One thing I don’t like about story-telling is them stories sitting still every time the teller goes on a smoke break; wouldn’t it be nice if the story can script itself without waiting for me to exhale?
Imagine, upon reading the definition of truth value object (Conceptual Mathematics, page 337), finding that the textbook ends then and there—no more spilled-ink to make sense of. There’s, within this doomsday scenario, enough momentum built, definition-after-definition beginning (where shall we) with terminal objects such as 1 (Conceptual Mathematics, pp. 225 – 9), into the story to keep characters moving in the heads heading home long after the curtain dropped on
true: 1 –> Ω
What if I reverse an arrow
–>
into the arrow
<–
How about an arrow from Ω
X <– Ω
instead of the arrow to Ω
X –> Ω
Reversing an arrow
–>
into the arrow
<–
doesn’t cost whole lot of consciousness and that’s all(?) it takes to get to the definition of 0 (empty set):
there is, for each set X, one function 0 –> X
from the definition of 1 (singleton set):
there is, for each set X, one function 1 <– X
Just as I was about to write-off 1 as bland (after all there is just one function to 1; Conceptual Mathematics, page 302), points i.e. functions from 1
1 –> X
are all over the place spicing up the story (Conceptual Mathematics, page 19, pp. 230 – 5).
With this preamble in place, let’s indulge in some head-scratching: the definition of truth value object is beaming with arrows (waiting to be reversed?) such as
(i) a given part true: T –> Ω
(ii) every part g: U –> X
(iii) exactly one arrow f: X –> Ω
or, shall we say, loaded with heavyweights such as
Beginning with belongs to, what do we get if we reverse arrows in
x belongs to y
I’d have never guessed
determined by
to be the opposite of
Then again it’s not like I suspected the initial object
0
to be the opposite of the terminal object
1
or SUM to be the opposite of PRODUCT (Conceptual Mathematics, page 254, 284). I have nothing to show even after looking in every gyri and sulci of my big male-brain for the opposite of PART (Sets for Mathematics, pp. 37 – 8).
That’s all the self-indulgent blather my shrink can stand this Sunday 
We say
f: A –> Y
is determined by
g: A –> X
whenever
f = pg (Conceptual Mathematics, pp. 45 – 9, 68 – 80, 370 – 1).
Note to self: One of these days you need to stop talking words or at least start drawing diagrams! Until then I need not highlight the fact that science is preoccupied in every step of its long-winding road to reality with DETERMINATION. Watching bodies-in-motion one wonders about distances traveled by the moving bodies
f: Moving bodies –> Distances
and thinks of noting down the duration of motion
g: Moving bodies –> Durations
and hopefully finds that there is a function
h: Durations –> Distances
(assigning to each duration of motion the distance traveled by moving bodies in that duration) such that
f = hg
in which case we say
f is determined by g
Here is
1 = {•}
a singleton set with one element (•).
Here’s a part
g: 1 –> 1
of 1 i.e. g(•) = (•).
This part g is the inverse image of another part
v: 1 –> W
with W = {true, false} and v (•) = true…
Now this is beginning to read a lot like Ramayana in three sentences, which is all the excuse I need to go back—way back to my very good friend function
f: A –> B
Let’s say we have a function
color: A –> B
with A = {Rayudu} and B = {Black, Brown} telling us the hair color of people in A. For example
color(Rayudu) = Black
which says something like: the hair color of Rayudu is black.
A recurring nightmare of mine [in mathematics] is inversion—fortunately, for now and here, it’s not that scary; it simply asks: who’s that in A who got black hair?
Rayudu
Rayudu, not unlike any other answer, is boring; it’s how we got to ‘Rayudu’ from ‘Black’ that’s worth a penny or two.
But, first, who’s Rayudu?
Rayudu is a part of the domain set A i.e.
Rayudu: 1 –> A
What, while we are at it, is black?
Black is a part of the codomain set B i.e.
Black: 1 –> B
Going from the color Black to the person who’s color is black is, as far as I can tell, going from a given part
Black: 1 –> B
to a part
Rayudu: 1 –> A
along the function
color: A –> B
The question then is: given
color: A –> B, color(Rayudu) = Black
and
Black: 1 –> B
how did we end up with this
Rayudu: 1 –> A
What’s so special about Rayudu?
It’s a part of A; but so is
0: 0 –> A, 0 = {}
Rayudu is special in that any generalized element of A (Sets for Mathematics, page 16) i.e.
x: X –> A
Rayudu: 1 –> A
if and only if
fx: X –> B
Black: 1 –> B
which is just about all the confusion I can cook up out of
Rayudu: 1 –> A is the inverse image of Black: 1 –> B along f: A –> B
Conceptual Mathematics, page 336
Since we are least interested in Rayudu, his color, or for that matter, any other values of physical variables associated with him, let’s get off this tangent and get back to business pronto.
The part
g: 1 –> 1, g(•) = •
is the inverse image of the part
v: 1 –> W, v(•) = true
but along what?
Even more bluntly, only if I specify a function
f: 1 –> W
and say something like
g is the inverse of v along f
then somebody else can tell me if it’s true or false.
Given that
W = {true, false}
we have two functions from 1 to W
f: 1 –> W, f (•) = true
f’: 1 –> W, f’ (•) = false
Now the question we have staring into our pupils:
Is g the inverse image of v along f or along f’ or along both f and f’?
We can say
g is the inverse image of v along f
because the generalized element
x: 1 –> 1
is in (Conceptual Mathematics, page 335)
g: 1 –> 1
i.e. there exists a p such that
x = gp
and
fx: 1 –> W
is in
v: 1 –> W
i.e. there exists a q such that
fx = vq
and because
f’x: 1 –> W
is not in
v: 1 –> W
i.e. because there is exactly one function
f: 1 –> W
along which
g is the inverse image of v
Just in case you feeling bad about the other function
f’: 1 –> W, f’(•) = false
being left out of our story, here’s fairness-in-action: there’s another part of 1
g’: 0 –> 1, 0 = {}
that is the inverse image of v along f’ (but not along f).
That’s enough pillow-fights and here comes the hardcore definition (Lawvere element):
An object W together with a given part v: 1 –> W is called a truth value object (for the category of sets) if and only if for every part g of any set X there is exactly one function f: X –> W for which g is the inverse image of v along f (Conceptual Mathematics, page 337).
Here’s the part I like in the above definition:
for every part g… there is exactly one function f
As much as I just do things, here’s the plan: redo this exercise
- for D (dot) in the category of graphs
- for U (= 0 –> 1) in the category of two-stage variable sets (Sets for Mathematics, pp. 114 – 9)
- for 0: 0 –> 1 in the category of labeled sets
The above 3 cases have something in common with the case of 1 = {•} of the present post which is the one dot (in visual depiction) of the objects
1, D, U, and 0
which means… nothin
Thing[s] Place[s] Time[s]
1 1 1 (being)
1 1 2 (stasis)
1 2 1 (motion)
1 2 2 (result of motion)
2 1 1 (superposition)
2 1 2 (substitution)
2 2 1 (configuration)
But what do you mean by
things: points or bodies?
places: boundaries or neighborhoods?
times: instants or durations?
Diving into all that feels like simulated-drowning For now here we go drop • drip •
MOTION: 1 thing in 2 places at 1 time
The concept of motion as the presence of a thing in one place at one time, in another place at another time describes only the result of motion and not motion.
While I’m paraphrasing Professor F. William Lawvere (see Continuously Variable Sets; Algebraic Geometry = Geometric Logic) I might as well share another observation from the same paper which I found fascinating:
(Please don’t expect physicists possessed by God particles to do the janitorial work of clarifying MATTER or MOTION; they are too busy tickling themselves to headline news.)
Distancing myself from the verbal description of the material situations with which we started, I see a big-picture:
constant vs. variable
To get a feel for that all too familiar ever-elusive variable:
I am a different person today than I was yesterday, yet I am still the same person.
Sets for Mathematics, page 16
we begin with an acknowledgment of how challenging it is to grasp CHANGE conceptually (definitely not fun like turning it into a lullaby for the masses and the enlightened alike).
Let’s start with, say, then and now and throw some things in. Some things that were may not be anymore: people die, vehicles move out of parking spaces, etc. Some things do survive, though not forever, both yesterday and today. Standing in the shoes of a thing that’s in today we can see the thing that it was yesterday. Straightening this bipolar rendering of variation, we have
history: Now –> Then
specifying for each thing x in Now the thing that x was back Then (see Truth Values for Two-Stage Variable Sets; Sets for Mathematics, pp. 114 – 119).
I promise to return or better yet stick with the category of two-stage variable sets at least until I get to its truth value object!
That’s—on the face of it—one billboard of a title for a puny blogpost. Those in the know know that I ain’t proving – nothing before the next pole-shift. In the meantime, how does one go about telling if any given two are one and the same or are two different ones?
Given two numbers
x = 1
and
y = 2
we say x ≠ y because there’s a digit (units place) that’s different. In other words, two numbers are different if they differ in at least one digit. Marching ahead, two pictures are different if there is at least one pixel that takes different values in the two pictures.
Then there are, what else, functions from a domain set A to a codomain set B. Given two functions
f: A –> B
g: A –> B
we say
f ≠ g
if there is at least one point
x: 1 –> A, where 1 = {•}
at which the two functions differ
f(x) ≠ g(x)
So far so good!
What was I thinking when I was saying
1 + 1 = 2
I guess
1 + 1 ≠ 0
1 + 1 ≠ 1
1 + 1 ≠ 3
.
.
.
All of the above goes on to say that our good sum
2
is unique—not as in unique with no qualifier whatsoever but as in unique with respect to the summands
1 + 1
Some such intrigue spins out and about products also provided one sleeps long and often enough with
In sober terms, 2 is the one and only one number, with respect to the factors (1, 2), amongst all the gazillion numbers, that qualifies to be on the right-hand side of ‘=’ in
1 x 2 =
(which for some reason reminds me of a hard lesson that I learned rather late in life: always walk on that side of your lady that’s closest to the cars swerving onto the sidewalk
For good or bad we find ourselves multiplying things other than numbers – finding products such as
mass x acceleration
F = ma is one product that’s too much to bite into now, but thankfully there are objects—you guessed it: sets—about whose products I don’t mind rambling on and on…
Product of two sets
1 = {•}
and
2 = {•, •}
is a set P and a pair of functions
p1: P –> 1
p2: P –> 2
satisfying:
for every set Q and every pair of functions
q1: Q –> 1
q2: Q –> 2
there is exactly one function
f: Q –> P
for which
q1 = p1f
and
q2 = p2f
(Conceptual Mathematics, page 237).
There’s something scary about this definition of product: if you say that a set
2 = {•, •}
along with a pair of functions
p1: 2 –> 1
p2: 2 –> 2
is a product of the two sets
1 = {•}
and
2 = {•, •}
then I have to check something for every set, which wouldn’t be much work if there were one or two sets but there are way too many sets
0 = {}
1 = {•}
2 = {•, •}
.
.
.
for me to check. It would be nice if I could get away with checking one or two…
Be careful, I was advised, what you wish for—you might get it!
I got my wish fulfilled in the guise of an exercise which my good friend Fidel suggested:
Exercise 1 (Conceptual Mathematics, page 250)
Consider the diagram of graphs
B1
^
|
P —–> B2
Suppose it satisfies the fragment of the definition of product in which we test against only the two figure-types X = A (arrow) and X = D (dot). Prove that the diagram actually is a product, i.e. that the product property holds for all graphs X.
Switching gears from the category of graphs to the category of sets we cook up another exercise:
Consider the diagram of sets
B1
^
|
P —–> B2
(Here B1, B2, and P are sets.) Suppose it satisfies the fragment of the definition of product in which we test against only X = 1 (singleton set). Prove that the diagram actually is a product, i.e. that the product property holds for all sets X.
If you must blame, then blame the above exercise on the fact that the terminal object in the category of sets i.e.
1 = {•}
plays the same role of telling two [maps] apart i.e. separating (Conceptual Mathematics, page 215) that the two objects
A = • –> • (arrow)
D = • (dot)
together play in the category of graphs.
Oftentimes I wonder
Why define?
Where on earth mathematical definitions come from?
At times words such as clear, clarity, and clarification sound like a way out of the Gordian knot of ‘why?’
It’s only a couple of days ago when I had trouble spelling-out for myself how to calculate composites given a collection of maps that an answer to
Why define composition?
presented itself in the guise of
HOW
in a resounding reaffirmation of DO, of acts, of practice in the name of theory.
Definition of CATEGORY (Conceptual Mathematics, page 21)
A category C consists of the DATA:
(1) OBJECTS, COb = {A, B, C…}
(2) MAPS, CMp = {A –f–> B, B –g–> C, A –1A–> A…}
(3) For each map f, one object as DOMAIN of f and one object as CODOMAIN of f.
Let’s pause.
Given COb and CMp, we can readily find the domain object (in COb) of a map (in CMp) with the help of a structural map
dom: CMp –> COb
assigning to each map
f: A –> B
in CMp its domain object
A
in COb. The story of CODOMAIN is not much different: we can readily find the codomain object (in COb) of a map (in CMp) with the help of another structural map
cod: CMp –> COb
assigning to each map
f: A –> B
in CMp its codomain object
B
in COb. The above two structural maps (dom, cod) remind me of irreflexive graphs (Conceptual Mathematics, page 141).
Inching along (akin to a caterpillar) ~~~
(4) For each object A an IDENTITY MAP, which has domain A and codomain A.
In sync with the insanity that worked: we can readily find the identity map (in CMp) of an object (in COb) with the help of yet another structural map
id: COb –> CMp
assigning to each object
A
in COb its identity map
1A: A –> A
in CMp. The above three structural maps (dom, cod, id) remind me of reflexive graphs (Conceptual Mathematics, page 145).
(5) For each pair of composable maps
A –f–> B –g–> C
with
dom (g) = cod (f) [= B]
A –gf–> C (read ‘gf’ as ‘g after f’)
Original headache which got me into all of this: how to find composite map (readily or not so readily)?
For starters, composite maps are maps; so they can be found in CMp, which is to say that any map
comp: X –> CMp
that we may think of using to find the composite map of a composable pair of maps has CMp as its codomain. What about the domain X?
Let’s take a min to take stock: since we are worried about finding the composite map of a composable pair of maps we need to worry about how we might go from maps in
CMp
to pairs of composable maps.
Here’s a flicker of hope: pairs of composable maps such as
A –f–> B –g–> C
are figures of shape 3 (ordinal number) i.e.
• –> • –> •
in
CMp
So graph maps
3 –> CMp
from
• –> • –> •
to
CMp
give us all composable pairs of maps in CMp.
Summing up
comp: CMp3 –> CMp
in pointing to composite maps such as
A –gf–> C
in CMp for each composable pair of maps
A –f–> B –g–> C
in CMp3 (where CMp3 is the map object [Conceptual Mathematics, page 313] of all composable pairs of maps i.e. all maps from 3 to CMp) sounds like a definition of COMPOSITION to me for the time being.
Disclaimer: Please don’t bet your life on any of the above… it may very well be one of those not even wrong!
I am always at someplace—at one of the many places out-there (I guess it has something to do with me being classical, which I’ll save for later). More often than not I find myself wanting to be at some place other than the place I’m at. Fortunately there are flights taking people like me from one place to another such as
f: Rajahmundry -> Hyderabad
Sometimes a flight may not go far enough; say, to San Diego from Hyderabad. Under such circumstances one looks for composable pairs of flights such as
g: Hyderabad -> Amsterdam, h: Amsterdam -> San Diego
with
origin (h) = destination (g)
This is what I used to do back in the day working at a travel agency – looking at graphs consisting of places (depicted as dots) and flights (depicted as arrows between dots) – looking for composable pairs of arrows such as
A –x–> B –y–> C
Given a set of flights such as
P = {A –x–> B, B –y–> C, E –z–> D…}
how does one go about finding composable pairs of arrows in P?
This question takes me back… back to a classic intuition. If I want to find the weight of something, then I use something (such as a stone) that has weight; if I want to find the length of something, then I use something (such as a stick) that has length. In a similar vein, maps from a graph Q (with 2 arrows and 3 dots as shown below)
• –> • –> •
to the graph P tell us all about composable pairs of arrows in P i.e. all about Q-shaped figures in P (Conceptual Mathematics, pp. 82 – 3), and along the way the definition of composition (Functorial Semantics, pp. 33 – 4) which I have yet to wrap my brain around… it might be a good idea to get to work with Exercise 2 (Conceptual Mathematics, page 252).
Number: Stop rewriting history! Losers don’t get to do that.
Me: um… err… what did I do
Number: Don’t act naïve. How would you feel if I called you a monkey?
Me: Monkey is one of my gods. In India, as you probably know, everything is a God.
Number: We are not talking about polytheism and you are not funny. I am not going to sit silent while you badmouth me—what were you saying the other day:
Me: An endomap • –> •
Number: Whatever! We the numbers own this place—been here from the very beginning and you make it seem like we are bad guys who colonized pristine space. But for our quantitative culture… what have you gotten with all your thinking about quality???
Me: What if I say: In the beginning there was number.
Number: That’s little better, but you must practice
And that is the end of the story. I am going to immediately institute a moratorium on your ridiculous ramblings. In the meantime
Me: Whew!!! I guess we have to begin with godsy numbers. I hope and pray that this is not going to hurt the feelings of maps.
Map: Nae… we all in this together.
Me: Thank you! Let’s start with two numbers
0, 1
where to next (watching our steps every step of the way)…
You: We can readily notice that
0 < 1
Me: Yes, yes there is a category which has
0, 1
as objects and the relation
0 < 1
as a map.
You: I need, before I can think of 0 as an object, to know what you have in mind for the identity map from 0 to 0.
Me: How about
0 = 0
You: Why not just one ‘less than or equal to’ instead of the two: <, =
Me: That sounds like a good idea to me. So we have two objects
0, 1
their identity maps
0 < 0
1 < 1
and a map from 0 to 1
0 < 1
You: There is no map from 1 to 0 since 1 is not ‘less than or equal to’ 0. It looks like there is at most one map between the two objects 0 and 1 in each one of the four pairings [(0, 0); (0, 1); (1, 0); (1, 1)].
Me: This reminds me of posets, parts of an object, degrees of truth (Conceptual Mathematics, pp. 339 – 57) and many things of such nature… I’m in a rush to catch a train to my hometown…
mañana numeros 
Posina Venkata Rayudu
