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.
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.
Unique up to Unique Isomorphism
This is a story about how ‘a’ gets to be ‘the’. Mathematics is, in a sense, about understanding something known until all traces of surprise are extinguished (yes, I don’t like surprises ;)
We all know that
and if we are asked ‘what, if any, are the numbers which when multiplied with 2 give 2?’ we readily answer: 1. In a similar vein, if we are given a map
f: A –> B
and asked ‘what, if any, are the maps which when pre-composed with f give f?’ we, going by the (first or left) identity law of composition of maps
f · 1A = f
(where ‘·’ denotes composition; Conceptual Mathematics, p. 17), answer that the identity map
1A: A –> A
is one such map. Let us now consider a pair of maps
pA: P –> A, pB: P –> B
(with a common domain P). What, if any, are the maps to the common domain P which upon pre-composition with the pair of maps (pA, pB) give the pair (pA, pB)? Once again, we find that the identity map
1P: P –> P
is one such map satisfying both
pA · 1P = pA and pB · 1P = pB
Now, if
P, pA: P –> A, pB: P –> B
is the product of A and B, then, going by the definition of product (Conceptual Mathematics, p. 217), there is exactly one map
q: Q –> P
from the common domain Q of any pair of maps to the factors A, B
qA: Q –> A, qB: Q –> B
to the common domain P of the product projections
pA: P –> A, pB: P –> B
which satisfies both
pA · q = qA and pB · q = qB
Since the pair of product projections
pA: P –> A, pB: P –> B
qualifies as ‘any pair of maps to the factors A, B’ in the above definition of product, there is exactly one map
p: P –> P
which satisfies both
pA · p = pA and pB · p = pB
But we already know, from the identity law we talked about a min ago, of a map
1P: P –> P
which satisfies both
pA · 1P = pA and pB · 1P = pB
So, if there’s going to be only one map
p: P –> P
satisfying both
pA · p = pA and pB · p = pB
then that one map must be
1P: P –> P
This is the key to the following exercise in which we are asked to show that products are unique up to unique isomorphism.
Exercise 12 (Conceptual Mathematics, p. 217):
If (P, pA: P –> A, pB: P –> B) and (Q, qA: Q –> A, qB: Q –> B) are both products of A and B, then the unique map f: P –> Q which satisfies both qA · f = pA and qB · f = pB is an isomorphism.
If
(Q, qA: Q –> A, qB: Q –> B)
is a product of A and B, then, given any pair of maps to the factors
pA: P –> A, pB: P –> B
with a common domain P, there is exactly one map
f: P –> Q
which satisfies both
qA · f = pA and qB · f = pB
Now, if
(P, pA: P –> A, pB: P –> B)
is a product of A and B, then, given any pair of maps to the factors
qA: Q –> A, qB: Q –> B
with a common domain Q, there is exactly one map
g: Q –> P
which satisfies both
pA · g = qA and pB · g = qB
Next, composing these two maps we get two endomaps
g · f: P –> Q –> P
and
f · g: Q –> P –> Q
Let’s look at the endomap on P
g · f: P –> P
which we can compose with the pair of projection maps
pA: P –> A, pB: P –> B
to get a pair of maps
pA · (g · f): P –> A, pB · (g · f): P –> B
Let’s see what these maps are:
pA · (g · f) = (pA · g) · f = qA · f = pA
pB · (g · f) = (pB · g) · f = qB · f = pB
Since there is only one map
1P: P –> P
which satisfies both
pA · 1P = pA
pB · 1P = pB
the above endomap (g · f) must be the identity 1P. Similar reasoning reveals that the other composite (f · g) is the identity 1Q. Thus the product (P, pA: P –> A, pB: P –> B) of A and B is unique up to unique isomorphism.
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).
Equalizer of idempotents
Exercise: If
f = m ∙ e
is an idempotent satisfying
f ∙ f = f
and
e ∙ m = 1P
(where ‘∙’ denotes composition), then the section
m: P –> A
is an equalizer of the parallel pair of maps
f, 1A: A –> A
(Conceptual Mathematics, p. 293).
Definition: An equalizer of a parallel pair of maps
f, g: A –> B
is a map
i: X –> A
satisfying
f ∙ i = g ∙ i
and every map
j: Y –> A
satisfying
f ∙ j = g ∙ j
is uniquely included in
i: X –> A
i.e. there is exactly one map
k: Y –> X
such that
j = i ∙ k
To show that
m: P –> A
is an equalizer of
f, 1A: A –> A,
given
f = m ∙ e
f ∙ f = f
e ∙ m = 1P,
we have to first show that
f ∙ m = 1A ∙ m
i.e.
f ∙ m = m
f ∙ m = (m ∙ e) ∙ m = m ∙ (e ∙ m) = m ∙ 1P = m
Next, we have to show that if any map
j: Y –> A
satisfies
f ∙ j = j
then there is exactly one map
k: Y –> P
satisfying
j = m ∙ k
First, we need a map from Y to P. Given
j: Y –> A
and
e: A –> P
we can take the composite map
e ∙ j: Y –> A –> P
as the map k from Y to P i.e.
k = e ∙ j
and see if
m ∙ k = j
i.e. if
m ∙ (e ∙ j) = j
m ∙ (e ∙ j) = (m ∙ e) ∙ j = f ∙ j = j
(since we are given f ∙ j = j).
Finally, we have to show that
k: Y –> P
satisfying
m ∙ k = j
is unique i.e. if there is another map
k’: Y –> P
satisfying
m ∙ k’ = j
then
k’ = k
Given
m ∙ k’ = j = m ∙ k
post-composing with
e: A –> P
on both sides of
m ∙ k’ = m ∙ k
we find
e ∙ (m ∙ k’) = e ∙ (m ∙ k)
(e ∙ m) ∙ k’ = (e ∙ m) ∙ k
1P ∙ k’ = 1P ∙ k
k’ = k
Thus the section
m: P –> A
of the splitting of an idempotent
f = m ∙ e: A –> P –> A
is an equalizer of
f, 1A: A –> A
Tailpiece: An equalizer of two idempotents
f, f’: A –> A
with a common section i.e.
f = m ∙ e
f’ = m ∙ e’
is the common section
m: P –> A
True or false? (For some reason, I’m not my usual verbose self :( I’ll back soon to spice-up the story all the way to cohesion vs. quality of reflexive graphs vs. idempotents via points and pieces functors constructed in terms of equalizers and coequalizers.)
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!
The WordPress.com stats helper monkeys prepared a 2014 annual report for this blog.
Here’s an excerpt:
A San Francisco cable car holds 60 people. This blog was viewed about 240 times in 2014. If it were a cable car, it would take about 4 trips to carry that many people.
Exercise 30 (Conceptual Mathematics, p. 151)
If S, d, c is a given bipointed object
in a category X, then for each object X of X, the graph of ‘X fields’ on S is actually a reflexive graph, and for each map f: X –> Y in X, the induced maps on sets constitute a map of reflexive graphs.
Given a space S representing a room and a map
t: S –> T
(where T is the temperature line) specifying the temperature at each point in the room, the ‘temperature field’ has reflexive graph structure when two points
d: 1 –> S
c: 1 –> S
in the space S are distinguished (1 is a one-point space). Since the domain set of the function
t: S –> T
is same as the codomain set of
d: 1 –> S
we can compose them to get a point on the temperature line T
1 – d –> S – t –> T = 1 – td –> T
which is the temperature at the point
d: 1 –> S
in the room S. So, corresponding to each temperature field in the room S i.e. corresponding to each function
t: S –> T
we have a temperature
td: 1 –> T
at the point
d: 1 –> S
in S. In other words, the function
d: 1 –> S
induces a function
d*: TS –> T1
(where TS is the set of all possible temperature fields on S and T1 is the set of all temperature values on T) with
d* (t: S –> T) = td: 1 –> T
giving the temperature td at the point d in S for a given temperature field t on S. Along the same lines, the second distinguished point
c: 1 –> S
induces another function
c*: TS –> T1
with
c* (t) = tc
giving the temperature tc at point c in S. Furthermore, the unique function
s: S –> 1
(mapping all points in S to the only point in 1) induces another function
s*: T1 –> TS
mapping each point on the temperature line to a temperature field which has that temperature at every point in S.
The above temperature-field or T-field consisting of two sets (TS and T1) and three functions
d*: TS –> T1
c*: TS –> T1
s*: T1 –> TS
has the structure of reflexive graphs as shown below.
A reflexive graph consists of two sets and three functions
p: A –> B
q: A –> B
r: B –> A
such that the function r is the common section of p and q. In other words,
pr = qr = 1B
Now we have to show that the three functions
d*: TS –> T1
c*: TS –> T1
s*: T1 –> TS
corresponding to the T-field satisfy
d*s* = c*s* = 1T1
d*s* (u) = d* (us) = usd = u11 = u
since sd = 1 – d –> S – s –> 1 = 11 and d*s* = 1T1
c*s* (u) = c* (us) = usc = u11 = u
since sc = 1 – c –> S – s –> 1 = 11 and c*s* = 1T1
Thus for each physical quantity, such as temperature T, we obtain a reflexive graph. Transformations from one physical quantity into another induce transformations of the corresponding reflexive graphs. For example, given a transformation
f: T –> V
we can pre-compose it with the T-field
t: S –> T
to obtain a V-field
ft: S –> V
on the space S. Pre-composing the V-field with a point
d: 1 –> S
gives the value of the V-field
(ft)d: 1 –> V
at that point in space S. Alternatively, we could first find the value of the T-field
t: S –> T
at the point
d: 1 –> S
i.e.
td: 1 –> T
and post-compose it with the given transformation of fields
f: T –> V
to obtain the value
f(td): 1 –> V
of the V-field at the point
d: 1 –> S
in the space S. The associativity of composition of functions i.e.
(ft)d = f(td)
preserves the reflexive graph structure. In addition to the above equation corresponding to
d: 1 –> S
we have two more equations
(ft)c = f(tc)
f(us) = (fu)s
corresponding to
c: 1 –> S
s: S –> 1
all of which together constitute a structure-preserving transformation of reflexive graphs. Thus fields (such as temperature T on a space S with two distinguished points and a retraction of the space to a one-point space) along with their transformations (such as from T to V) can be construed as a category of reflexive graphs.
Representable Functor
A set-valued functor
Q: A –> S
is called representable if there is an object A in the domain category A and an element q in the set Q(A) such that, for any object B of the category A, the function
A(A, B) –> Q(B)
(from the set A(A, B) of functions from A to B to the set Q(B), which is the value of the functor Q at B) assigning to each map
b: A –> B
(in the set A(A, B)) the element
Q(b) (q)
(where
Q(b): Q(A) –> Q(B)
is the function to which the map b is assigned to by the functor Q) is an isomorphism of sets (Sets for Mathematics, p. 248).
Consider a set
W = {you, me}
There are two elements in W. Next consider another set
1 = {·}
There are two functions from 1 to W. They are
you: 1 –> W
with you(·) = you
and
me: 1 –> W
with me(·) = me
Let
W1 = {you, me}
be the map set of functions from 1 to W. Note that both sets W and W1 have the same number of elements i.e.
|W| = |W1| = 2
which means that there is an isomorphism
W = W1
which is an opposed-pair of functions
f: W –> W1
with
f(you) = you: 1 –> W
f(me) = me: 1 –> W
and
g: W1 –> W
with
g(you: 1 –> W) = you(·)
g(me: 1 –> W) = me(·)
satisfying
gf = 1W
fg = 1W^I
All of this may seem like an excessively elaborate discourse on a rather trivial matter: the number of elements of a set W is equal to the number of functions from the singleton set 1 to the set W.
I have two folders in my head
1. profound
2. all else
I put everything that looks trivial such as functions from the terminal set
1 –> W
in the profound folder. Of course, I didn’t recognize either DISCRETE SUBCATEGORY or REPRESENTABLE FUNCTOR in
W = W1
(Conceptual Mathematics, p. 314) on my own :(
Saving discrete subcategory for later, let’s return to our representable functor. A simple example is, as you guessed, our “you, me” story, little generalized. More specifically, the identity functor
1: S –> S
(from the category S of sets to the category S) is a representable functor. What do I need to do in order to convince you that
1: S –> S
is indeed a representable functor?
We need an object
A
of the domain category S of sets and an element
q
of the value of the set-valued functor 1: S –> S at A i.e. of the set
1(A)
Simply put, we need a set
A
and an element
q: 1 –> A
What do these duo i.e. this set
A
and this element
q: 1 –> A
have to do with representable functor?
For any set
B
the function
S(A, B) –> 1(B)
from the set of functions (from A to B)
S(A, B) = BA
(to the value of our identity functor at the set B i.e.) to the set
1(B) = B
assigning to each function
b: A –> B
(in the set BA of functions from A to B) the element
1(b) (q)
i.e. (since 1(b: A –> B) = b: A –> B) the element
b(q)
is an isomorphism of sets.
Thanks to our “you, me” story, we take a guess and take a singleton set
1 = {·}
as our A which has only one element ‘·’ which is our element ‘q’ both of which will make the assignment to each function
b: 1 –> B
in the set
S(1, B) = B1
the element
1(b)(q) = b(q) = b(·) = b
in the set
1(B) = B
an isomorphism
B1 = B
Next, let’s see if the set-valued functor on the category of dynamical systems, which assigns to each dynamical system its set of states, is a representable functor.
Exercise 5.17c (Sets for Mathematics, p. 109)
First, S is the category of sets, whose objects and morphisms are sets and functions, respectively. Next, a functor
P: V –> W
is an assignment of objects and morphisms of the codomain category W to objects and morphisms, respectively, of the domain category V in a way respectful of the domain, codomain, identity, and composition structure of the category.
Let’s look at the case of the functor of our present concern, which has the category S of sets as both domain and codomain category i.e. a functor
P: S –> S
The functor P assigns sets to sets, which we denote as
POb: SOb –> SOb
and functions to functions, which we denote as
PMp: SMp –> SMp
These two functions
POb: SOb –> SOb
PMp: SMp –> SMp
together are required to satisfy, in order to constitute a functor
P: S –> S
the following four conditions corresponding to preserving (i) domain, (ii) codomain, (iii) identity, and (iv) composition.
(i) Preserving Domain
SOb –POb–> SOb
^ ^
| |
domain domain
| |
SMp –PMp–> SMp
The commutativity of the above diagram stated as
domain o PMp = POb o domain
guarantees the preserving of domain (where ‘o’ denotes composition). This commutativity equation is read as: the domain set (in top-right SOb) of the function (in bottom-right SMp) to which a function (in bottom-left SMp) is assigned to by PMp is same as the set (in top-right SOb) to which the domain set (in top-left SOb), of the function (in bottom-left SMp), is assigned to by POb.
(ii) Preserving Codomain
SOb –POb–> SOb
^ ^
| |
codomain codomain
| |
SMp –PMp–> SMp
The commutativity of the above diagram stated as
codomain o PMp = POb o codomain
guarantees the preserving of codomain. This commutativity equation is read as: the codomain set (in top-right SOb) of the function (in bottom-right SMp) to which a function (in bottom-left SMp) is assigned to by PMp is same as the set (in top-right SOb) to which the codomain set (in top-left SOb), of the function (in bottom-left SMp), is assigned to by POb.
(iii) Preserving Identity
SMp –PMp–> SMp
^ ^
| |
identity identity
| |
SOb –POb–> SOb
The commutativity of the above diagram stated as
identity o POb = PMp o identity
guarantees the preserving of identity. This commutativity equation is read as: the identity function (in top-right SMp) of the set (in bottom-right SOb) to which a set (in bottom-left SOb) is assigned to by POb is same as the function (in top-right SMp) to which the identity function (in top-left SMp), of the set (in bottom-left SOb), is assigned to by PMp.
(iv) Preserving Composition
PMp (g o f) = PMp (g) o PMp (f)
The function (PMp (g o f)) to which the composite function (g o f) is assigned to by PMp is same as the composite of the functions (PMp(g) and PMp(f)) to which the functions (g and f, respectively) are assigned to by PMp.
Now that we know what it takes to be a functor, let’s see if what we are given i.e. the functions
(-)T(A) = AT
(-)T(f) = f T
together constitute a functor
(-)T: S –> S
The function
(-)TOb: SOb –> SOb
assigns to each set
A
(in the domain set SOb) the map set
AT
of T-shaped figures in A i.e. the functions
a: T –> A
from [a fixed set] T to A.
The function
(-)TMp: SMp –> SMp
assigns to each function
f: A –> B
(in the domain set SMp) the induced function
f T: AT –> BT
where AT and BT are map sets whose elements are T-shaped figures in A and B, respectively.
The function
f T: AT –> BT
assigns to each element (a T-shaped figure in A)
a: T –> A
in the domain map set
AT
the element
T —a–> A —f–> B
i.e. a T-shaped figure in B
fa: T –> B
in the codomain set
BT
Thus
f T(a) = fa
for all
a: T –> A
in the domain map set
AT
of the function
f T: AT –> BT
Now we have to check to see if the object function
(-)TOb (A) = AT
and the morphism function
(-)TMp (f: A –> B) = f T: AT –> BT
together constitute a functor
(-)T: S –> S
i.e. preserve (i) domain, (ii) codomain, (iii) identity, and (iv) composition.
(i) Preserving Domain
SOb –(-)TOb–> SOb
^ ^
| |
domain domain
| |
SMp –(-)TMp–> SMp
domain o (-)TMp = (-)TOb o domain
LHS
domain o (-)TMp (f: A –> B) = domain (f T: AT –> BT) = AT
RHS
(-)TOb o domain (f: A –> B) = (-)TOb (A) = AT
(ii) Preserving Codomain
SOb –(-)TOb–> SOb
^ ^
| |
codomain codomain
| |
SMp –(-)TMp–> SMp
codomain o (-)TMp = (-)TOb o codomain
LHS
codomain o (-)TMp (f: A –> B) = codomain (f T: AT –> BT) = BT
RHS
(-)TOb o codomain (f: A –> B) = (-)TOb (B) = BT
(iii) Preserving Identity
SMp –(-)TMp–> SMp
^ ^
| |
identity identity
| |
SOb –(-)TOb–> SOb
identity o (-)TOb = (-)TMp o identity
LHS
identity o (-)TOb (A) = identity (AT) = 1AT: AT –> AT
RHS
(-)TMp o identity (A) = (-)TMp (1A: A –> A) = 1AT: AT –> AT
where
1AT (a: T –> A) = T —a–> A –1A–> A = 1A o a = a: T –> A
(iv) Preserving Composition
(-)TMp (g o f) = (-)TMp (g) o (-)TMp (f)
where
f: A –> B, g: B –> C, and g o f: A –> C
LHS
(-)TMp (g o f: A –> C) = (gf)T: AT –> CT
where
(gf)T(a: T –> A) = T —a–> A —gf–> C = (gf)a: T –> C
RHS
(-)TMp (g) o (-)TMp (f)
= gT: BT –> CT o f T: AT –> BT
= AT —f T–> BT —gT–> CT
= gT o f T: AT –> CT
where
gT o f T (a: T –> A) = gT (fa: T –> B) = g(fa): T –> C
Thanks to the associative law
of composition
T —a–> A —f–> B —g–> C
the morphism component
(-)TMp: SMp –> SMp
of the functor
(-)T: S –> S
preserves composition. Thus with (i) domain, (ii) codomain, (iii) identity, and (iv) composition preserved by the assignments
(-)T(A) = AT
(-)T(f) = f T
we do have a functor
(-)T: S –> S
Science of Knowing 