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 savingtheworld for today ;)
What we need to take back from this excursion is functions such as
c_{P}: 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
c_{P}: 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 1valued function)
1
^

c_{X}

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 — 1_{1} –> 1
^ ^
 
c_{X} c_{Y}
 
X — f –> Y
with
1_{1} ^{o} c_{X} = c_{Y} ^{o} f
Is this a functor
C: S –> F
from the category of sets S to the category of functions F?
Let’s check.
A functor
C: S –> F
is a pair of functions
C_{Ob}: S_{Ob} –> F_{Ob}
C_{Mp}: S_{Mp} –> F_{Mp}
assigning objects (functions in F_{Ob}) to objects (sets in S_{Ob}) and morphisms (commutative squares in F_{Mp}) to morphisms (functions in S_{Mp}) in a way respectful of domain, codomain, identity, and composition as follows:
1. Respecting domain:
S_{Mp} — C_{Mp} –> F_{Mp}
 
dom_{S} dom_{F}
 
v v
S_{Ob} — C_{Ob} –> F_{Ob}
with
C_{Ob} ^{o} dom_{S} = dom_{F} ^{o} C_{Mp}
where
dom_{S}: S_{Mp} –> S_{Ob}
assigns to each morphism (function)
f: X –> Y
in the category of sets S its domain object (set)
X
and
dom_{F}: F_{Mp} –> F_{Ob}
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:
S_{Mp} — C_{Mp} –> F_{Mp}
 
cod_{S} cod_{F}
 
v v
S_{Ob} — C_{Ob} –> F_{Ob}
with
C_{Ob} ^{o} cod_{S} = cod_{F} ^{o} C_{Mp}
where
cod_{S}: S_{Mp} –> S_{Ob}
assigns to each morphism (function)
f: X –> Y
in the category of sets S its codomain object (set)
Y
and
cod_{F}: F_{Mp} –> F_{Ob}
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:
S_{Mp} — C_{Mp} –> F_{Mp}
^ ^
 
id_{S} id_{F}
 
S_{Ob} — C_{Ob} –> F_{Ob}
with
C_{Mp} ^{o} id_{S} = id_{F} ^{o} C_{Ob}
where
id_{S}: S_{Ob} –> S_{Mp}
assigns to each object (set)
X
in the category of set S its identity morphism (function)
1_{X}: X –> X
and
id_{F}: F_{Ob} –> F_{Mp}
assigns to each object (function)
Y
^

f

X
in the category of functions F its identity morphism (commutative square)
Y — 1_{Y} –> Y
^ ^
 
f f
 
X — 1_{X} –> X
4. Respecting composition:
C_{Mp} (g ^{o} f) = C_{Mp} (g) * C_{Mp} (f)
which says that the value of the functor at the composite of two morphisms
C_{Mp} (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
C_{Mp} (g) * C_{Mp} (f)
(note that * denotes composition in the category of functions F).
So if
C: S –> F
with
C_{Ob} (X) =
1
^

c_{X}

X
and
C_{Mp} (f: X –> Y) = (1_{1} ^{o} c_{X} = c_{Y} ^{o} f) =
1 — 1_{1} –> 1
^ ^
 
c_{X} c_{Y}
 
X — f –> Y
wants to be a functor, then it must satisfy
C_{Ob} ^{o} dom_{S} = dom_{F} ^{o} C_{Mp}
C_{Ob} ^{o} cod_{S} = cod_{F} ^{o} C_{Mp}
C_{Mp} ^{o} id_{S} = id_{F} ^{o} C_{Ob}
C_{Mp} (g ^{o} f) = C_{Mp} (g) * C_{Mp} (f)
Let’s start with checking ‘respecting domain’:
C_{Ob} ^{o} dom_{S} = dom_{F} ^{o} C_{Mp}
C_{Ob} ^{o} dom_{S} (f: X –> Y) = C_{Ob} (X) = c_{X}: X –> 1
dom_{F} ^{o} C_{Mp} (f: X –> Y) = dom_{F} (1_{1} ^{o} c_{X} = c_{Y} ^{o} f) = c_{X}: X –> 1
Next we check ‘respecting codomain’:
C_{Ob} ^{o} cod_{S} = cod_{F} ^{o} C_{Mp}
C_{Ob} ^{o} cod_{S} (f: X –> Y) = C_{Ob} (Y) = c_{Y}: Y –> 1
cod_{F} ^{o} C_{Mp} (f: X –> Y) = cod_{F} (1_{1} ^{o} c_{X} = c_{Y} ^{o} f) = c_{Y}: Y –> 1
Next we check ‘respecting identity’:
C_{Mp} ^{o} id_{S} = id_{F} ^{o} C_{Ob}
C_{Mp} ^{o} id_{S} (X) = C_{Mp} (1_{X}: X –> X) = (1_{1} ^{o} c_{X} = c_{X} ^{o} 1_{X}) =
1 — 1_{1} –> 1
^ ^
 
c_{X} c_{X}
 
X — 1_{X} –> X
id_{F} ^{o} C_{Ob} (X) = id_{F} (c_{X}: X –> 1) = (1_{1} ^{o} c_{X} = c_{X} ^{o} 1_{X}) =
1 — 1_{1} –> 1
^ ^
 
c_{X} c_{X}
 
X — 1_{X} –> X
Finally we check ‘respecting composition’:
C_{Mp} (g ^{o} f) = C_{Mp} (g) * C_{Mp} (f)
C_{Mp} (f: X –> Y) = (1_{1} ^{o} c_{X} = c_{Y} ^{o} f) =
1 — 1_{1} –> 1
^ ^
 
c_{X} c_{Y}
 
X — f –> Y
C_{Mp} (g: Y –> Z) = (1_{1} ^{o} c_{Y} = c_{Z} ^{o} g) =
1 — 1_{1} –> 1
^ ^
 
c_{Y} c_{Z}
 
Y — g –> Z
C_{Mp} (g) * C_{Mp} (f) = (1_{1} ^{o} c_{X} = c_{Z} ^{o} g ^{o} f) =
1 — 1_{1} –> 1
^ ^
 
c_{X} c_{Z}
 
X — gf –> Z
C_{Mp} (g ^{o} f) = C_{Mp} (gf: X –> Z) = (1_{1} ^{o} c_{X} = c_{Z} ^{o} g ^{o} f) =
1 — 1_{1} –> 1
^ ^
 
c_{X} c_{Z}
 
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
c_{X}: X –> 1
in F and each function
f: X –> Y
in S to the commutative square (1_{1} ^{o} c_{X} = c_{Y} ^{o} f) =
1 — 1_{1} –> 1
^ ^
 
c_{X} c_{Y}
 
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
we looked at in the context of
where the functor
discrete: S –> F
sends each set
X
in S to its identity function
1_{X}: X –> X
in F. The identity function
1_{X}: 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 1_{X}) somewhat like the unique selves each one of us are. At the other extreme is the property
c_{X}: X –> 1
based on which (i.e. on the values of c_{X} 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
c_{X}: 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) = c_{X}: X –> 1
With these we have to show that
A –> X = points (f –> c_{X}) ^{o} points (c_{X}) –> X
Since
points (c_{X}: X –> 1) = X
we have to show
A –> X = points (f –> c_{X}) ^{o} X –> X
But what is
points (f –> c_{X})
The functor
points: F –> S
sends the commutative square (f –> c_{X})
B –> 1
^ ^
 
f c_{X}
 
A –> X
in F to the function
A –> X
in S, i.e.
points (f –> c_{X}) = 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
1_{X}: X –> X
we have
A –> X = A –> X ^{o} X – 1_{X} –> 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
Unity and Identity of Adjoint Opposites (UIAO)
OK! But what about the functor
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 11 functions
like: B → A
with
like (Sita) = Rama
and
hate (Sita) = Ravana
We can, with these two sections (11 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 = 1_{B}: B → B
love º hate = 1_{B}: 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 = e_{0}: A → A
hate º love = e_{1}: 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
e_{0}: A → A
e_{1}: A → A
we find
e_{0} º e_{0} = e_{0}
e_{0} º e_{1} = e_{0}
e_{1} º e_{0} = e_{1}
e_{1} º e_{1} = e_{1}
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
e_{j} º e_{i} = e_{j}
is something to sleep with. One not so thoughtful question is if the above equation (reproduced below)
e_{j} º e_{i} = e_{j}
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.
e_{j} º e_{i} = e_{i}
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 
s_{1}  s_{2}
 v 
B
where we have a pair of sections
s_{1}: B → A
s_{2}: B → A
with a common retract
r: A → B
That structure (i.e. the structure opposite to the above UAIO) is
A
 ^ 
r_{1}  r_{2}
 s 
v  v
B
which is the structure of reflexive graphs (Conceptual Mathematics, page 145) consisting of a pair of retracts
r_{1}: A → B
r_{2}: 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 = 1_{D}: D → D
t º p = 1_{D}: 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 = 1_{D}
t º p = 1_{D}
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 = e_{0}: A → A
p º t = e_{1}: A → A
The above two idempotents, as you would expect, satisfy:
e_{0} º e_{0} = e_{0}
e_{1} º e_{1} = e_{1}
We might as well check:
e_{0} º e_{0} = p º s º p º s
= p º 1_{D} º s
= p º s
= e_{0}
Next:
e_{1} º e_{1} = p º t º p º t
= p º 1_{D} º t
= p º t
= e_{1}
What about
e_{0} º e_{1} = ?
e_{1} º e_{0} = ?
Let’s see
e_{0} º e_{1} = p º s º p º t
= p º 1_{D} º t
= p º t
= e_{1}
Next we calculate
e_{1} º e_{0} = p º t º p º s
= p º 1_{D} º s
= p º s
= e_{0}
Collecting all four
e_{0} º e_{0} = e_{0}
e_{0} º e_{1} = e_{1}
e_{1} º e_{0} = e_{0}
e_{1} º e_{1} = e_{1}
and summing up as
e_{j} º e_{i} = e_{i}
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
e_{j} º e_{i} = e_{j}
In any case, now that we have a reflexive graph
A
 ^ 
s  t
 p 
v  v
D
satisfying
s º p = 1_{D}
and
t º p = 1_{D}
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
G_{A}
 ^ 
s_{G}  t_{G}
 p_{G} 
v  v
G_{D}
to a reflexive graph G’
G’_{A}
 ^ 
s_{G’}  t_{G’}
 p_{G’} 
v  v
G’_{D}
is a pair of set maps
f_{A}: G_{A} → G’_{A}
f_{D}: G_{D} → G’_{D}
satisfying three equations
f_{D} º s_{G} = s_{G’} º f_{A}
f_{A} º p_{G} = p_{G’} º f_{D}
f_{D} º t_{G} = t_{G’} º f_{A}
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 = 1_{D} and t º p = 1_{D}
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…
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 outthere, 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
 how we think about the things that are
and
 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 selffoundations. If you are still a nonbeliever, then start with
and then make friends with
 Small family of objects (Conceptual Mathematics, pp. 150 – 1, 369 – 71; Sets for Mathematics, pp. 154 – 5, 248 – 50)
along with
 Separating (Conceptual Mathematics, page: 215, 245)
and not to be forgotten are
 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 nonisomorphic 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} = 1_{Furniture} and f ^{1} f = 1_{Animals}
where
f ^{1}: Furniture → Animals
with
f ^{1} (chair) = cat and f ^{1} (table) = dog
In general, in the category of sets, all sets with 1 point are isomorphic, and so is the case with 2point sets, 3point 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
1_{1}: 1 → 1
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
1 –p–> A
 
1_{1} f
 
v v
1 –q–> 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 nonisomorphic 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 nonisomorphic 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):
The functor
discrete: Sets → Functions
mapping each object (set)
X
in the domain category of sets to its identity function
1_{X}: 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) = 1_{X}: 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 1_{X} we do have
X ––> A = X – 1_{X} –> 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
1_{X}: 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 (f → discrete (X)) º pieces (discrete (X)) → X
where º denotes composition.
With
f: A → B
pieces (f) = B
discrete (X) = 1_{X}: X → X
pieces (discrete (X)) = X
pieces (f → discrete (X)) = B → X
we have to show that
B → X = B → X º X → X
and taking the endomap on X as identity 1_{X} we do have
B ––> X = B ––> X – 1_{X} –> X
B → X = B → X
I’m terribly sorry: I don’t think I can get to marrying discrete objects to selffoundations 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 bitesized blogposts, which are not living up to the original idea of making otherwise challenging concepts (e.g. selffoundations 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 fixedstate. These fixedstates are also called points since they can be listed using maps from the terminal object
1_{1}: 1 → 1
(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
f ◦ id: A → B
f ◦ sq: 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.
f ◦ id = f ◦ sq
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 f ◦ id = f ◦ sq) 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 = f ◦ sq). In other words, there’s a unique determination
g: Y → Z
such that
f = g ◦ f^{*}
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
(Conceptual Mathematics, page 294).
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
f ◦ id = f ◦ sq
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
(Conceptual Mathematics, page 294).
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 longlost electrical engineering? In any case, we are about to undress one.
Any map
f: A → B
induces a map
i: π_{0}A → π_{0}B
making the diagram
π_{0}A –– i ––> π_{0}B
^ ^
 
c_{A} c_{B}
 
A –– f ––> B
commute.
First note that π_{0}A and π_{0}B are spaces of components of A and B, respectively, which means that the two maps
c_{A}: A → π_{0}A
and
c_{B}: B → π_{0}B
are universal components maps.
Going back to the diagram
π_{0}A –– ? ––> π_{0}B
^ ^
 
c_{A} c_{B}
 
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 –– c_{B} ––> π_{0}B
to get the map
A –– c_{B} º f ––> π_{0}B
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 –– c_{B} º f ––> π_{0}B
and the universal components map for A i.e.
A –– c_{A} ––> π_{0}A
we realize (based on the definition of space of components) that there is exactly one map
i: π_{0}A → π_{0}B
from the space of components (π_{0}A) of A to π_{0}B (a discrete object as far as A is concerned), which when composed with
c_{A}: A → π_{0}A
gives
c_{B} º f: A → π_{0}B
i.e.
i º c_{A} = c_{B} º f
which says that the diagram
π_{0}A –– i ––> π_{0}B
^ ^
 
c_{A} c_{B}
 
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. slowthinking) 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 structurepreserving 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 respectingstructure and scientificmethod.
Many, if not all, mathematical calculations (e.g. solving equations for unknowns) involve structurepreserving morphisms, which raises the following question:
Is mathematical knowing [invariably] structurepreserving?
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 respectingstructure is the [basic / fundamental] method of knowing / the scientific method? Phrased differently, what does the fact that many (all?) calculations [of unknowns] are structurepreserving 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 α^{n} x = α^{m} y. If moreover y is connected to a third state z, for example if α^{k} y = α^{p} 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
α^{n} x = α^{m} 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
α^{n} x
which is same as the state
α^{m} y
I got myself into after pressing that α button m times, having started at the state
y
Since we ended up together
α^{n} x = α^{m} 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
α^{n} x = α^{m} y
Any number k of button (α) pressing will only move them, but together
α^{k} (α^{n} x) = α^{k} (α^{m} y)
α^{k + n} x = α^{k + m} y
We are also told that y is connected to z, which means
α^{k} y = α^{p} z
for some natural numbers k and p.
If we now press that α button m times, we find
α^{m} (α^{k} y) = α^{m} (α^{p} z)
α^{m + k} y = α^{m + p} z
Since
k + m = m + k
α^{k + n} x = α^{k + m} y = α^{m + k} y = α^{m + 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
a^{n} (ON) = a^{m} (OFF)
Let’s say that the bulb initially was OFF and you press the button once to get to
a^{1} (OFF) = ON
Let’s now say that the bulb initially was ON and I press the button twice to get to
a^{2} (ON) = a^{1} (a^{1} (ON)) = a^{1} (OFF) = ON
Since
a^{1} (OFF) = a^{2} (ON)
we can safely say that the two states
ON, OFF
are connected.
Is our light bulb
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: 1 → 1)
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 = {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 1point 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 :(