects?

I’m glad you asked, which brings us to Exercise 6. 14 (Sets for Mathematics, page 119):

pieces –| discrete –| points

The functor

discrete: Sets → Functions

mapping each object (set)

X

in the domain category of sets to its identity function

1X: X → X

(an object) in the codomain category of functions is left adjoint to the functor

points: Functions → Sets

mapping each object (function)

f: A → B

in the domain category of functions to its set of points

A

(an object) in the codomain category of sets.

Let’s now show that

discrete is left adjoint to points

in terms of functions and their determinations.

First consider an object (a set)

X

of the domain category Sets of the left adjoint

discrete: Sets → Functions

to the functor

points: Functions → Sets

Next, consider functions

X → points (f)

(on X) whose type

points (f)

is given as a value of the points functor at an object (function)

f: A → B

of the category of functions.

Now we have to show that every such function

X → points (f)

is uniquely determined by a function

X → points (discrete (X))

i.e. there is a unique

discrete (X) → f

such that

X → points (f) = X → points (discrete (X)) º points (discrete (X) → f)

where º denotes composition.

With

f: A → B

points (f) = A

discrete (X) = 1X: X → X

points (discrete (X)) = X

points (discrete (X) → f) = X → A

we have to show that

X → A = X → X º X → A

and taking the endomap on X as identity 1X we do have

X ––> A = X – 1X –> X ––> A

X → A = X → A

Moving along…

The functor

discrete: Sets → Functions

mapping each object (set)

X

in the domain category of sets to its identity function

1X: X → X

(an object) in the codomain category of functions is right adjoint to the functor

pieces: Functions → Sets

mapping each object (function)

f: A → B

in the domain category of functions to its set of pieces

B

(an object) in the codomain category of sets.

Let’s now show that

discrete is right adjoint to pieces

in terms of figures and their inclusions.

First consider an object (a set)

X

of the domain category Sets of the right adjoint

discrete: Sets → Functions

to the functor

pieces: Functions → Sets

Next, consider figures

pieces (f) → X

(in X) whose shape

pieces (f)

is given as a value of the pieces functor at an object (function)

f: A → B

of the category of functions.

Now we have to show that every such figure

pieces (f) → X

is uniquely included in a figure

pieces (discrete (X)) → X

i.e. there is a unique

f discrete (X)

such that

pieces (f) → X = pieces (fdiscrete (X)) º pieces (discrete (X)) → X

where º denotes composition.

With

f: A → B

pieces (f) = B

discrete (X) = 1X: X → X

pieces (discrete (X)) = X

pieces (fdiscrete (X)) = B → X

we have to show that

B → X = B → X º X → X

and taking the endomap on X as identity 1X we do have

B ––> X = B ––> X – 1X –> X

B → X = B → X

I’m terribly sorry: I don’t think I can get to marrying discrete objects to self-foundations as I intended.  More importantly, the above adjoint situation involving discrete objects didn’t come out as clearly as I hoped for (assuming I got it right)…  Making it all brief, I think it’s time to bid farewell to this practice of bite-sized blog-posts, which are not living up to the original idea of making otherwise challenging concepts (e.g. self-foundations alluded to in the title) relatively easy to understand…  I’ll see you again if and when I have something substantial to say in readily accessible examples.

Goodbye!

Let us consider couple of numbers—none of the kind that bought Ramanujan lasting fame—simple

-1, 1

seemingly devoid of any depth to delve into.  Given the set

A = {-1, 1}

of thus chosen numbers, we think of, say, squaring them i.e.

sq: A → A

with

sq (-1) = 1 and sq (1) = 1

If I were to picture this, then it would look like couple of dots denoting the numbers -1 and 1, and couple of arrows—one going from -1 to 1 and the other going from 1 to 1—denoting the process of squaring.

What we have here is a dynamical system

sq: A → A

with two states ‘-1’ and ‘1’, one of which i.e.  ‘1’ is a fixed-state.  These fixed-states are also called points since they can be listed using maps from the terminal object

11: 11

(where 1 = {•}) of the category of dynamical systems.  So we say our dynamical system

sq: A → A

has 1 point.

Points, for some inexplicable reason, remind me of pieces.  Now that we remember piece, we ask: are there any pieces (also called connected components) in our dynamical system?

If, for now, you don’t mind taking my word for it, then there is 1 piece in our dynamical system

sq: A → A

First, the idea of piece or connected component is, fortunately, not that far removed from the common understanding of connected.  Remember the two states

-1, 1

are connected i.e. there’s an arrow connecting -1 to 1 and another from 1 to 1 (depicting the dynamic of squaring).  Next, the idea of the number of connected components is based on the fact that if I map ‘-1’ to state ‘p’ which is a fixed state (of some codomain dynamical system), then the state ‘1’ to which ‘-1’ goes as a result of squaring must also be mapped to the same state ‘p’ (all of which follows straight from the definition of morphism in the category of dynamical systems).  Then there is the realization that though every state in a piece gets mapped to the single state of a point, there’s nothing in the definition of the morphism (of dynamical systems) that stops me from mapping every state of every piece into one state of one point.  However all such maps and for that matter every map that maps states of a dynamical system into dynamical systems consisting of more (or less) number of points than the number of pieces in the domain dynamical system is determined by the map from the domain dynamical system to a codomain dynamical system which has as many points as the number of pieces in the domain dynamical system.

But where’s the coequalizer [with which you lured me] in all this?

Going back to our beloved dynamical system

sq: A → A

we realize that we can think of the endomap

sq: A → A

(which is a structure consisting of one component object i.e. set A and one component map i.e. sq: A → A; plz see Conceptual Mathematics, pp. 149 – 51) as an irreflexive directed graph (which is a structure consisting of two component objects and two component maps).

One immediate question, given that there’s only one component object and only one component map in our endomap

sq: A → A

and given that we need two component objects and two component maps to have the above mentioned irreflexive directed graph structure, where are we going to get that which we don’t have, but do need?

First, let’s specify that graph structure: it consists of two component objects

Arrows, Dots

and two component maps

source: Arrows → Dots

target: Arrows → Dots

Taking both component objects as A i.e.

Arrows = A

Dots = A

and the component map source as identity and the other component map target as our endomap sq, we arrive at a irreflexive directed graph (incarnation of our endomap; see Conceptual Mathematics, pp. 143 – 4)

id: A → A

sq: A → A

with

id (-1) = -1 and id (1) = 1

and

sq (-1) = 1 and sq (1) = 1

Whenever we see a parallel pair of maps

id, sq: A → A

we can think of looking for maps from the codomain (of the parallel pair) such as

A – id, sq –> A – f –> B

Now, of course, we can compose to get another parallel pair

fid: A → B

fsq: A → B

Even though id and sq are clearly different maps, it is possible that upon composing with the map f we might end up with equality i.e.

fid = fsq

Let’s say we look at all such maps

f: A → X

(i.e. all maps which when composed with id and sq result in the same map fid = fsq) and find amongst these a special one

f*: A → Y

(satisfying f*id = f*sq, of course) determining every

f: A → Z

(satisfying f ◦ id = fsq).    In other words, there’s a unique determination

g: Y → Z

such that

f = gf*

for every map f satisfying f ◦ id = f ◦ sq.  This map

f*: A → Y

is called the coequalizer of the parallel pair of maps

id, sq: A → A

Let us now look to see how the coequalizer

f*: A → Y

looks like.

First, it must satisfy

f*id = f*sq

Taking

Y = 1 (= {•}, a singleton set)

and

f*: A → 1

as

f* (-1) = • and f* (1) = •

we find that

f*id = f*sq

since

f*id (-1) = f* (-1) = •                                      f*sq (-1) = f* (1) = •

f*id (1) = f* (1) = •                                        f*sq (1) = f* (1) = •

Now that

f*: A → 1

satisfies

f*id = f*sq

we have to see if every

f: A → Z

satisfying

fid = fsq

f*: A → 1

For now, if you take my word for it,

f*: A → 1

is the coequalizer of the parallel pair of maps

id, sq: A → A

And now recollect that the above parallel pair of maps is a dynamical system with two states

A = {-1, 1}

and a dynamic

sq (-1) = 1, sq (1) = 1

and most important of all, for the present purposes, is a dynamical system with one piece, which is given as the singleton set 1 of its coequalizer

f*: A → 1

Or that’s how I understood “when parallel maps in the category of sets are understood as source/target structure, the coequalizer becomes the ‘set of components’ of the graph” (Conceptual Mathematics, page 294). And I found myself on page 294 as I was trying to construct the connected components functor on page 359; and yes, I skipped the Exercise 4 of Session 32… it scared me!

Every now and then I run into these ‘induced maps,’ and every time I bump into one of those I think of checking it out… any relation with those induced currents from my long-lost electrical engineering?  In any case, we are about to undress one.

Any map

f: A → B

induces a map

i: π0A → π0B

making the diagram

π0A –– i ––> π0B

^                      ^

|                       |

cA                     cB

|                       |

A    –– f ––>    B

commute.

First note that π0A and π0B are spaces of components of A and B, respectively, which means that the two maps

cA: A → π0A

and

cB: B → π0B

are universal components maps.

Going back to the diagram

π0A –– ? ––> π0B

^                      ^

|                       |

cA                     cB

|                       |

A    –– f ––>    B

(which hopefully will turn out to be commutative at least by the end of this post) we notice that we can compose the two maps

A –– f ––> B –– cB ––> π0B

to get the map

A –– cB º f ––> π0B

Now going back (I like going back and forth) to the definition of space of components, while simultaneously looking at the above map (reproduced below)

A –– cB º f ––> π0B

and the universal components map for A i.e.

A –– cA ––> π0A

we realize (based on the definition of space of components) that there is exactly one map

i: π0A → π0B

from the space of components (π0A) of A to π0B (a discrete object as far as A is concerned), which when composed with

cA: A → π0A

gives

cB º f: A → π0B

i.e.

i º cA = cB º f

which says that the diagram

π0A –– i ––> π0B

^                      ^

|                       |

cA                     cB

|                       |

A    –– f ––>    B

is commutative.

I think we just did, somewhat haphazardly, the first part of Exercise 2 (Conceptual Mathematics, page 359), and (in case you haven’t noticed) my muse went missing… writing (i.e. slow-thinking) feels strained :(

From: Fred E.J. Linton, To: Venkata Rayudu Posina, Date:  Mar 13, 2014, Subject: Re: Morphisms and The Scientific Method

Actually, Posina,

There are quite a few mathematical problems that don’t seem to involve structure-preserving morphisms at all. Finding solutions to differential equations, for example, or finding roots of polynomials, or determining for a given group, with given presentation, whether or not a given word is in the generators and their inverses is equal to the group identity.

So: I would answer “no” to the question you raise; and I would question the validity of the “fact” you posit two paragraphs later (below).

Cheers, — Fred Linton

—— Original Message ——

Received: 11 Mar 2014, From: Venkata Rayudu Posina, To: Fred E.J. Linton, Subject: Morphisms and The Scientific Method

Dear Professor Fred E. J. Linton,

I am doing fine and wish the same for you.  Sometime ago you helped me with a question I had about the method of mathematics.  If I may, I have a question about the relationship between respecting-structure and scientific-method.

Many, if not all, mathematical calculations (e.g. solving equations for unknowns) involve structure-preserving morphisms, which raises the following question:

Is mathematical knowing [invariably] structure-preserving?

Respecting structure, when thought of as ‘do not tear‘, is reminiscent of the ‘do not disturb’ of physical knowing (e.g. the act of measurement, in order to result in a true measure, is required to not disturb the system under measurement).

Given the ideal ‘do not disturb’ of physical measurements and the actual ‘do not tear’ of mathematical calculations, can we conclude that respecting-structure is the [basic / fundamental] method of knowing / the scientific method?  Phrased differently, what does the fact that many (all?) calculations [of unknowns] are structure-preserving say about ‘how we know?’

(Please forgive me if all this is gibberish.)

I eagerly look forward to your corrections and clarifications.

Thanking you,

Yours sincerely,

posina

Exercise 1: Suppose that there exist n, m so that αn x = αm y.  If moreover y is connected to a third state z, for example if αk y = αp z, show that x is connected to z (Conceptual Mathematics, page 358).

Hint: The addition of natural numbers is commutative (a + b = b + a).

We are given

αn x = αm y

which means that if you were to start at the state

x

then the state you will be in after pressing the button

α

(Conceptual Mathematics, page 161) n times is

αn x

which is same as the state

αm y

I got myself into after pressing that α button m times, having started at the state

y

Since we ended up together

αn x = αm y

it makes sense to call the states at which we started

x, y

connected.  Don’t you agree?

Connected should feel like married; I mean marriage with no divorce.  Wait! Scarier than ‘Till Death Us Do Part.’  There’s no separating states

x, y

once they got together

αn x = αm y

Any number k of button (α) pressing will only move them, but together

αkn x) = αkm y)

αk + n x = αk + m y

We are also told that y is connected to z, which means

αk y = αp z

for some natural numbers k and p.

If we now press that α button m times, we find

αmk y) = αmp z)

αm + k y = αm + p z

Since

k + m = m + k

αk + n x = αk + m y = αm + k y = αm + p z

i.e.

x is connected to z

Summing it all up, if x is connected to y and y is connected to z then x is connected to z.

Moral: If you must mess with two, then make sure that the two don’t know each other or else you’ll get caught sooner or later ;)

Nope!  We barely got started.

We talked about what it takes for states of a dynamical system to be connected.  Now what does it take for a dynamical system to be connected.  For hors d’oeuvre,  the dynamical system must have at least one state, and every two states must be connected.

Let’s say we are given a dynamical system, say, a light bulb switching between states

S = {ON, OFF}

every time I press the button

a: S → S

i.e.

a(ON) = OFF

a(OFF) = ON

Are the two states

ON, OFF

connected?

Connected, if only we can find some

n, m

such that

an (ON) = am (OFF)

Let’s say that the bulb initially was OFF and you press the button once to get to

a1 (OFF) = ON

Let’s now say that the bulb initially was ON and I press the button twice to get to

a2 (ON) = a1 (a1 (ON)) = a1 (OFF) = ON

Since

a1 (OFF) = a2 (ON)

we can safely say that the two states

ON, OFF

are connected.

Is our light bulb

S, a: S → S

connected?

It has more than one state i.e. two states

S = {ON, OFF}

and since the two states are connected, it (light bulb / dynamical system) is connected.

Of course, not every dynamical system is connected even if every state is connected (e.g. a system in which every state is a fixed state).  If a state x is connected to itself as in

a(x) = x

then we call them loners points.  If a state x is connected to some other state y, then those states and for that matter all the states that are connected surely deserve a name and that name is piece or connected component.

Now given any dynamical system X, we can ask:

Does X have any points?

Does X have any pieces?

Our light bulb has 0 points and 1 piece.  We know how to count points in a dynamical system X using maps

T → X

from the terminal dynamical system

T = (1, t: 11)

to the dynamical system X.

How about counting connected components?

First note that points are connected pieces with just one state.  Next note that there is exactly one map from any connected component to a point, which sends all the states (no matter how many there are) in the connected piece to the only state in the point.  Next, note that even though there is no map from a point to a connected piece (which is not a point), the two objects are “same” when viewed as pieces (both have exactly one connected component).  As a result of which any map from a dynamical system to a discrete system (consisting of fixed states only) is determined by a map from the dynamical system to a discrete system which has as many points as there are pieces in the domain dynamical system (see the definition of space of components on page 359 of Conceptual Mathematics).

Now recollect that we said that our light bulb

S, a: S → S

S = {ON, OFF} with a(ON) = OFF and a(OFF) = ON

has 1 connected component.  How did we come up with that number?  First consider a map from (S, a) to a discrete dynamical system with 1 point.  Next consider any map from the same (S, a) to a discrete dynamical system with any number of points.  Every such map is uniquely determined by the map from (S, a) to 1-point discrete dynamical system.  In general, if a dynamical system X has n connected components, then maps from X to discrete dynamical systems are uniquely determined by maps from X to a discrete dynamical system Y which has n points.  This universal property (of Y with respect to X) is what gives us the number of pieces of X in terms of the points of Y.

How I wish I can draw diagrams here :(