# Algebraic Topology – Introduction

Hey! welcome to the first post of ‘algebraic topology’. First off all some heads up, I am going to assume that the reader is familiar with two topics: Topology & Group theory. If this is not the case, I recommend you to pause here and come back to this series when you are familiar with those topics since without pre-knowledge, you won’t understand me at all.

OK, if you kept reading it means that you are familiar with the topics above and then I shall begin. What is algebraic topology? If you remember topology classes, then you are probably familiar with the term homeomorphism. If two topological spaces are Homeomorphic, it means that they are the same topolical structre in a way. An improtant problem is to determine whether two topological spaces are homeomorphic or not. In topology class, (At least in my class) we see ways to prove that two topological spaces are not homeomorphic, using topological properties, (for example, connectivty). Sometimes, we need more tools in order to prove it. This is where algebraic topology comes to help, It turns out that we can match each topological space with a group, and learn about the space using the group. Before I start with this match, I need to talk about someting bigger and more general in order to trully understand what is going on here.

## Categories

A Category $\mathcal{C}$ is a collection of objects, where for every pair of objects $A,B\in\mathcal{C}$ there exists a set $\text{Mor}(A,B)$.

Wow, that is super abstract, lets try understand that carefully. First of all, $\text{Mor}$ is a short for ‘morphism’, does that sound familiar to you? if not, let me just write some words here: homomorphism, Isomorphism, monomorphism

How about now? this gives us some kind of clue that this set of morphisms, $\text{Mor}(A,B)$ is kind of an abstract way to name the set of property presevring maps between structures. And now we can also think of object as actual structure that we know: Vector space, Group, Field, Ring, Topological space and so on. What we are trying to do here is to generalize all of those cases into this kind of global thing.

There is an action on the objects, for any given objects $A,B,C\in\mathcal{C}$ and $f\in\text{Mor}(A,B),g\in\text{Mor}(B,C)$ we define:

g\circ f\in \text{Mor}(A,C)

It kind of feels like function compostion, although it might be actually composition in specific cases, here we treat this action as some abstract thing that takes two elements from two different sets, and returns a third element of a another set.

This action is associative: if $A,B,C,D\in\mathcal{C}$, $f\in\text{Mor}(A,B),g\in\text{Mor}(B,C),h\in\text{Mor}(C,D)$ Then:

h\circ(g\circ f)=(h\circ g)\circ f

Also, for any given $A\in\mathcal{C}$, there is $1_A\in\text{Mor}(A,A)$. This element is kind of special: for every $A,B\in\mathcal{C}$ and for every $f\in\text{Mor}(A,B)$:

1_B\circ f= f

and:

f\circ 1_A=f

That means that this element is acting like kind of a unit.

Some famous examples of categories are:

• Topolical spaces with the action of continous maps.
• Group with homomorphisms.
• Vector Spaces and linear transformations.

Our goal is To deal with two main categories – one which contains topological spaces, and the other contains groups. However, we still need a way to map objects from one category to the other.

## Functors

Given two categories $\mathcal{C}, \mathcal{D}$. a Functor from $\mathcal{C}$ to $\mathcal{D}$ is a map $F$ that matches an object $A\in\mathcal{C}$ with an object $F(A)\in\mathcal{D}$. Moreover, for any pair of objects $A,B\in\mathcal{C}$ and morphism $\varphi\in\text{Mor}(A,B)$ it matches a morphism $F(\varphi)\in\text{Mor}(F(A),F(B))$ such that:

• $\varphi\in\text{Mor}(A,B)$, $\psi\in\text{Mor}(B,C)$ then $F(\psi\circ\varphi)=F(\psi)\circ F(\varphi)$.
• $F(1_A)=1_{F(A)}$.

Again, the functor we are interested in is the one from the category of topological spaces to the category of groups.

Before I proceed, I want to notify that I will start writing $\varphi\in\text{Mor}(A,B)$ as $\varphi:A\to B$ or $A\overset{\varphi}{\to}B$, Those are just symbols that are easier to use.

## Generalized Isomorphism

Now we let $\mathcal{C}$ be a category, $\varphi:A\to B$ is called an Isomorphism if there is $\psi:B\to A$ such that $\varphi\circ\psi=1_B$ and $\psi\circ\varphi=1_A$.

As you might be thinking now, this is the ‘global’ definition of isomorphism, where all the types of isomorphisms you have met so far, are derived from this definition. Let’s play with this definition a bit and prove a statement:

I’ll show that a composition of isomorphisms is also an isomorphism. Let $A\overset{\varphi}{\to}B\overset{\psi}{\to}C$ be two isomorphisms. Then, there exists $B\overset{\varphi^{\prime}}{\to}A$ such that:

\varphi^\prime \circ\varphi = 1_A
\varphi\circ\varphi^\prime=1_B

Moreover, there exits: $C\overset{\psi^{\prime}}{\to}B$ such that:

\psi^\prime \circ\psi=1_B
\psi\circ\psi^\prime = 1_C

We want to to show that $\psi\circ\varphi$ is also an isomorphism. Observe that:

(\varphi^\prime\circ\psi^\prime)\circ(\psi\circ\varphi)=\varphi^\prime\circ(\psi^\prime\circ\psi)\circ\varphi

However, $\psi^\prime\circ\psi=1_B$ then we get

=\varphi^\prime\circ1_B\circ\varphi=\varphi^\prime\circ\varphi=1_A

It works! with a similar process we get that:

(\psi\circ\varphi)\circ(\varphi^\prime\circ\psi^\prime)=1_C

And we proved the statement.

When there is an isomorphism $A\overset{\varphi}{\to}B$ we say that $A,B$ are isomorphic and we write $A\cong B$. It is easy to see that the realtion $\cong$ is an equavilence relation, I won’t prove it here, try it yourself, it is really not that hard and I bet you have already proved similar statements in a group theory class or topology class or some other class. Remember, we are trying to generalize what we have studied before, so that makes sense that the proof will be similar to other related statements in different topics.

Now I want to present an improtant stamenent that will make things clearer, justify the use of functors and motivate us to work with functors. $\mathcal{C},\mathcal{D}$ be two categories. $F:\mathcal{C}\to\mathcal{D}$ is a functor. Let $A,B\in\mathcal{C}$ be isomorphic objects with an isomorphism $\varphi$. Then $F(A)\overset{F(\varphi)}{\to}F(B)$ is also in isomorphism.

Before I’ll go into the proof, Let’s understand first what this theorem says. It basically means that we can translate an isomorphism from one category to anoter. In our desired case, If we have a functor from the category of topological spaces to the category of groups then we can conclude that if two topological spaces are homeomorphic, so the corresponding groups are isomorphic as well! We can also conclude that if the corresponding groups are not isomorphic, then the spaces aren’t homeomorphic! This theorem gives us a tool to check if two topological spaces are not homeomorphic, which is exactly what we were looking for!

Let’s prove this theorem: since $\varphi$ is an isomorphism, there exits $B\overset{\varphi^{\prime}}{\to}A$ such that:

\varphi ^\prime \circ \varphi=1_A
\varphi\circ\varphi^\prime=1_B

Then:

F(\varphi^\prime)\circ F(\varphi)=F(\varphi^\prime\circ\varphi)=F(1_A)=1_{F(A)}
F(\varphi)\circ F(\varphi^\prime)=F(\varphi\circ\varphi^\prime)=F(1_B)=1_{F(B)}

So we have found $B\overset{F(\varphi)}{\underset{F(\varphi^{\prime})}{\underset{\leftarrow}{\rightarrow}}}A$ which proves the theorem!

A short way to state this theorem is: $A\cong B \Rightarrow F(A)\cong F(B)$. I find this theorem actually very intuitive, however, the opposite is not always true: $F(A)\cong F(B)\not\Rightarrow A\cong B$.

## Summary

We have met a really abstract topic here, which may seems a little unrelated to topology, however, we will see how we use everything here for our own specific needs, as I mentioned a few times in this post.

In the next post I will start talking about actually topology and introduce the term of Homotopy.