## Category Theory for the Mathematically Inclined Person(10/7/21)

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Category Theory
For
The Mathematically Inclined Person

A Book for Those who like math and categories but don’t know much about them

Written By

Kasra Mossayebi

2021

# Introduction

In a subject as abstract and unique as category theory, it is often hard to become acclimated to the common techniques and tools in category theory, even though they be extremely common in what has already been studied due to the level of abstraction. To help cure, or at least minimize the massive amounts of confusion that occur that encountering a learning curve like abstraction bring, it’s best to learn the basic intuition behind concepts before generalizing and formalizing them, for abstracting without a general idea of what you’re abstracting leads only to thinking about definitions instead of thinking about how these definitions relate to what they generalize and how they can be used.

In category theory, the most important idea is obviously a category. A category consists of three parts:

Objects - Objects are essentially the "elements" of a category, analogous to the elements of a set

Morphisms - Morphisms are structure preserving "maps" or "functions" between objects in a category

Composition of Morphisms - Composing morphisms, like composing functions, can be expressed as $$g\circ f$$ = g(f(x)),

but there are also other ways to compose morphisms, so it is important to a category to specify composition

Firstly, it’s good to think of what the word "category" means. A category is defined as a division of people or things that share a common characteristic. In category theory, that definition still rings true, but these categories of mathematical objects all share the same mathematical "structure". For example, a monoid is a set with the added structure of an associative binary operation(one taking two inputs and giving one output, for example multiplication, *(a,b) = a*b = c), and an identity. This means that there can exist a category with objects being monoids. However, there are two parts of the category that we have still not provided: morphisms and composition of morphisms. Well, the first is actually easy because there already exists a name for structure preserving morphisms between monoids. These are called (monoid) homomorphisms. The structure preserved by these homomorphisms is, as some may have already guessed, the binary operation of the monoid, along with the condition that the identity is preserved, in other words the identity is mapped to the identity. Now, we can define the composition of morphisms the same we define normal function composition, the composition of two morphisms f:A $$\rightarrow$$ B, g:B $$\rightarrow$$ C, g $$\circ$$ f:A $$\rightarrow$$ C with g $$\circ$$ f = g(f). The way we just described a category’s properties was woefully inefficient, and category theory helps with this. A profound realization made by category theory is the fact that many of the properties of mathematical systems(such as binary operations satisfying certain properties) can be described in the far more simple and efficient terms of diagrams using arrows. For us, these arrows are morphisms, and these diagrams are called commutative diagrams. However, we want to change some of our normal notation for functions, firstly, when we have a function f:A$$\rightarrow$$ B, we also have a rule x$$\mapsto$$ f(x).Now, we want to remove all unnecessary parenthesis so we will instead write $$x \mapsto f(x)$$ as $$x\mapsto fx$$, so this would also mean that composition $$\circ(g,f) = g \circ f \mapsto gf$$. A typical diagram would look like

First thing you might notice are the arrows going from A to B to C and from A to B. Here, A,B, and C are all objects in a category, and the arrows are morphisms. Here, we say that this diagram commutes if $$g\circ f$$ = h. We can visualize this:

Commutes if

Though such a diagram might seem slightly trivial in terms of usefulness, however thanks to the fact that we have already stated that many properties of mathematical systems can be simplified in terms of diagrams using arrows. For example, we have already defined a monoid as a set with a binary operation and an identity satisfying some certain properties. However, there is a far more efficient way to define a monoid in terms of a commutative diagram. Firstly, we recognize that commutative diagrams need not be triangles. Realistically, they can be almost any shape, e.g. squares, pentagons. Note however, they also do not need to be a shape in the sense of a polygon. As a simple example, a commutative diagram in the shape of a square:

Commutes if

Hopefully, from those two examples anyone can try to understand what would make a diagram of a non-polygonal shape commute. Either way, we will only need square diagrams to describe monoids. However before we start, we must describe a few functions,

$$*_M$$:$$M \times M \rightarrow M$$, $$\mu:1 \rightarrow M$$,

Author’s Note: as of current(10-17-21) I haven’t worked on the book for a month thanks to school, but I’ll try to do some more during my free time. I am now studying some AG from Vakil

# Notes

This was a beamer slideshow that [badly] converted into a more text like format. This was also meant for a "maths talk" I did on how to introduce homeomorphisms in a more interesting way.

# Introducing Continuity

Continuity: The idea of continuity isn't well defined before analysis, and the definition changes. For example many people are told that a function is continuous if it can be graphed without picking up your pen. Topology offers us a much clearer definition of continuity:

Definition: A function f from a topological space X to Y is continuous if for every $V \subseteq Y$, $f^{-1}(V)$ is open in X

# Donut = Cup?

Why Donut = Cup? The infamous meme that claims topologist think that a donut is the same as a mug or cup. Well, the reality is that that is true, except possibly not in the way you would expect. The topological properties of the cup/mug and a torus(donut) are the same, so we introduce a concept called a homeomorphism.

Homeomorphisms: A homeomorphism is very similar to a isomorphism from algebra, but where instead of preserving algebraic properties, the homeomorphism preserves topological properties like genus and compactness.

Definition A homeomorphism is a function $f:X \mapsto Y$, where X and Y are topological spaces, such that:

1. f is bijective

2. f is continuous

3. $f^{-1}$ is continuous

# Defining a donut

Introduction: In this next section, I will introduce the quotient topology, so that we can create the donut(torus), and examine some of its topological properties(and maybe even analyze what we can tell about a mug or cup from it). The idea of an equivalence class(and relation) will be an important part in understanding the quotient topology:

Definition: A binary relation, $\sim$, on a set X is said to be an equivalence relation if it satisfies certain conditions(for all $x,y,z \in X$):

1. $x \sim x$

2. $x \sim y$ iff $y \sim x$

3. if $x\sim y$ and $y\sim z$, $x\sim z$

Equivalence Class

Definition: For some $x \in X$, an equivalence class is said to be the subset of X given by := $\{y \in X \mid x \sim y\}$ We can also define the set of all equivalence classes on X as: $(X/ \sim) = \{[x] \mid x\in X\}$

Now, we can define the open set in ($X/\sim$) :

Definition: For some $U \subseteq (X/\sim)$, where X is a topological space and $\sim$ is an equivalence relation on it, U is open iff: $\bigcup_{[x] \subseteq U} [x] \subseteq X$ is open in X

Here, we have created the quotient topology! We can confirm that this is indeed a topology

Creating a Torus: To create a torus, we use the equivalence relation $(x,y)\sim (x+1,y) \sim (x, y+1)$ in the Cartesian plane. This is wonderful, but now we can analyze some of the topological properties of a torus.

Example Property: A torus is topologically compact, and therefore so is a mug. A compact space is one where any covering of the space has a finite subcover.

# End Notes

We can use the idea of a quotient map to also build a quotient topology, but for the purpose of creating a torus, I used the method using equivalence classes:

Definition A surjective map $p:X \mapsto Y$, where X and Y are topological spaces, is said to be a quotient map if for $V \subseteq$ Y, V is open in Y iff $f^{-1}(V)$ is open in X

using this idea, we can show that given a set A and a space X, there exists only one topology on A, relative to p that is, and this topology is called the topology induced by p

# References

References Topology Second Edition. Munkres, James. Pearson Publishing Quotient Topology Supplement, http://www.math.ucsd.edu/$\sim$bprhoades/190w16/quotient.pdf

# Initial Notes

These notes were also written in December 2020 just as I started to figure out how to use $\LaTeX$, and started to write things related to math. This was actually used in a Youtube video I made(now privatized however because it was pretty terrible)

# Defining the Topology

Our study of topological spaces shall begin with the definition of a topology/topological space. The definition of the topology was under debate during the early 20th century, due to the fact that we wanted a definition that was both vague enough that all special case were included, but also be limited enough so that it was useful and our standard theorems still held true. For mathematicians, it is always a big debate how general a definitions should be, as it can affect so much in mathematics. Though eventually a common definition was agreed upon, though there are different forms of it, which we may or may not cover in this lecture or future lectures. The definition may seem quite vague, but I believe the reader should be able to grasp it after some time and practice. We define the topology $\mathcal{T}$ on a set X as the collection of subsets of X such that: (1) Both $\varnothing$ and X are within $\mathcal{T}$ (2) The union of the elements of any sub-collection of $\mathcal{T}$ is also in $\mathcal{T}$ (3) The intersection the elements of any finite sub-collection of $\mathcal{T}$ is also in $\mathcal{T}$ Any set X for which a topology $\mathcal{T}$ has been defined is called a topological space.

Well, properly a topological space is actually the ordered pair (X,$\mathcal{T}$), but $\mathcal{T}$ will be omitted and our space will just be notated by X, unless otherwise stated.

# Open and Closed Sets

In our study of topology, the open and closed set with be an integral part of our discussion, so I believe it is best for both the reader and I to define it as early as possible. If X is a topological space, we call the set U of X an open set if U belongs to the collection $\mathcal{T}$. Now, we can reach another way of defining the topological space,

A topology $\mathcal{T}$ can be defined as the the set X with a collection of open sets such that both X and $\varnothing$ are open, and so that the arbitrary unions of open sets and finite intersections of open sets are both open.

We also have the closed set, which can be simply defined as a set who's complement is an open set, or in mathematical terms, A subset C of a topological space X is defined to be closed if: $(X-C) \in \mathcal{T}$ Theorem A.2.1 Let X be a topological space, then the following holds true: (1) X and $\varnothing$ are closed (2) Finite unions of closed sets are closed (3) Arbitrary intersections of closed sets are closed proof: (1) is easy to prove, as $X-X = \varnothing$, which is open, implying X is closed. $X - \varnothing = X$, which is open, therefore implying $\varnothing$ is also closed. (2) Given an indexed family of closed sets $\{A_a\}_{a \in J}$ :

$X - \bigcap_{a\in J}A_a = \bigcup_{a \in J}(X - A_a)$

Since the sets $X - A_a$ are open, the right side of our equation represents the arbitrary union of open sets, and therefore one can clearly see that $\cap A_a$ is closed

consider the equation:

$X - \bigcup_{i=1}^n A_i = \bigcap_{i=1}^n(X-A_i)$

Here, we have $A_i$ which is open for i = 1,2,....,n. Our right hand side is the the finite intersection of open sets and is therefore open. Therefore, $\cap A_i$ is closed

Now we see the student's dilemma, how can a set be both open and closed? Well, this is just one of those things that are hard to think about and accept, but you must. The confusion will clear as you continue your journey through topology, so do not be too worried dear student.

We can discuss both the interior and closure of a set, the interior is not so important, but as you will see later the closure will be of much use to us: We define the interior of a subset C of the topological space X to be the intersection of all open sets contained in C:

Int C = $(\bigcup_{A \subset C}A) \bigcap \mathcal{T}$

Here, Int means the interior of C. Now, we want to move on the the closure of the subset C, which, as said earlier, is much more useful than the interior, We define the closure of the subset C of the topological space X is defined as the intersection of all closed sets containing C:

Cl C = $\bigcap_{C \subset A, (X-A) \in \mathcal{T}} C$

Again, here Cl means closure of set C. There are actually more notations for the closure of the set, one of which is $\overline{A}$, and we shall use this for this for efficiency's sake.

# Defining and Using the Basis to Create a Topology

Sometimes when we want to define a topology, going through a classic approach as we described is quite inefficient, or even impossible, so we will use an idea called the basis to make our topology in a more efficient manner. A basis $\mathcal{B}$ of a topological space X is defined as the collection of subsets of X(called basis elements) such that: (1) For each $x \in X$, there exist a basis element B containing x (2) If $x \in B_1\cap B_2$, then there is a basis element $B_3$ containing x such that $B_3 \subset B_1\cap B_2$ If $\mathcal{B}$ satisfies these two conditions, there is a Topology $\mathcal{T}$ generated by $\mathcal{B}$ defined as such: As subset U of X is said to be open(an element of $\mathcal{T}$) if for each $x \in U$ there exists a basis $B\in \mathcal{B}$ such that $x\in B$ and $B \subset U$. One must also remember that the basis elements themselves are elements of $\mathcal{T}$.

Just to make sure, we will make sure that the collection generated by $\mathcal{B}$ is indeed a topology on X. We check our first condition, when U is an empty set, it vacuously satisfies the condition of openness. Similarly, for X there very clearly exists a basis element such that for all $x \in X$, $x \in B$ and $B \subset X$. Now, let's take an indexed family $\{U_a\}_{a\in J}$, and show that $U = \bigcup_{a \in J} U_a$ does indeed belong to $\mathcal{T}$.

Given a $x \in U$, there exists an index a so that $x \in U_a$. Since $U_a$ is open, there exist a basis element B such that $x \in B \subset U_a$, then $x \in B$ and $B \subset U$, showing us that the set U is open.

Finally, we want to show that for any two $U_1$ and $U_2$ in $\mathcal{T}$, $U_1 \cap U_2 \subset \mathcal{T}$. Given $x \in U_1 \cap U_2$, we want to assign a basis element $B_1$ such that $x \in B_1 \subset U_1$, and a basis element $B_2$ such that $x \in B_2 \subset U_2$. The second condition allows us to create a third basis so that $x\in B_3 \subset B_1 \cap B_2$. We can therefore we can say $B_3 \subset U_1 \cap U_2$, and since x is within $B_3$, we have proven that $U_1 \cap U_2 \in \mathcal{T}$ by definition.

Additionally, we shall seek to prove that any finite intersection of open sets in $\mathcal{T}$ is also in $\mathcal{T}$. The proof for n=1 is trivial; we should induct this, so we assume that this is true for n-1 and prove it is true for n. We start with:

$(U_1 \bigcap U_2 \bigcap...\bigcap U_{n-1})\bigcap U_n = (U_1 \bigcap U_2 \bigcap...\bigcap U_n$)

By hypothesis, $U_1 \bigcap...\bigcap U_{n-1}$ belongs to $\mathcal{T}$; By the result we just proved, the union of $(U_1 \bigcap...\bigcap U_{n-1}$ and $U_n$ also belongs to $\mathcal{T}$. The following lemma may make the result easier to understand: Lemma A.3.1: For the basis $\mathcal{B}$ of the topology $\mathcal{T}$ on the set X, $\mathcal{T}$ is defined as the collection of all unions of elements of $\mathcal{B}$ proof: As we said earlier, the collection of elements of $\mathcal{B}$ is also a part of $\mathcal{T}$. One can clearly see that the union of the elements of $\mathcal{B}$ is also in $\mathcal{T}$. Given $U \in \mathcal{T}$, we can find an element $B_x \in B$ such that we have $x \in B_x \subset U$. It can therefore be concluded that $U = \bigcup_{x \in U}B_x$, in other words, U is equal to a union of the elements of $\mathcal{B}$.

Now, we want to be as clear as possible in mathematics, so I'd like to remove any confusion regarding the use of the term basis. In linear algebra, a basis is a way of describing a given vector as the linear combination of of basis vectors. This is unique, unlike a basis in topology, which is as one may be able to tell from the previous lemma, not unique. So far, we have shown two different ways to create a topology out of its basis, but now we will go in the opposite direction and get the basis from the topology! Lemma A.3.2: Let X be a topological space. Suppose we have a collection $\mathcal{C}$ of the open sets U of X, so that for every U and $x \in U$ , there exist an element $C \in \mathcal{C}$ such that $x \in C \subset U$. The collection $\mathcal{C}$ is the basis for the topology $\mathcal{T}$ on X proof: We must first show our collection $\mathcal{C}$ is a basis. The first condition is very clearly satisfied, as for a given $x \in X$, since X is an open set, there is by hypothesis an element C of $\mathcal{C}$, such that $x \in C \subset X$. To check the second condition, let x belong to $C_1 \bigcap C_2$, where of course $C_1$ and $C_2$ are elements of $\mathcal{C}$. Since both $C_1$ and $C_2$ are open sets, $C_1 \bigcap C_2$, and therefore, by hypothesis, there exists an element $C_3$ such that $x \in C_3 \subset C_1 \bigcap C_2$

Now that we've achieved that, we want to prove that the basis $\mathcal{T}$' produced by $\mathcal{C}$ is the same as $\mathcal{T}$. One should note that if $U \in \mathcal{T}$ and $x\in U$, then by hypothesis there exists an element $C \in \mathcal{C}$, such that $x\in C \subset U$. It therefore follows that U belongs to the topology $\mathcal{T}$' by definition. Now, let's have some basis W belonging to the topology $\mathcal{T}$', then W is a union of elements of $\mathcal{C}$, as one can see using the lemma we just proved. Since each element of $\mathcal{C}$ also belongs to $\mathcal{T}$, and $\mathcal{T}$ is a topology, $W \in \mathcal{T}$.