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This document is draft.

Table of Contents

• Degrees of endomorphisms
• Definition
  • Terminological notes
• Groupoid of endomorphisms
• Category of elements
  • Categories of binary relations
  • Projection of a morphism
  • Category of elements
    • Category of elements of endomorphisms
  • Image of an object or a set
• Limiting and ancestry
  • Limiting a morphism
  • Endomorphism series
  • Ancestry
  • Ancestry morphisms and ancestry categories

I suggest to add this (or at least a part of this) to programs of math faculties of Universities as a sub-course of Category Theory course. Probably, this is especially important for students specializing in Math Logic of Computer Science, it is also important for algebraists. This article quite suits to be used as a studybook.

Category of pseudomorphisms is used in my algebraic theory of formulas (expressions) which is developed by me for purposes of math logic and computer science.

This is a generalization of the concept of dependencies and categories of dependencies as defined in my earlier articles on this topic.

This variant of the article is severely revised compared with old variant. Some materials are removed as unnecessary.

In this article I use a little generalized definition of category which allows the same morphism to have the same source and destination. (In other words it does not require the sets of morphisms for different pairs of objects to be disjoint.)

Moreover I use the more general notion of precategory instead of category. (Precategories are simply categories without identity morphisms.)

Degrees of endomorphisms

An endomorphism is such a morphism $f$ that $f\in Mor\left(A,A\right)$ for some object $A$. We call $A$ an object of the endomorphism $f$.

Below we will need the concept of degrees of an endomorphism.

Let $U$ is an endomorphism of an object $A$ of some precategory. Then $U^i$ is defined for any $i=1,2,3,\dots$ and $U^i$ is also an endomorphism of the object $A$.

For categories additionally we can define: $U^0=1_A$.

Obvious  $U^j\circ U^i=U^{i+j}$ for any $i,j=0,1,2,\dots$ ($i,j=0$ is defined only for these precategories which are categories).

Definition

Let we have some precategory (original precategory) and let the set of its morphisms is partially ordered (by relation $\subseteq$). Then we can define a new precategory which I call category of pseudomorphisms™. (This can be generalized for the case of $\subseteq$ being arbitrary reflexive and transitive relation of morphisms.)

Below we will also consider the case when $\subseteq$ is the equality relation ($=$). In this case the precategory of pseudomorphisms will be instead called precategory of intermorphisms.

When the original precategory is a category (In this case I will call it original category.), the precategories of pseudomorphisms and intermorphisms will be categories and I will call them category of pseudomorphisms and category of intermorphisms.

Also the original precategory may be a semigroup (a precategory with only one object).

Let $U$ and $V$ are endomorphisms of the original category. By definition $f$ is a pseudomorphism from $U$ to $V$ ($f\in Pseud\left(U,V\right)$) if and only if $f$ is a morphism from the object of $U$ to the object of $V$ and $f\circ U\subseteq V\circ f$.

The precategory of pseudomorphisms is defined as follows:

• Objects of this category are endomorphisms of the original category, and the set of morphisms (pseudomorphisms) from $U$ to $V$ is $Pseud\left(U,V\right)$.
• Composition of morphisms (pseudomorphisms) of the category of pseudomorphisms is the same as in the original category.

To prove correctness of the above definition, we need to prove that:

1. Composition of pseudomorphisms is a pseudomorphism.
2. Composition of pseudomorphisms is associative.

◄

Composition of pseudomorphisms is a pseudomorphism.

Let $\left(U,V\right)f$ and $\left(V,W\right)g$ are pseudomorphisms. Then their composition is $\left(U,W\right)\left(g\circ f\right)$.

We have:

• $f\circ U\subseteq V\circ f$;
• $g\circ V\subseteq W\circ g$;

$$g\circ f\circ U\subseteq g\circ V\circ f\subseteq W\circ g\circ f\text{.}$$

So $\left(U,W\right)\left(g\circ f\right)$. is a pseudomorphism.

Associativity.

Directly follows from associativity of composition of morphisms in the original category.

►

Proposition  If the original precategory is a category then the precategory of pseudomorphisms is also a category

◄  Let $U$ is an endomorphism. Then its identity pseudomorphism is $U^0$.  ►

In this case precategory of pseudomorphisms is also called category of pseudomorphisms.

In the case when $\subseteq$ is the equality relation of morphisms of the original precategory, I will call pseudomorphisms intermorphisms™ and the (pre)category of pseudomorphisms (pre)category of intermorphisms. (Formerly I was instead calling these homomorphisms but it was a terminological glitch because in category theory the term homomorphism is used as a synonym of morphism.)

Remark  See also math encyclopedias about intertwiners and intertwining if you are curious about terminology.

Theorem  Let $U$ and $V$ are endomorphisms. Then $f\in Pseud\left(U,V\right)$ if and only if $f\in Pseud\left(U^n,V^n\right)$.

◄

The reverse implication is obvious.

Let now $f\in Pseud\left(U,V\right)$. That $f\in Pseud\left(U^0,V^0\right)$ (if the original precategory is a category) is obvious.

We will prove $f\in Pseud\left(U^n,V^n\right)$ for $n>0$ by induction. For $n=1$ it is given to be true. Let it is true for $n=k$, that is $f\circ U^k\subseteq V^k\circ f$. Then $$f\circ U^{k+1}=f\circ U^k\circ U\subseteq V^k\circ f\circ U\subseteq V^k\circ V\circ f=V^{k+1}\circ f\text{.}$$

►

Terminological notes

The full names for the introduced (pre)categories (for relations $\subseteq$ and $=$ correspondingly) are:

• (pre)category of endomorphisms and pseudomorphisms™;
• (pre)category of endomorphisms and intermorphisms™.

In practice it is more convenient to use shorter names:

• (pre)category of pseudomorphisms™;
• (pre)category of intermorphisms™.

Groupoid of endomorphisms

Theorem

If $U$ and $V$ are endomorphisms of the original category, then the following statements are equivalent:

1. $f$ is an isomorphism from $U$ to $V$ in the sense of the category of pseudomorphisms.
2. $f$ is an isomorphism from $U$ to $V$ in the sense of the category of intermorphisms.
3. $f$ is an intermorphism from $U$ to $V$ and $f$ is an isomorphism from $U^0$ to $V^0$ in the sense of the original category.

◄

$\text{(2)}\Rightarrow \text{(1)}$

Obvious.

$\text{(2)}\Rightarrow \text{(3)}$

If $f$ is an isomorphism from $U$ to $V$ in the category of intermorphisms then $f^{-1}\circ f=1_{U^0}$, $f\circ f^{-1}=1_{V^0}$ where the reverse is understood in the sense of the category of intermorphisms. So $f$ is also an isomorphism of the original category.

$\text{(3)}\Rightarrow \text{(2)}$

Let $f$ is an intermorphism from $U$ to $V$ ($f\circ U=V\circ f$) and $f$ is an isomorphism of the original category. Then reverse isomorphism $f^{-1}$ exists. So $f^{-1}\circ f$ and $f\circ f^{-1}$ are identity morphisms.

Multiplying the formula above with $f^{-1}$ at both left and right sides we get $U\circ f^{-1}=f^{-1}\circ V$. So $f^{-1}$ is an intermorphism from $V$ to $U$. It is the reverse of the intermorphism $f$. And so $f$ is an isomorphism in the sense of the category of intermorphisms.

$\text{(1)}\Rightarrow \text{(2)}$

Let $f$ is an isomorphism from $U$ to $V$ in the category of pseudomorphisms ($f\circ U\subseteq V\circ f$). Then for the reverse (in the sense of the category of pseudomorphisms) morphism $f^{-1}$ holds $f^{-1}\circ f=1_{U^0}$, $f\circ f^{-1}=1_{V^0}$ and $f^{-1}\circ V\subseteq U\circ f^{-1}$. Multiplying the last formula with $f$ at both left and right sides, we get $V\circ f\subseteq f\circ U$. Comparing this formula with the above, we get $f\circ U\subseteq V\circ f$. So $f$ is an intermorphism from $U$ to $V$. Multiplying the last formula with $f^{-1}$ at both left and right sides we get $U\circ f^{-1}=f^{-1}\circ V$, so $f^{-1}$ is an intermorphism from $V$ to $U$. It is the reverse intermorphism of $f$, so it is an isomorphism in the category of intermorphisms.

►

We can call morphisms conforming to the equivalent statements of the above theorem isomorphisms between endomorphisms (or more specifically e.g. isomorphisms between systems of formulas, when an endomorphism will represent a system of formulas).

BTW, it defines the groupoid of endomorphisms for any given category. (There are no requirement for morphisms to be partially ordered.) I recall that groupoid is such a category all morphisms of which are isomorphisms.

Category of elements

In this section we will descend from morphisms to relations of individual elements.

We will consider (pre)categories of binary relations. Morphisms of these categories are binary relations and objects are sets. There are several different (pre)categories of binary relations having different sets of morphisms $Mor\left(A,B\right)$ for given objects (sets) $A$ and $B$.

Categories of binary relations

The following categories of binary relations are important:

• $Mor\left(A,B\right)$ is the set of all binary relations which are subsets of $A\times B$ (category of all binary relations);
• $Mor\left(A,B\right)$ is the set of all binary relations whose domain is exactly $A$ and whose image is a subset of $B$ (category of binary relations defined on entire source);
• the previous with additional requirement that the relations are monovalued (the famous category $\mathbf{Set}$ of functions between sets).

Identity morphisms for all these categories are the identity functions on the object (that is on the set).

Projection of a morphism

I will call projection of a precategory a function $Pr$ from the set of morphism to the set of binary relations, distributive regarding composition.

I will call projection of a category a projection of a precategory which maps identity morphisms to subsets of equality relation.

For example, any functor to the category of binary relations is essentially a projection.

Problem for the reader  Generalize the definition of projection for the case of arbitrary partially ordered semigroup instead of the semigroup of binary relations.

Normally, $Pr$ is a monotonous function (regarding $\subseteq$), but we do not need this property:

Having projection also allows to define for any morphism $f$ its set theoretic domain and image:

• $imf=imPrf$;
• $domf=domPrf$.

Remark  In category theory the words image and domain can be used in other sense, as synonyms of source and destination.

We will call a morphism monovalued if and only if its projection is monovalued.

Also having projection allows to restrict a (pre)category to a sub(pre)category corresponding to a subcategory of the category of all binary relations (see above).

Category of elements

I will call source elements pairs $\left(A,a\right)$, where $A$ is an object of the original precategory, and there exists such morphism $f$ of the original category having $A$ as its source that $a\in domf$.

I will call destination elements pairs $\left(B,b\right)$, where $B$ is an object of the original precategory, and there exists such morphism $f$ of the original category having $B$ as its destination that $b\in imf$.

I will call elements these pairs which are either source elements or destination elements (or both).

The precategory of elements is the precategory:

• whose objects are all elements;
• whose morphisms (element morphisms) from element $\left(A,a\right)$ to element $\left(B,b\right)$ are such morphisms $f$ of the original category from $A$ to $B$ that $\left(a,b\right)\in Prf$;
• whose composition of morphisms is the same as in the original category.

To prove correctness of the above definition we need to prove that composition of two element morphisms is an element morphism.

◄

If $f$ and $g$ are element morphisms from an element $\left(A,a\right)$ to an element $\left(B,b\right)$ and from an element $\left(B,b\right)$ to an element $\left(C,c\right)$ correspondingly, then $\left(a,c\right)\in Pr\left(g\circ f\right)=Prg\circ Prf$ because $\left(a,b\right)\in Prf$, $\left(b,c\right)\in Prg$. So $g\circ f$ is an element morphism from an element $\left(A,a\right)$ to an element $\left(C,c\right)$.

►

Proposition  If the original precategory is a category then the precategory of elements is a category.

◄  I will prove it for source elements (for destination elements the proof is analogous). Let $\left(A,a\right)$ is a source element. Then $a\in domf$ for some morphism $f$ which has $A$ as its source. Then $a\in domPrf$. $f\circ 1_A=f$; $Prf\circ Pr{1}_A=Prf$. Because $Pr{1}_A$ is a subset of identity relation, $Pr{1}_A=I_{domPrf}$; $\left(a,a\right)\in Pr{1}_A$. Consequently $1_A$ is an identity morphism of $\left(A,a\right)$.  ►

Proposition  For categories source and destination elements are the same.

◄  Having a source element $\left(A,a\right)$ we have $1_A$ being identity morphism of $\left(A,a\right)$. So it is also a destination element. The reverse is analogous.  ►

So in the case if the original precategory is a category, I will call the precategory of elements category of elements.

There exists a prefunctor from the precategory of elements to the original precategory, which maps element $\left(A,a\right)$ to $A$ and $f$ to itself. It is really a prefunctor because it preserves composition.

If the original precategory is a category, it is a functor, because it preserves identities.

By this (pre)functor in an obvious way (as inverse image) to every sub(pre)category of the original (pre)category corresponds some sub(pre)category of the (pre)category of elements. So we can speak about (pre)categories of isomorphic, pseudomorphic, intermorphic, etc. elements.

When two elements are related with a morphism of such a precategory, I will call second of these elements isomorphic, pseudomorphic, intermorphic, etc. image of the first one.

Proposition  If two object of the category of elements $\left(A,a\right)$ and $\left(B,b\right)$ are isomorphic, then objects $A$ and $B$ are also isomorphic.

◄  Because any functor maps an isomorphism to isomorphism.  ►

Remark

The above proposition reflects the fact that an element encapsulates the structure of the object (like as in object oriented programming an instance encapsulates the structure of the class), not just an element of a set. So every element contains all information about the structure of the object.

Two elements are isomorphic means that these elements belong to isomorphic objects and are in equivalent places of that objects.

Category of elements of endomorphisms

For the precategory of pseudomorphisms (or category of intermorphisms) we will define $Pr$ simply as $Pr$ of the original category restricted to the set of pseudomorphisms (or intermorphisms).

So for the (pre)category of pseudomorphisms is also defined the (pre)category of elements. I will call this (pre)category (in relation to the original (pre)category) the (pre)category of elements and pseudomorphisms (or the (pre)category of elements and intermorphisms in the case of intermorphisms instead of pseudomorphisms).

• Objects of this (pre)category (elements of endomorphisms) are certain pairs $\left(U,a\right)$ where $U$ is an endomorphism of the original (pre)category.
• Morphisms of this (pre)category are morphisms of the original (pre)category. $f$ is a morphism from $\left(U,a\right)$ to $\left(V,b\right)$ if and only if $f\in Pseud\left(U,V\right)$ and $\left(a,b\right)\in Prf$.

So we can speak about (pre)categories of isomorphic, pseudomorphic, intermorphic, etc. elements of endomorphisms. When we will come to study of endomorphisms representing systems of expressions, we will so speak about isomorphic, pseudomorphic, intermorphic, etc. formulas.

Image of an object or a set

A relation $f$ applied to a set $A$ is by definition the set $$f\left(A\right)=\left\{b|\exists a:\left(a,b\right)\in f\right\}\text{.}$$

By definition a morphism $f$ applied to set $A$ is $f\left(A\right)=\left(Prf\right)\left(A\right)$

Obvious  (For our original category) image of an object by composition of morphisms is the image of the image of this object, that is $\left(g\circ f\right)\left(A\right)=g\left(f\left(A\right)\right)$.

Obvious  $imf=f\left(domf\right)$ for a morphism $f$ of our original category.

Obvious  $f\left(A\cup B\right)=f\left(A\right)\cup f\left(B\right)$ for sets $A$ and $B$.

A set $A$ is called closed regarding a morphism (or a binary relation) $f$ if and only if $f\left(A\right)\subseteq A$.

Limiting and ancestry

Limiting a morphism

As a generalization of limiting a binary relation $f$ to a Cartesian product $A\times B$ of sets $A$ and $B$ $$f\left[A\times B\right]=f\cap \left(A\times B\right)\text{,}$$ let rectangular limiting a morphism $f$ to a direct product of sets $A$ and $B$ is an operation on the set of morphisms conforming to the following axioms:

• $Pr\left(f\left[A\times B\right]\right)=\left(Prf\right)\cap \left(A\times B\right)$;
• $f\left[A_1\times B_1\right]\left[A_2\times B_2\right]=f\left[\left(A_1\cap A_2\right)\times \left(B_1\cap B_2\right)\right]$;
• $Prf\subseteq A\times B\Rightarrow f\left[A\times B\right]=f$.

Then we can define limiting and square limiting a morphism $f$ to a set $A$ corresponding as:

• $f{|}_A=f\left[A\times imf\right]$;
• $f\square A=f\left[A\times A\right]$.

Additionally I will require that square limiting of an endomorphism is also an endomorphism.

Remark  In fact we will need limiting operation only for endomorphisms.

Proposition

• $\left(f{|}_A\right){|}_A=f{|}_A$;
• $\left(f\square A\right)\square A=\left(f\square A\right){|}_A=\left(f{|}_A\right)\square A=f\square A$;

for any morphism $f$ and a set $A$.

◄  Easy to prove taking into account that $dom\left(f{|}_A\right)\subseteq A$, $dom\left(f\square A\right)\subseteq A$, $im\left(f\square A\right)\subseteq A$.  ►

Proposition  $f\left(A\right)=im\left(f{|}_A\right)$ that is image of a set by a morphisms is image of the morphism limited to the set.

◄  $im\left(f{|}_A\right)=im\left(\left(Prf\right){|}_A\right)=\left(Prf\right)\left(A\right)=f\left(A\right)$.  ►

Proposition  If $f$ is closed on a set $A$ then $f\square A=f{|}_A$.

◄  From proved above $im\left(f{|}_A\right)\subseteq A$; also $dom\left(f{|}_A\right)\subseteq A$. So $\left(f{|}_A\right)\square A=f{|}_A$. Taking in account a proposition above we produce what we need to prove.  ►

Endomorphism series

For a binary relation $U$ I will define $U^0=I_{domU\cup imU}$ (identity relation on the set $domU\cup imU$).

For a binary relation $U$ I will define:

• $S_1\left(U\right)=U\cup U^2\cup U^3\cup \dots$;
• $S\left(U\right)=U^0\cup S_1\left(U\right)=U^0\cup U^1\cup U^2\cup \dots$.

For a precategory with projection we can define:

• $S_1\left(U\right)=S_1\left(PrU\right)$;
• $S\left(U\right)=S\left(PrU\right)$.

Proposition

The following statements are equivalent for an endomorphism $U$ and a set $A$:

1. $A$ is closed regarding $U$.
2. $A$ is closed regarding $S\left(U\right)$.
3. $A$ is closed regarding $S_1\left(U\right)$.

◄

$\text{(2)}\iff \text{(3)}$

Obvious.

$\text{(3)}\iff \text{(1)}$

Obvious.

$\text{(1)}\iff \text{(3)}$

It is enough to prove that if a set if closed regarding $U$ then it is closed regarding $U^n$ for any positive whole $n$. This can be easily proved by induction.

►

Proposition  $\left(S\left(U\right)\right)\left(A\right)$ is closed regarding $U$ and regarding $S\left(U\right)$ (for any set $A$ and endomorphism $U$).

◄  It is enough to prove for the case when $U$ is a binary relation. We need to prove that $U\left(\left(S\left(U\right)\right)A\right)\subseteq \left(S\left(U\right)\right)A$. But it is obvious.  ►

Ancestry

By definition exclusive ancestry™ and ancestry™ of a set $A$ by an (endo)morphism $U$ (of the original precategory) are correspondingly the (endo)morphisms

• ${Fam}_1\left(U,A\right)=U\square \left(S_1\left(U\right)\right)\left(A\right)$;
• $Fam\left(U,A\right)=U\square \left(S\left(U\right)\right)\left(A\right)$.

(Recall that square limiting of an endomorphism is an endomorphism.)

Remark  $Fam$ is an abbreviation of the English word family.

Ancestry and exclusive ancestry of an element of a pseudomorphism are defined correspondingly as:

• ${Fam}_1\left(U,a\right)={Fam}_1\left(U,\left\{a\right\}\right)$;
• $Fam\left(U,a\right)=Fam\left(U,\left\{a\right\}\right)$.

Example

Oriented graphs are morphisms of a category whose objects are sets of points, composition of two orgraphs is simply their set theoretic union, identity morphism of a set of points is a discrete graph (graph without vertices).

Ancestry of a point of an (oriented) graph is the subgraph which can be reached from this point. (Inclusive ancestry will contain this point, and exclusive ancestry does not necessarily contain this point.)

Proposition  $\left(Fam\left(U,A\right)\right)\left(B\right)=U\left(B\right)$ for any set $B\subseteq S\left(U\right)\left(A\right)$.

◄  $\left(Fam\left(U,A\right)\right)\left(B\right)=\left(\left(PrU\right)\cap \left(S\left(U\right)\left(A\right)\times S\left(U\right)\left(A\right)\right)\right)\left(B\right)=S\left(U\right)\left(A\right)\cap \left(PrU\right)\left(B\right)=S\left(U\right)\left(A\right)\cap U\left(B\right)=U\left(B\right)$ (I have taken in account that $\left(S\left(U\right)\right)\left(A\right)$ is closed regarding $A$.)  ►

Proposition  ${\left(Fam\left(U,A\right)\right)}^n\left(B\right)=U^n\left(B\right)$ for any set $B\subseteq S\left(U\right)\left(A\right)$ and positive whole $n$ (for categories also when $n=0$).

◄  It can be easily proved by math induction, taking in account that $B\subseteq S\left(U\right)\left(A\right)\Rightarrow U\left(B\right)\subseteq S\left(U\right)\left(A\right)$  ►

Consequence  ${\left(Fam\left(U,A\right)\right)}^n\left(A\right)=U^n\left(A\right)$.

Let $\left(U,a\right)$ is an element of an endomorphism. I will call the corresponding detached element the pair $\left(Fam\left(U,a\right),a\right)$.

I will call a head element such element of an endomorphism which does not change being detached. Equivalently: a head element of an endomorphism $U$ is such element of this endomorphism whose ancestry is $U$.

Remark  When a system of an expression and all its subexpressions is represented as an endomorphism, head element corresponds to the main expression (as opposed to subexpressions).

BTW, now we can define two way connected endomorphism: an endomorphism is two way connected if and only if each its element is head.

Ancestry morphisms and ancestry categories

Remark  The terminology defined in this subsection is yet unstable and may vary in future works.

Remark  Ancestry morphisms (see below) are important for the theory of formulas where e.g. two formulas (expressions) being ancestry isomorphic means that they have the same structure of subformulas (subexpressions).

I will call the ancestry (pre)category™ of isomorphisms (pseudomorphisms, intermorphisms, etc.) the following category:

• Objects are head elements of endomorphisms.
• Morphisms from an object $\left(U,a\right)$ to an object $\left(V,b\right)$ are are are triples $\left(f,a,b\right)$ where $f$ is an isomorphism (pseudomorphism, intermorphism, etc.) between $U$ and $V$.
• Composition of morphisms is naturally induced by composition of pseudomorphisms.

I leave to the reader the trivial problem to prove that it is really a category.

This category can be intersected with the category of elements of endomorphisms. I call the resulting category strict ancestry category of isomorphisms (pseudomorphisms, intermorphism, etc.)

So we can speak about (strictly) ancestry isomorphic, pseudomorphic, intermorphic, etc. head elements of endomorphisms.

This can be generalized for any elements of endomorphisms calling them (strictly) ancestry isomorphic (pseudomorphic, intermorphic, etc.) if and only if the corresponding detached elements are (strictly) ancestry isomorphic (pseudomorphic, intermorphic, etc.) In this version of the article I leave a problem to the reader to define the corresponding categories.

Related Links

• Category theory pages on this site.
• Theory of Dependencies (on this site).
• Definition of category with multiple domains and ranges.
• www.mathematics21.org
• Theory of Formulas (on this site).
• Journal of post-Axiomatic Mathematics and Logic.

License

See here about licensing of this text etc.

Related Web Categories

See also my suggestions for new mathematics classification categories.

• /Science/Math/Algebra/Category_Theory/
• /Science/Math/Publications/Online_Texts/
• http://www.3apes.com/directory/index.cgi?page=category_theory

Copyright © 2005 Victor Porton

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