The hard problem of consciousness is explicating how moving matter becomes thinking matter. Harder yet is the problem of spelling out the mutual determinations of individual experiences and the experiencing self. Determining how the collective social consciousness influences and is influenced by the individual selves constituting the society is the hardest problem. Drawing parallels between individual cognition and the collective knowing of mathematical science, here we present a conceptualization of the cognitive dimension of the self. Our abstraction of the relations between the physical world, biological brain, mind, intuition, consciousness, cognitive self, and the society can facilitate the construction of the conceptual repertoire required for an explicit science of the self within human society.
Category Archives: Note to myself
Functorial Semantics for the Advancement of the Science of Cognition
Cognition involves physical stimulation, neural coding, mental conception, and conscious perception. Beyond the neural coding of physical stimuli, it is not clear how exactly these component processes constitute cognition. Within mathematical sciences, category theory provides tools such as category, functor, and adjointness, which are indispensable in the explication of the mathematical calculations involved in acquiring mathematical knowledge. More specifically, functorial semantics, in showing that theories and models can be construed as categories and functors, respectively, and in establishing the adjointness between abstraction (of theories) and interpretation (to obtain models), mathematically accounts for knowing-within-mathematics. Here we show that mathematical knowing recapitulates–in an elementary form–ordinary cognition. The process of going from particulars (physical stimuli) to their concrete models (conscious percepts) via abstract theories (mental concepts) and measured properties (neural coding) is common to both mathematical knowing and ordinary cognition. Our investigation of the similarity between knowing-within-mathematics and knowing-in-general leads us to make a case for the development of the basic science of cognition in terms of the functorial semantics of mathematical knowing.
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
pA: P –> A, pB: P –> B
(to the two factors A, B) such that given any pair of maps
fA: F –> A, fB: 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
pA · f = fA and pB · f = fB
(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
pA: P –> A and pB: P –> B
such that for every pair of functions
fA: F –> A and fB: F –> B
there is exactly one function
f: F –> P
(to the product set P) satisfying both
pA · f = fA and pB · f = fB
The key to calculating products is the 1-1 correspondence in the definition of product:
for each pair of functions
fA: F –> A and fB: 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 pA and pB 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
pA: A x B –> A and pB: A x B –> B
Now we have to find out the product set A x B, along with its two projection maps pA and pB. 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
fA: F –> A, fB: 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
pA: A x B –> A, pB: A x B –> B
gives the pair of functions we started with i.e.
pA · f = fA, pB · f = fB
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
pA: A x B –> A, pB: 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
fA: F –> A, fB: 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 pA, pB gives the pair of functions we started with i.e.
pA · f = fA
pB · f = fB
With F = 1, we have
pA · (a, b1) = a and pA · (a, b2) = a
pB · (a, b1) = b1 and pB · (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
pA: A x B –> A, pB: 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 pA, pB structuring the product set
A x B = {(a, b1), (a, b2)}
as
pA (a, b1) = a and pA (a, b2) = a
pB (a, b1) = b1 and pB (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).
Groundhog Day
I wish I can remember how it ends… until then, it’s yet another glitch in the matrix: is a given adjoint of a functor
discrete: S –> F
left or right adjoint? As if to compound my confusion, left adjoint coincides with right:
pieces = points
in the definition of quality type (see Axiomatic Cohesion), which I’m trying to understand.
Let’s start with the functor
discrete: S –> F
from the category S of sets to the category F of functions (simply because I have a vague feeling that this is how I went about telling left from right). The category S of sets has sets and functions as its objects and morphisms, respectively, while the category F of functions has functions and commutative squares as its objects and morphisms, respectively (Conceptual Mathematics, pp. 144-5). The functor
discrete: S –> F
assigns to each object (set)
A
(in S) its identity function
1A: A –> A
(an object in F) i.e.
discrete (A) = 1A: A –> A
and to each morphism (function)
f: A –> B
(in S) a morphism
discrete (f: A –> B) = discrete (A) –> discrete (B)
= <f, f>: 1A –> 1B
(in F) which is a commutative square
A –– f ––> B
^ ^
| |
1A 1B
| |
A –– f ––> B
(satisfying 1B · f = f · 1A, where ‘·’ denotes composition).
Now, we are told that the functors
points: F –> S
pieces: F –> S
are adjoint functors of the functor
discrete: S –> F
(see [or is it more like do] Exercise 6.14 on page 119 of Sets for Mathematics). And our job is to figure out which one is left adjoint and which one is right adjoint and, of course, why?
Let’s start with the functor
points: F –> S
which assigns to each object (function)
f: A –> B
(in F) its domain set
A
(in S) i.e.
points (f: A –> B) = A
and to each morphism from an object
f: A –> B
to an object
f’: A’ –> B’
i.e. to each commutative square
B –– w ––> B’
^ ^
| |
f f’
| |
A –– v ––> A’
(satisfying f’ · v = w · f) a function
v: A –> A’
(in S) i.e.
points (f’ v = w f) = points (f) –> points (f’)
= v: A –> A’
Looking at the definition of adjoint functor (Conceptual Mathematics, pp. 374-5), we realize that we could call
discrete: S –> F
left adjoint to
points: F –> S
if we can find a natural correspondence
d: discrete (X) –> f
————————————
p: X –> points (f)
for every object X in S and f in F.
In other words, every function
p: X –> points (f)
(in S; whose type is given as a value of the points functor) is determined by the function
nX: X –> points (discrete (X))
and the determination
points (d: discrete (X) –> f)
is unique i.e. for every
p: X –> points (f)
(in S) there is a unique
d: discrete (X) –> f
(in F) such that
p = points (d) · nX
In (yet) other words, there is a natural transformation
n: 1S –> points · discrete
from the identity functor
1S: S –> S
(on S) to the composite functor
points · discrete: S –> S
(an endofunctor on S), whose components are
nX: X –> points (discrete (X))
Summing up, so far, we say
discrete: S –> F
functor is left adjoint to
points: F –> S
functor if there is a natural transformation
n: 1S –> points · discrete
What if, instead, the functor
discrete: S –> F
is right adjoint to
points: F –> S
functor? Well, then we would expect to see a natural correspondence
p’: points (f) –> X
————————————
d’: f –> discrete (X)
for every object X in S and f in F.
In other words, every figure
p’: points (f) –> X
(in S; whose shape is given as a value of the points functor) is included in the figure
n’X: points (discrete (X)) –> X
and the inclusion
points (d’: f –> discrete (X))
is unique i.e. for every
p’: points (f) –> X
(in S) there is a unique
d’: f –> discrete (X)
(in F) such that
p’ = n’X · points (d’)
In (yet) other words, there is a natural transformation
n’: points · discrete –> 1S
from the composite functor
points · discrete: S –> S
(an endofunctor on S) to the identity functor
1S: S –> S
(on S), whose components are
n’X: points (discrete (X)) –> X
Summing up, so far, we say
discrete: S –> F
functor is right adjoint to
points: F –> S
functor if there is a natural transformation
n’: points · discrete –> 1S
Summing it all, if there’s a natural transformation
n: 1S –> points · discrete
we say
discrete is left adjoint to points
and if there is a natural transformation
n’: points · discrete –> 1S
we say
discrete is right adjoint to points
Let’s see: since
points · discrete (X) = X
and, of course,
1S (X) = X
and since we can take the identity function
1X: X –> X
as components of both the natural transformations i.e. with
nX: X –> points (discrete (X)) = 1X: X –> X
n’X: points (discrete (X)) –> X = 1X: X –> X
we have both the natural transformations
n: 1S –> points · discrete
n’: points · discrete –> 1S
which means
discrete is both left and right adjoint of pieces
But it’s clearly not the case:
discrete is left adjoint to pieces
(do Exercise 6.14 on page 119 of Sets for Mathematics). Are we doomed? No, it’s just intermission and everything that could possibly go wrong for the protagonist goes wrong half-way through Tollywood movies…
Happy New Year!
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Here’s an excerpt:
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Science of Knowing 