Robert, List: I realize that *in general* the same term "can be used without problem in a field other than the one you are used to." My question was really whether there is any significance to the *specific* use of the same term, "category," for 1ns/2ns/3ns by Peirce and for "a mathematical object" by modern mathematicians.
I sincerely appreciate the additional attempt to explain below, but I am afraid that I still cannot detect an answer despite reading it several times, because it remains couched entirely in the unfamiliar vocabulary of mathematical category theory--"morphism," "composition," "functor," "natural transformation," etc. I do not have the requisite acquaintance with that particular system of signs to interpret it successfully. I am also still wondering if your analysis somehow logically *requires* the order of the interpretant trichotomies to be Ii→Id→If, rather than the other way around. Regards, Jon Alan Schmidt - Olathe, Kansas, USA Professional Engineer, Amateur Philosopher, Lutheran Layman www.LinkedIn.com/in/JonAlanSchmidt - twitter.com/JonAlanSchmidt On Thu, May 7, 2020 at 5:06 AM Robert Marty <robertmarty...@gmail.com> wrote: > Jon Alan, Helmut, Edwina, List > > JAS > "Unfortunately I am not adept enough with mathematical category > theory to make heads or tails of Robert's exposition below. It still seems > to me that "category" means something quite different in that context than > it does for Peirce when he is writing about 1ns, 2ns, and 3ns. Am I > wrong? If so, I would appreciate some further explanation of how they > relate to each other." > > > > RM > I'd love to. The term "category" can be used without problem in a > field other than the one you are used to. In the category theory it refers > to a mathematical object in the usual sense chosen by Peirce as" "a > construction independent of its real existence". An industrial property law > allows the same name to be used for products that are in very remote areas > of the economy such as Corona for a beer or for a virus (which is not > without danger, hence Covid-19!). You see the comparison ...😉 > > Nb: I noted that you are a professional engineer; your training should > allow you to understand the following where it is only definitions. No > theorem, no specific technique; definitions, I stress that point. > > > > 1 - https://en.wikipedia.org/wiki/Category_(mathematics) > > *Definition of category* > > There are many equivalent definitions of a category.[2] > <https://en.wikipedia.org/wiki/Category_(mathematics)#cite_note-2> One > commonly used definition is as follows. A *category* *C* consists of > > · a class <https://en.wikipedia.org/wiki/Class_(set_theory)> ob( > *C*) of *objects* > > · a class hom(*C*) of *morphisms > <https://en.wikipedia.org/wiki/Morphism>*, or *arrows*, or *maps*, > between the objects. Each morphism *f* has a *source object a* and a *target > object b* where *a* and *b* are in ob(*C*). We write *f*: *a* → *b*, and > we say "*f* is a morphism from *a* to *b*". We write hom(*a*, *b*) (or hom > *C*(*a*, *b*) when there may be confusion about to which category hom(*a*, > *b*) refers) to denote the *hom-class* of all morphisms from *a* to *b*. > (Some authors write Mor(*a*, *b*) or simply *C*(*a*, *b*) instead.) > > · for every three objects *a*, *b* and *c*, a binary operation > hom(*a*, *b*) × hom(*b*, *c*) → hom(*a*, *c*) called *composition of > morphisms*; the composition of *f* : *a* → *b* and *g* : *b* → *c* is > written as *g* ∘ *f* or *gf*. (Some authors use "diagrammatic order", > writing *f;g* or *fg*.) > > such that the following axioms hold: > > · (associativity <https://en.wikipedia.org/wiki/Associativity>) > if *f* : *a* → *b*, *g* : *b* → *c* and *h* : *c* → *d* then *h* ∘ (*g* ∘ > *f*) = (*h* ∘ *g*) ∘ *f*, and > > · (identity <https://en.wikipedia.org/wiki/Identity_(mathematics)>) > for every object *x*, there exists a morphism 1*x* : *x* → *x* (some > authors write *idx*) called the *identity morphism for x*, such that for > every morphism *f* : *a* → *x* and every morphism *g* : *x* → *b*, we > have 1*x* ∘ *f* = *f* and *g* ∘ 1*x* = *g*. > > From these axioms, one can prove that there is exactly one identity > morphism for every object. Some authors use a slight variation of the > definition in which each object is identified with the corresponding > identity morphism. > > > > *( ! **) By extraordinary the example at the top right is interpretable > with* > > > > A = 3ns ; B = 2ns ; C = 1ns ; f = involvesβ ; g = involvesα ; g o f = β o > α ; let's name C this category > > But also > > > > A = O ; B = S ; C = I ; f = det1 ; g = det2 ; g o f = det2 o det1 ; > let's name *D* this category > > > > *Remember these two interpretations, * *they will* *serve ....* > > > > 2 - https://en.wikipedia.org/wiki/Functor > > *Definition** of functor* > > Let *C* and *D* be categories > <https://en.wikipedia.org/wiki/Category_(mathematics)>. > A *functor* *F* from *C* to *D* is a mapping that > <https://en.wikipedia.org/wiki/Functor#cite_note-FOOTNOTEJacobson2009p._19,_def._1.2-3> > associates to each object X {\displaystyle X} X in *C* an object F ( X ) > {\displaystyle F(X)} f(X) in *D*, > > · associates to each morphism f : X → Y {\displaystyle f\colon > X\to Y} f : XàY in *C* a morphism F ( f ) : F ( X ) → F ( Y ) > {\displaystyle F(f)\colon F(X)\to F(Y)} f(X)àY in *D* such that the > following two conditions hold: F ( i d X ) = i d F ( X ) {\displaystyle > F(\mathrm {id} _{X})=\mathrm {id} _{F(X)}\,\!} > > · F(idX) = idF(X) for every object X {\displaystyle X} in *C*,F ( > g ∘ f ) = F ( g ) ∘ F ( f ) {\displaystyle F(g\circ f)=F(g)\circ F(f)} > > > > · F(g o f) = F(g) o F(f) for all morphisms f : X → Y > {\displaystyle f\colon X\to Y\,\!} f ; XàY and g : Y → Z {\displaystyle > g\colon Y\to Z} g : YàZ in *C*. > > That is, functors must preserve identity morphisms > <https://en.wikipedia.org/wiki/Morphism#Definition> and composition > <https://en.wikipedia.org/wiki/Function_composition> of morphisms. > > __________________________ > > *Exercise 1* 😉 > > Build a functor of *(C)* in (*D)*). > *Graphic Hint:* for this it is necessary to connect the elements of C to > those of D by 3 arrows avoiding any intersection > > *Exercice 2* > > Build all the functors of *(C)* in (*D*) > > *Answer: *there are exactly 10 funtors (C*C*) in (*D*) > > > > 3 – https://en.wikipedia.org/wiki/Natural_transformation > > *Definition of natural transformation of functors* > > If F {\displaystyle F} F and G {\displaystyle G} G are functors > <https://en.wikipedia.org/wiki/Functor> between the categories *C > {\displaystyle C} C* and *D* D {\displaystyle D} , then a *natural > transformation* η {\displaystyle \eta } µ from F {\displaystyle F} F to G > {\displaystyle G} G is a family of morphisms that satisfies two > requirements. > *3.1* - The natural transformation must associate, to every object X > {\displaystyle X} X in *C {\displaystyle C} C* , a morphism > <https://en.wikipedia.org/wiki/Morphism> > > η X : F ( X ) → G ( X ) {\displaystyle \eta _{X}:F(X)\to G(X)} µX : F(X) > à G(X) between objects of *D {\displaystyle D} D*. The morphism η X : F > ( X ) → G ( X ) {\displaystyle \eta _{X}:F(X)\to G(X)} µX η X > {\displaystyle \eta _{X}} is called the *component* of η {\displaystyle > \eta } µ at X {\displaystyle X} X. > > *3.2-* Components must be such that for every morphism f : X → Y > {\displaystyle f:X\to Y} f : XàY in *C {\displaystyle C} C *we have: > > µY o F(f) = G(f) o µXη Y ∘ F ( f ) = G ( f ) ∘ η X {\displaystyle \eta > _{Y}\circ F(f)=G(f)\circ \eta _{X}} > > The last equation can conveniently be expressed by the commutative diagram > <https://en.wikipedia.org/wiki/Commutative_diagram> > > X F(X) ---------µX--------à G(X) > | | | > f | F(f) | | G(f) > v v v > Y F(Y)---------- µY --------à G(Y) > > *Nb*: *I am obliged to rewrite all the diagrams and even the letters that > are images in Wikipedia and I put **µ** in place of "eta".* > > If µ η {\displaystyle \eta } is a natural transformation from F > {\displaystyle F} to G {\displaystyle G} F to G, we also write µ : F àG η > : F → G {\displaystyle \eta :F\to G} η : F ⟹ G {\displaystyle \eta > :F\implies G} . > > *Exercise* 3: Choose from the 10 functors found in Exercise 2 two > functors for which there is a natural transformation. > > *Graphic Hint*: To do this, you have to link the elements of F to the > elements of G by 3 arrows, avoiding any intersection. > > *Exercice 4* : find all possible natural transformations and make sure > they get the lattice of the sign classes! > > ! > > That all ! 😊 > > Best regards, > > Robert > >>
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