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22
_chapters/01_set.md
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@ -131,11 +131,11 @@ Note that the question "Who is my child?" is *NOT* a straightforward function, b
The Identity Function
---
For every set $G$, no matter what it represents, we can define the function that does nothing, or in other words, a function which maps every element of $G$ to itself. It is called *the identity function* of $G$ or $idG: G → G$.
For every set $G$, no matter what it represents, we can define the function that does nothing, or in other words, a function which maps every element of $G$ to itself. It is called *the identity function* of $G$ or $ID_{G}: G → G$.
![The identity function](../01_set/function_identity.svg)
You can think of $idG$ as a function which represents the set $G$ in the realm of functions. Its existence allows us to prove many theorems, that we "know" by intuition, formally.
You can think of $ID_{G}$ as a function which represents the set $G$ in the realm of functions. Its existence allows us to prove many theorems, that we "know" by intuition, formally.
Functions and Subsets
---
@ -314,7 +314,7 @@ If we apply the first function $g$ to some element from set $Y$, we will get an
We can define a function that is the equivalent to performing the operation described above.
That would be a function such that, if you follow the arrow $h$ for any element of set $Y$ you will get to the same element of the set $G$ as the one you will get if you follow the $g$ and then follow $f$.
Let us call it $h: Y → G$. We may say that $h$ is the *composition* of $g$ and $f$, or $h = f \bullet g$ (notice that the first function is on the right, so it's similar to $b = f(g(a)$).
Let us call it $h: Y → G$. We may say that $h$ is the *composition* of $g$ and $f$, or $h = f \circ g$ (notice that the first function is on the right, so it's similar to $b = f(g(a)$).
![Functional composition](../01_set/functions_compose.svg)
@ -327,7 +327,7 @@ Composition of relationships
To understand how powerful composition is, consider the following: one set being connected to another means that each function from the second set can be transferred to a corresponding function from the first one.
If we have a function $g: P → Y$ from set $P$ to set $Y$, then for every function $f$ from the set $Y$ to any other set, there is a corresponding function $f \bullet g$ from the set $P$ to the same set. In other words, every time you define a new function from $Y$ to some other set, you gain one function from $P$ to that same set for free.
If we have a function $g: P → Y$ from set $P$ to set $Y$, then for every function $f$ from the set $Y$ to any other set, there is a corresponding function $f \circ g$ from the set $P$ to the same set. In other words, every time you define a new function from $Y$ to some other set, you gain one function from $P$ to that same set for free.
![Functional composition connect](../01_set/morphism_general.svg)
@ -351,7 +351,7 @@ By the way, diagrams that are "zoomed out" that show functions without showing s
Composition and external diagrams
---
Let's look at the diagram that demonstrates functional composition in which we showed that successive application of the two composed functions ($f \bullet g$) and the new function ($h$) are equivalent.
Let's look at the diagram that demonstrates functional composition in which we showed that successive application of the two composed functions ($f \circ g$) and the new function ($h$) are equivalent.
![Functional composition](../01_set/functions_compose.svg)
@ -390,7 +390,7 @@ Invertible functions are called *isomorphisms*. When there exists an invertible
Isomorphism means "same form" in Greek (although actually their form is the only thing which is different between two isomorphic sets).
More formally, two sets $R$ and $G$ are isomorphic (or $R ≅ G$) if there exist functions $f: G → R$ and its reverse $g: R → G$, such that $f \bullet g = idR$ and $g \bullet f = idG$ (notice how the identity function comes in handy).
More formally, two sets $R$ and $G$ are isomorphic (or $R ≅ G$) if there exist functions $f: G → R$ and its reverse $g: R → G$, such that $f \circ g = ID_{R}$ and $g \circ f = ID_{G}$ (notice how the identity function comes in handy).
Isomorphism and identity
---
@ -421,17 +421,17 @@ Composing two isomorphisms into another isomorphism is possible by composing the
Informally, we can see that the two morphisms are indeed reverse to each other and hence form an isomorphism. If we want to prove that fact formally, we will do something like the following:
Given that if two functions are isomorphic, then their composition is equal to an identity function, proving that functions $g \bullet f$ and $f' \bullet g'$, are isomorphic is equivalent to proving that their composition is equal to identity.
Given that if two functions are isomorphic, then their composition is equal to an identity function, proving that functions $g \circ f$ and $f' \circ g'$, are isomorphic is equivalent to proving that their composition is equal to identity.
$g \bullet f \bullet f' \bullet g' = id$
$g \circ f \circ f' \circ g' = id$
But we know already that $f$ and $f'$ are isomorphic and hence $f\bullet f' = id$, so the above formula is equivalent to (you can reference the diagram to see what that means):
But we know already that $f$ and $f'$ are isomorphic and hence $f\circ f' = id$, so the above formula is equivalent to (you can reference the diagram to see what that means):
$g \bullet id \bullet g' = id$
$g \circ id \circ g' = id$
And we know that anything composed with $id$ is equal to itself, so it is equivalent to:
$g \bullet g' = id$
$g \circ g' = id$
which is true, because $g$ and $g'$ are isomorphic and isomorphic functions composed are equal to identity.

2
_chapters/02_category.md
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@ -510,7 +510,7 @@ Identity and isomorphisms
The reason the identity law is required is by far the more obvious one. We need to have a morphism that does nothing? It's because morphisms are the basic building blocks of our language, we need the identity morphism to be able to speak properly. For example, once we have the concept of identity morphism defined, we can have a category-theoretic definition of an *isomorphism* (which is important, because the concept of an isomorphism is very important for category theory):
Like we said in the previous chapter, an isomorphism between two objects ($A$ and $B$) consists of two morphisms --- ($A → B$. and $B → A$) such that their compositions are equivalent to the identity functions of the respective objects. Formally, objects $A$ and $B$ are isomorphic if there exist morphisms $f: A → B$ and $g: B → A$ such that $f \bullet g = id_{B}$ and $g \bullet f = id_{A}$.
Like we said in the previous chapter, an isomorphism between two objects ($A$ and $B$) consists of two morphisms --- ($A → B$. and $B → A$) such that their compositions are equivalent to the identity functions of the respective objects. Formally, objects $A$ and $B$ are isomorphic if there exist morphisms $f: A → B$ and $g: B → A$ such that $f \circ g = ID_{B}$ and $g \circ f = ID_{A}$.
And here is the same thing expressed with a commuting diagram.

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_chapters/02_category/application_external.svg
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4
_chapters/03_monoid.md
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@ -205,7 +205,7 @@ There are abelian groups that are not cyclic, but, as we shall see below, the co
Group isomorphisms
---
We already mentioned group isomorphisms, but we didn't define what they are. Let's do that now --- an isomorphism between two groups is an isomorphism ($f$) between their respective sets of elements, such that for any $a$ and $b$ we have $f(a \bullet b) = f(a) \bullet f(b)$. Visually, the diagrams of isomorphic groups have the same structure.
We already mentioned group isomorphisms, but we didn't define what they are. Let's do that now --- an isomorphism between two groups is an isomorphism ($f$) between their respective sets of elements, such that for any $a$ and $b$ we have $f(a \circ b) = f(a) \circ f(b)$. Visually, the diagrams of isomorphic groups have the same structure.
![Group isomorphism between different representations of S3](../03_monoid/group_isomorphism.svg)
@ -407,7 +407,7 @@ The functions that represent the monoid elements have the same set as source and
And the same is valid for the addition monoid --- number functions can be combined using functional composition.
$+2 \bullet +3 \cong +5$
$+2 \circ +3 \cong +5$
So, basically the functions that represent the elements of a monoid also form a monoid, under the operation of functional composition (and the functions that represent the elements that form a group also form a group).

40
_chapters/07_natural_transformations.md
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@ -37,34 +37,50 @@ This is simply because (though it may seem contradictory) *isomorphic objects ar
![Isomorphic categories](equal_categories.svg)
However, they are equivalent.
However, these categories they are equivalent.
**Parmenides:** This category surely cannot be equal to the other one --- it has two objects!
**Parmenides:** This category surely cannot be equal to the other one --- it has a different amount of objects!
**Heraclitus:** Who cares bro, they are isomorphic.
To understand it better, let's go back to the functor between a given map and the area it represents. In order for this functor to be invertible (and the categories --- isomorphic) the the map should represent the area completely: there should be arrow for each road and a point for each little place.
![Isomorphic categories](isomorphic_map.svg)
Such a map is necessary if your goal is to know about all *places*, however, like we said, when working with category theory, we are interested not in *places*, but in the *routes* that connect them i.e. we focus not on *objects* but on *morphisms*
Suppose, that there are two intersections that are positioned in such a way that there are routes from one and to the other and vice-versa
![Equivalent categories](equivalent_map.svg)
Or suppose, (a better example, since, in reality, all intercections are linked) that you are graphing a process, in which there are two steps that are always preceded by the same steps and are also always succeeded by the same set of steps. Are they really different steps, or just versions of the same step.
![Equivalent categories](equivalent_process.svg)
and so it is more apt to define equality with a concept that would only require for the objects to be equal *up to a unique isomorphism*. This concept is called **equivalence**.
Equivalence of orders
---
To understand what I mean, let's go back to the functor between a given map and the area it represents. In order for this functor to be invertible (i.e. in order for the two categories to be isomorphic) the the map should represent the area completely, like there should be arrow for each road and a point for each little place.
Now, we will present the definition of order equivalence. But let's start with order isomorphisms first. In the chapter about orders we presented a definition of order isomorphisms that is based on *set* isomorphisms, but, since we know about functors, we will present a new definition, based on functors:
![Isomorphic categories](isomorphic_map.svg)
Given two orders $A$ and $B$, an *order isomorphism* consists of two functors $F: A \to B$ and $G: B \to A$. Furthermore, those two functors should be such that composing them leads us back where we started. More formally, when for all objects $a$ of $A$ and $b$ of $B$ we have $b = F(G(b))$ and $a = G(F(a))$ or alternatively (if we are using the identity morphisms), such that $ ID_{B} = F \circ G$ and $ ID_{A} = G \circ F$.
But, like we said, in category theory, we are not interested in places, but in the *routes* that connect them.
![isomorphic orders](isomorphic_orders.svg)
![Isomorphic categories](isomorphic_map.svg)
The equivalence of orders is the same, except you replace equality with isomorphism:
So in other words, when we focus on objects, we would define two categories as equal if there exists an isomorphism between them. But when we focus on morphisms, then it is more apt to define equality with a concept that would only require for the objects to be equal *up to a unique isomorphism*. This concept is called **natural equivalence**.
...such that $ ID_{B} \boldsymbol{\cong} F \circ G$ and $ ID_{A} \boldsymbol{\cong} G \circ F$.
One way to understand natural equivalence relies on the concept of *equivalence classes* that we covered on the chapter about orders.
![Equivalent orders](equivalent_orders.svg)
So suppose, that there are two intersections that are positioned in such a way that for every road that goes to and from one of them, there is an identical road to the other one (maybe the one of these intercection was meant to replace the other one but it wasn't closed). Then suppose that you have a map which displays these two intercections as one and the same intercection.
By the way, remember the concept of equivalence classes that we covered in the orders chapter? Turns our that two orders are equivalent precisely when their equivalence classes orders are isomorphic.
![Orders with isomorphic equivalence classes](equivalent_order_classes.svg)
Equivalence of categories
---
Now that we warmed our minds up with order equivalences, we are ready to tackle the little more-complex *categorical equivalences.*
Natural transformations

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