What is reality? Reality is not fiction. Then, what is fiction? Anything / everything can happen in fiction. So is the case with violence. Anything can happen in violent situations. So is with inconsistent. Everything can be proved in inconsistent systems (Tarski). Violence, in view of these considerations, is inconsistenthappenings. Nonviolence, in stark contrast, is consistentbecoming. By virtue of being not where anything / everything happens (e.g. me walking on water… not happening ;) reality is nonviolent.
Violence appears as a means under conditions characterized by loss of contact with reality (c.f. Bibi). Thinking of solving problems using violent means is diagnostic of breakdown of the boundary separating reality from fiction. Thinking of finding solutions using violent methods is reason enough for referral to mental health professional.
Reality is nonviolent. So must knowing of reality be. I know that there are few books and some papers on my desk (aka bed). How did I know? I look and see the books and papers. This actofperception is nonviolent. Staring at someone on streets for too long would bring home the need to be nonviolent in knowing. Thus, in a meeting of means with ends, we find that nonviolence [as in respecting / preserving] is how we know reality.
Remember the [domain] identity law
A –1_{A}–> A –f–> B = A –f–> B
which can be read as
‘f is determined by 1_{A}’
and the function
f: A → B
is the unique determination of f by 1_{A}.
For the following we don’t need the entire identity map; an isomorphism, as we’ll see, will do.
Let’s say we are given an isomorphism
f: A → B
then every function (with the same domain A)
g: A → C
is uniquely determined by f.
First,
‘g is determined by f’
means that there is a function
h: B → C
satisfying
h f = g
Let’s see. What we are given is that
f: A → B
is an isomorphism, which means that there is exactly one function
f ^{1}: B → A
satisfying
f ^{1} f = 1_{A}
We can precompose this inverse
f ^{1}: B → A
with
g: A → C
to get a function from B to C
g f ^{1}: B → C
Is this the determination
h: B → C
that we are looking for? If so, then
h = g f ^{1}
should satisfy
h f = g
Let’s see.
g f ^{1} f = g
g 1_{A} = g
g = g
OK. So
g f ^{1}: B → C
is a determination of
g: A → C
by
f: A → B
Next question:
Is g f ^{1} the only determination of g by f ?
Let’s say we have another determination
h‘: B → C
satisfying
h‘ f = g
Precomposing with
f ^{1}: B → A
we get (since f ^{1} is an isomorphism)
h‘ f f ^{1} = g f ^{1}
h‘ 1_{B} = g f ^{1}
h‘ = g f ^{1}
which is the same as our earlier determination. So we can say that given an isomorphism
f: A → B
every function (on the domain A)
g: A → C
is determined by f (thanks to the identity f ^{1} f = 1_{A}) and uniquely so (thanks to the identity f f ^{1} = 1_{B}).
Now remember the other [codomain] identity law
A –f–> B –1_{B}–> B = A –f–> B
which can be read as
‘f is in 1_{B}’
and the figure
f: A → B
is the unique proof of inclusion of f in 1_{B}.
For the following, once again, we don’t need the entire identity map; an isomorphism, as we’ll see, will do.
Given an isomorphism
f: A → B
every figure (in the codomain B)
g: C → B
is uniquely included in f.
First, ‘g is in f’ means that there is a
h: C → A
satisfying
f h = g
Since f is an isomorphism we have exactly one
f ^{1}: B → A
satisfying
f f ^{1} = 1_{B}
Postcomposing this inverse
f ^{1}: B → A
with
g: C → B
gives us the
f ^{1} g: C → A
which fits the bill of the required
h: C → A
But is it really the one we are looking for i.e. does it satisfy
f h = g
i.e. is
f f ^{1} g = g
Since
f f ^{1} = 1_{B}
we have
1_{B} g = g
g = g
Is this
h = f ^{1} g
the only proof of inclusion of
g: C → B
in
f: A → B
Let’s say we have another proof
h = h‘
of ‘g is in f’ i.e.
g = f h‘
Postcomposing with
f ^{1}: B → A
we get (since f ^{1} is an isomorphism)
f ^{1} g = f ^{1} f h‘
f ^{1} g = 1_{A} h‘
f ^{1} g = h‘
So we can say that every figure
g: C → B
is in the isomorphism
f: A → B
and uniquely so.
This inclusion story is related to Exercise 2 (Conceptual Mathematics, page 235) and even more so to the definition of space of points (Conceptual Mathematics, page 360). The determination story with which we started is related to the definition of space of components (Conceptual Mathematics, page 359). If not, then this
(i) every function on the domain of an isomorphism is determined by the isomorphism and uniquely so
and
(ii) every figure in the codomain of an isomorphism is in the isomorphism and uniquely so
is how I tried to understand the ‘universal’ in those two definitions.
1. Solid stays.
2. Liquid flows.
3. Gas diffuses.
This becoming consistent with its Being is DHARMA. Hijacking ‘dharma’ to excuse caste and untouchability is insulting intellect.
I. Translation [without losing the meaning of the text that is translated] is AHIMSA or NONVIOLENCE (don’t misinterpret / lie).
II. Biological reproduction within any species [without mutating the meaning of genetic code] is nonviolence.
III. Medical treatment of a disease [without harming the patient] is nonviolence (do no harm).
IV. Physical measurement [without disturbing the system that is measured] is nonviolence (don’t disturb).
V. Mathematical morphism [without tearing apart the structure that is transformed] is nonviolence (don’t tear; Conceptual Mathematics, page 210).
This unadulterated childlike respect for the object of one’s study is the method of science–in its etymological sense–of knowing.
Cognitive science, in its pathological obsessivecompulsive glorification of heuristics and biases, continues to fail to recognize and [then] respect the method of knowing: test under conditions compatible with the exhibition of that which is [true] of the subject (cf. students in classrooms, thinking in minds, etc).
We have had enough–enough violation of the soul of science, enough silliness:
NOW is the time to get serious about Knowing!
In practice, of course, there’s violence–the violence of probing (e.g. poking sharp electrodes into the brains of awake primates). Hence the ‘paramo’ in
Ahimsa paramo Dharma
and the Gandhian guideline for knowing:
Nonviolence is the only means to know Truth.
Conflictofinterest Statement: I’m paid to study nonviolence at the Centre for Gandhian Studies, GITAM University.
Happy (don’t sleep ;) Shivaratri!
People belong to countries and countries are ruled by people.
How are we going to model this strange phenomenon?
First, we have to decide where we are going to model, which is no different from deciding the medium of expression (e.g. jotting an idea on a blank page).
Let’s choose the category of sets for recording our observation.
Why category of sets (Conceptual Mathematics, pp. 133 – 7, Sets for Mathematics pp. ix – x; cf. there’s nothing in the conclusion that’s not in the premise)?
It’s as simple and straightforward as using blank papers in printer; try reading text printed on a paper that already has some text.
Let’s start with a set of people
P
and a set of countries
C
Every person in the set P belongs to a country in the set C, which gives us a function
n: P –> C
For example,
n(Rayudu) = India
Every country in C has a person in P who is its ruler, which gives us another function
h: P <– C
An illustration,
h(America) = Obama
So our model of citizens or peopleincountries consists of a pair of sets
P, C
and a pair of functions
n: P –> C
h: P <– C
I’m glad you noticed that the arrows are opposite in direction. Our category is like the category of irreflexive graphs (Conceptual Mathematics, pp. 141 – 2) in that both have a pair of sets and a pair of functions. And that’s where the likeness stops; the pair of functions
source: Arrows –> Dots
target: Arrows –> Dots
of the category of irreflexive graphs are parallel, whereas the pair of functions
nationality, n: People –> Countries
head, h: People <– Countries
of our category of citizens are opposite.
Fulldisclosure: We are not building this model to understand people and their strange behaviors (e.g. voting). For sometime, I have been feeling that a category in between the category of irreflexive graphs and the category of reflexive graphs (Conceptual Mathematics, page 145) is missing in the Conceptual Mathematics textbook. You see, irreflexive graphs have two arrows
s: A –> D
t: A –> D
and reflexive graphs have three arrows
s: A –> D
i: A <– D
t: A –> D
In going from irreflexive graphs to reflexive graphs there is (in addition to the added arrow i) a change of direction (of the arrow i with respect to s, t). So I thought of cooking up a category with an oppositepair of arrows
n: P –> C
h: P <– C
as a bridge to ease the conceptual transit from the category of irreflexive graphs to the category of reflexive graphs.
Whew!
Let’s start in America (in the set C) and take
h: P <– C
to arrive at
h(America) = Obama
If we now board
n: P –> C
we land at
n(Obama) = America
which is where we started. Our entire trip can be summed up as
n(h(America)) = America
There’s nothing special about America (I mean, in our model); the same holds true of every country in C. That is, the nationality of the head of a country is that very country. In more general terms, we say
nh = 1_{C}
This is another reason why I like to think of our category as a conceptual bridge to reflexive graphs. You see, the three structural componentmaps of the category of reflexive graphs
s: A –> D
i: A <– D
t: A –> D
satisfy
si = 1_{D}
ti = 1_{D}
The two structural componentmaps
s: A –> D
t: A –> D
of irreflexive graphs, unlike those of reflexive graphs, are not required to satisfy any equations (Conceptual Mathematics, page 150). Our peopleincountries with two structural componentmaps
n: P –> C
h: P <– C
is like irreflexive graphs, but since the two componentmaps are opposite and, more importantly, satisfy
nh = 1_{C}
our category is like reflexive graphs also.
In any case, objects of our category have two component sets
P, C
and two component functions
n: P –> C
h: P <– C
satisfying one equation
nh = 1_{C}
We are assuming that there are no headless countries (sorry, no room for enlightened anarchy :( and, of course, no dual citizenship business.
Next order of business is defining maps between objects, composition of maps, and verifying that we do have a category for sure (Conceptual Mathematics, page 21)!
Then there’s the truth value object of our category that we would like to know all about.
Yes, of course, we are interested in finding the terminal object of our category, especially to see if it’s a separator (Conceptual Mathematics, page 215).
And yes, it all can wait!
Acting is–knowing full well–acting unawares!
The above acting got nothing to do with the action below (just showing off what’s left of my brief stint in Hollywood ;)
Actions (Conceptual Mathematics, pp. 218 – 9, 303), as if hellbent on confusing me, are twohanded: left, right (Sets for Mathematics, pp. 171 – 6). Thankfully there’s a more functional nomenclature (Conceptual Mathematics, pp. 370 – 1):
(i) algebraic
(ii) geometric
to talk about actions. Geometry, rightfully, brings figures to figural salience and it is these
that are called right (for reasons I have yet to fathom) actions. Algebra, on the other hand (i.e. left), highlights functions and it is these
that are called left (for, by now, obvious reasons?) actions.
I’m right handed; so as you’d expect, I think right actions are hot (Conceptual Mathematics, pp. 360 – 1):
Dynamical systems are right actions of the additive monoid of natural numbers (Conceptual Mathematics, pp. 136 – 8).
Reflexive graphs are right actions of a threeelement (one identity and two idempotents) monoid (Conceptual Mathematics, pp. 145 – 51; Sets for Mathematics, pp. 180 – 92).
Reversible graphs are right actions of a fourelement (one identity, one involution, and two idempotents) monoid (Sets for Mathematics, pp. 176 – 80).
Irreflexive graphs are right actions of a twoobject category (Conceptual Mathematics, pp. 141 – 5; pp. 360 – 1).
But, what exactly is a right action?
Actions are maps from products (Conceptual Mathematics, pp. 218 – 9) and a right action or a representation of a monoid M on a set S is a function
r: S x M –> S
satisfying
r(s, m.m’) = r(r(s, m), m’)
and
r(s, 1) = s
Let’s unpack.
Let’s start with
S = {a, b, c}
and
M = {1, p, q}
with the elements of the monoid M (Conceptual Mathematics, pp. 166 – 9) satisfying:
1.1 = 1
1.p = p
1.q = q
p.1 = p
p.p = p
p.q = p
q.1 = q
q.p = q
q.q = q
with ‘.’ denoting composition. We have to see how many of the 3^{9} (since S = 3 and S x M = 9) functions:
a b c
^

r

a, q b, q c, q
a, p b, p c, p
a, 1 b, 1 c, 1
satisfy
r(s, m.m’) = r(r(s, m), m’)
and
r(s, 1) = s
Once we fix
r(a, 1) = a
r(b, 1) = b
r(c, 1) = c
we are left with 3^{6} functions to check. We have to see how many, if any, of these 729 functions satisfy
r(s, m.m’) = r(r(s, m), m’)
For some not yet to be disclosed reasons, let’s try if
r(a, p) = a
r(b, p) = a
r(c, p) = c
r(a, q) = a
r(b, q) = c
r(c, q) = c
satisfies
r(s, m.m’) = r(r(s, m), m’)
Case 1: m = 1, m’ = p
r(s, (1.p)) = r(r(s, 1), p)
r(s, p) = r(s, p)
Case 2: m = 1, m’ = q
r(s, (1.q)) = r(r(s, 1), q)
r(s, q) = r(s, q)
Case 3: m = p, m’ = 1
r(s, (p.1)) = r(r(s, p), 1)
r(s, p) = r(s, p)
Case 4: m = p, m’ = p
r(s, (p.p)) = r(r(s, p), p)
r(s, p) = r(r(s, p), p)
4.1: s = a
LHS:
r(a, p) = a
RHS:
r(r(a, p), p) = r(a, p) = a
4.2: s = b
LHS:
r(b, p) = a
RHS:
r(r(b, p), p) = r(a, p) = a
4.3: s = c
LHS
r(c, p) = c
RHS:
r(r(c, p), p) = r(c, p) = c
Case 5: m = p, m’ = q
r(s, (p.q)) = r(r(s, p), q)
r(s, p) = r(r(s, p), q)
5.1: s = a
LHS:
r(a, p) = a
RHS:
r(r(a, p), q) = r(a, q) = a
5.2: s = b
LHS:
r(b, p) = a
RHS:
r(r(b, p), q) = r(a, q) = a
5.3: s = c
LHS:
r(c, p) = c
RHS:
r(r(c, p), q) = r(c, q) = c
Case 6: m = q, m’ = 1
r(s, (q.1)) = r(r(s, q), 1)
r(s, q) = r(s, q)
Case 7: m = q, m’ = p
r(s, (q.p)) = r(r(s, q), p)
r(s, q) = r(r(s, q), p)
7.1: s = a
LHS:
r(a, q) = a
RHS:
r(r(a, q), p) = r(a, p) = a
7.2: s = b
LHS:
r(b, q) = c
RHS:
r(r(b, q), p) = r(c, p) = c
7.3: s = c
LHS:
r(c, q) = c
RHS:
r(r(c, q), p) = r(c, p) = c
Case 8: m = q, m’ = q
r(s, (q.q)) = r(r(s, q), q)
r(s, q) = r(r(s, q), q)
8.1: s = a
LHS:
r(a, q) = a
RHS:
r(r(a, q), q) = r(a, q) = a
7.2: s = b
LHS:
r(b, q) = c
RHS:
r(r(b, q), q) = r(c, q) = c
7.3: s = c
LHS:
r(c, q) = c
RHS:
r(r(c, q), q) = r(c, q) = c
I think what we have done is show that our function
r: S x M –> S
with
r(a, q) = a r(b, q) = c r(c, q) = c
r(a, p) = a r(b, p) = a r(c, p) = c
r(a, 1) = a r(b, 1) = b r(c, 1) = c
is indeed a representation of our threeelement monoid M (= {1, p, q}) on our set S (= {a, b, c}) i.e. satisfies
r(s, (m.m’)) = r(r(s, m), m’)
What remains to be seen is that this representation is a reflexive graph. But, more importantly, as you can tell it’ll take lot more than one take for me to get a good feel for these right actions.
One thing that’s worth noting is that the condition
r(s, (m.m’)) = r(r(s, m), m’)
(that kept us busy) looks like the associative law:
I remember reading in Conceptual Mathematics (page 136) that associative law is all that we need to prove all that’s proved in the textbook. Isn’t it fascinating (in the sense of ‘in need of explanation’) that such an unremarkablelooking law has such vast reach?
I, not that anybody cares, don’t like this left adjoint, right adjoint business; where to begin: I have great trouble telling left from right without looking at my hands, which only compounds the trouble I have seeing the leftness of left adjoints such as
Discrete functor is left adjoint to Points functor
Free functor is left adjoint to Forgetful functor
Structure is left adjoint to Semantics
No less important: why bother?
Adjoints figure prominently in knowing—of the kind not necessarily limited to ‘I know I’m handsome cuz I see it all over the mirror’ or ‘I know sun rises in the east cuz I see it there every morning’, but also including this kind: earth goes around the sun, which took some serious thought. It’s great to have all this scientific knowledge, but to not have an explicit scientific account of knowing is like producing litter after litter without any declarative understanding of reproductive biology (somewhat like streetdogs ;) Adjoint functor is usually introduced in terms of undoing i.e. as the next best thing to inverse (Conceptual Mathematics, page 375). Inverse is, of course, intimately related to isomorphism, and going by the etymology i.e. ‘same shape’, we arrive at the idea of isomorphisms as instruments of knowing. Once we know that two objects—one known object K and the other unknown U—are isomorphic, we get to know the shape of the unknown U in terms of the shape of the known K:
U has same shape as that of K.
We are, of course, not completely in the dark when we don’t have an isomorphism. An onto function
r: K → U
tells us that the unknown codomain U is no bigger than known domain K; a 11 function
s: U → K
tells us that the unknown domain U is no bigger than known codomain K. This is the idea of getting to know an unknown object with maps from and to known objects. Adjoints, like the section and retract above, are used to get closer and closer to the unknown.
But how?
That remains to be seen ;)
Let’s start with something simple. We begin with our good friend: category of sets S and another category: terminal category T with one morphism. There is a functor
A: S → T
which takes every set in S to the only object in T and the every function in S to the only morphism (identity morphism) in T. This functor is left adjoint to another functor
I: S ← T
assigning the singleton set
1 = {•}
and the identity function
1_{1}: 1 → 1
in S to the only object and map, respectively, in T.
Now let’s see how one goes about showing that
A is left adjoint to I
or (even less symbols)
There is the ‘figures and their incidences’ way and then there is the ‘functions and their determinations’ route; I like both, but will start with figureandincidence.
Given an object in the codomain category of the left adjoint
A: S → T
Wait… there’s only one object T in the terminal category T.
That’s no problem… given the object T, every figure in T whose shape is given as a value of the functor A i.e. every figure
t: A(S) → T
is in the figure
e_{T}: A(I(T)) → T
and there’s only one proof of inclusion i.e. there’s only one
s: S → I(T)
for which
t = e_{T} ° A(s)
Since every set S in the category of sets S goes to the only object T in the terminal category T, we have
A(S) = T
and since there’s only one morphism
1_{T}: T → T
in the terminal category T, we have only one figure
t: A(S) → T
i.e.
1_{T}: T → T
whose inclusion in the figure
e_{T}: A(I(T)) → T
we have to worry about. Let’s look at the shape
A(I(T))
of the figure e_{T} i.e.
A(I(T)) = A(1) = T
since the functor
I: S ← T
sends the only object T in T to the singleton set 1 in S and then the functor
A: S → T
sends the set 1 in S to the object T in T. So the figure
e_{T}: A(I(T)) → T
turns out to be
1_{T}: T → T
which is same as our
t: A(S) → T
Now we have to show that
1_{T} is in 1_{T}
No less important is the requirement that there be exactly one proof
s: S → I(T)
of the inclusion i.e. only one s satisfying
t = e_{T} ° A(s)
Since
I(T) = 1
there’s exactly one function
s: S → 1
from every set S to the terminal set 1, and since
A(s) = 1_{T}
we have to show
1_{T} = 1_{T} ° 1_{T}
1_{T} = 1_{T}
Now that we have seen that the functor
A: S → T
is left adjoint to the functor
I: S ← T
i.e.
one simpleminded question that comes to my mind: is there a left adjoint
O: S ← T
to the left adjoint
A: S → T
i.e.
Is O ― A?
Let’s see what we find going through the same motions we went through…
Given an object S in the codomain category S of the left adjoint
O: S ← T
every figure in S whose shape is given as a value of the functor O i.e. every figure
s: O(T) → S
is in the figure
e_{S}: O(A(S)) → S
and there’s only one proof of inclusion i.e. there’s only one
t: T → A(S)
for which
s = e_{S} ° O(t)
But first, what does the functor
O: S ← T
do? Let’s say, O sends the only object T in T to the empty set
0 = {}
and the only [identity] map
1_{T}: T → T
in T to the only function on the empty set i.e.
1_{0}: 0 → 0
Next, what is
s: O(T) → S
Since O(T) = 0, there’s only one figure
s: 0 → S
Next, what is
e_{S}: O(A(S)) → S
Since O(A(S)) = O(T) = 0, once again there’s only one figure
e_{S}: 0 → S
Next, how many
t: T → A(S)
do we have? Once again there’s only one
t = 1_{T}: T → T
and this identity map goes to
1_{0}: 0 → 0
since
O(1_{T}) = 1_{0}
All that’s left to do is to show that
s = e_{S} ° O(t)
0 –s–> S = 0 –O(t)–> 0 –e_{S}–> S
0 → S = 0 → S
So we say
O ― A
I think, all said and done, what we have done is show that the functor which sends every set S in the category of sets S to the only object T in the terminal category T has both left adjoint (sending T in T to initial object 0 in S) and right adjoint (sending T in T to terminal object 1 in S) i.e.
(unless of course I got confused with left and right once again ;) which is an example of Unity and Identity of Adjoint Opposites. Little later let’s look at other ways of looking at adjoints, along with some other adjoints.
Knowing is getting to know that which we do not know in terms of that which we do know.
Let’s say that you know that your left hand has five fingers. Your left hand, for our present purposes, is the known K. You want to know how many fingers your right hand has. What do you do now?
In doing namaskaram, you put fingers of your right hand (unknown) in onetoone correspondence with the fingers of your left hand (known). Once you establish this correspondence between the known K and unknown U, you know that the unknown is isomorphic to the known i.e. U has the ‘same shape’ as that of K, and shape, in the present context, is the number of fingers. This is illustrative of what we mean by knowing or getting to know that which we do not know in terms of that which we do know.
Knowing, understood as above, can be analyzed into two ways:
Before we go into that which distinguishes geometric from algebraic (see Two general aspects or uses of maps, pp. 81 – 5; Geometry of figures and algebra of functions, pp. 369 – 70 in Conceptual Mathematics), let’s first note that which is common:
The moment one hears ‘map’, one asks:
 maps from where?
 maps to where?
Answers to these two questions take us to the two ways of knowing:
 maps K → U from a known K to unknown U
 maps K ← U to a known K from unknown U
Let’s say, we have an unknown set U. With a singleton set
1 = {•}
as the known K, we get to know about U (e.g. size of U) by way of maps from K (= 1) to U (Kshaped figures in U) i.e.
1 → U
Let’s [once again] say, we have an unknown set U. With a twoelement set
2 = {false, true}
as the known K, we get to know about U (e.g. answers to True or False questions) by way of maps to K (= 2) from U (Kvalued functions on U) i.e.
U → 2
and the eternal conflict between Being and knowing goes on a break ;)