When I was studying topology, I kinda disregarded the Quotient Topology part. It was short, but I didn't see much use of it inside the theory we were looking at during the course. Turns out I was right, but the conclusion - "so, disregard it" turned out to be quite wrong, since the knowledge is both aesthetic and useful.

Well, to begin with, I think it's a good idea to explain what is a

1 - Arbitrary union of open sets is still open.

2- Finite intersection of open sets is still open.

Turns out these two properties are enough to demonstrate a good amount of results. Just like when we leave $\mathbb{R}^n$ and go to metric spaces, having realized that (for a lot of purposes) all we need is the triangle inequality (plus some good conditions) to talk meaningfully of

So, what is a Topological Space?

i) $\emptyset \in \tau$; $X \in \tau$

ii) Given a family $A_{\alpha}$ in $\tau$, $\displaystyle \bigcup A_{\alpha} \in \tau$

iii) $A_1, A_2 \in \tau \Rightarrow A_1 \cap A_2 \in \tau$

The elements of $\tau$ are called

Examples:

i) Metric Space - A metric space is a topological space, where the open sets are precisely its open sets (note that the first "open sets" refers to "the open sets of the topology", whereas the second "open sets" refers to the "open set as defined in a metric space).

ii) Given a set $X$, the topology $\tau=\{\emptyset, X\}$ is called the

iii) Given a set $X$, the topology $\tau=\mathcal{P} (X)$ is called

For an example that is new to first-readers of topology:

iv) Given a set $X$, consider $\tau=\{\text{subsets of } X \text{ that have finite complement, and the empty set}\} $. This is a topology, called the

Essentially, this says we can "properly" separate points. It is obvious a metric space is a Hausdorff topological space. In fact, every topological space should be Hausdorff... or not?

Take our example (iv): Consider the set $\mathbb{Z}$ with the finite complement topology. Take the points $1$ and $2$. A neighbourhood $U_1$ of $1$ is, for example, $\{1,2,6,7,8,9,10,...\}$ and a neighbourhood $U_2$ of $2$ is, for example, $\{2,3,4,1210,1430,1431,1432,1433,1434,...\}$. But it doesn't matter what neighbourhoods of $1$ and $2$ you use, since they always have finite complement, there is a number $N_1$ such that any $n>N_1$ will be in $U_1$. Likewise, there is a number $N_2$ such that any $n>N_2$ will be in $U_2$. So, the intersection will always be non-empty ($\max\{N_1, N_2\}+1$ will be in the intersection, for example). Therefore, $\mathbb{Z}$ with the finite complement topology is

Well, now that we know what is a topological space, let's introduce one of the most fundamental concepts of Topology: Continuous Functions. As we will see, continuous functions with a continuous inverse will behave quite analogous to what a "isomorphism" is in linear algebra.

So, let's begin with the analysis definition of continuity:

There is, in this definition, an explicit use of distance. Here comes one that is easy to see to be equivalent, but quite more appealing:

In fact, this definition is automatically generalized to topological spaces, where there is no distance. Now, we have the following proposition:

Note that this can be rephrased more compactly with:

Demonstration: Let's prove $\Rightarrow$:

Take an open set $V$ of $Y$. We have to prove $f^{-1}(V)$ is open. So, take a point $p$ in $f^{-1}(V)$. By continuity, there is a neighbourhood $U$ of $p$ that goes into $V$. So, this implies $U \subset f^{-1}(V)$. Therefore, $p$ is an interior point. Since $p$ is arbitrary, $f^{-1}(V)$ is open.

For $\Leftarrow$,

let $p$ be a point of X. We will prove $f$ is continuous at $p$. Let $V$ be a neighbourhood of $f(p)$. By hypothesis, $f^{-1}(V)$ is open. Hence, there is a neighbourhood $U$ of $p$ in $f^{-1}(V)$. So, this neighbourhood will be taken to $V$. Therefore, $f$ is continuous at $p$. $\square$

Turns out that, for topological purposes, this property of continuity is so important that it is common to

This "Pt.1" serves as an introduction to "Pt.2", where I'll talk about quotient spaces, and adjunction spaces.

Well, to begin with, I think it's a good idea to explain what is a

*Topological Space.*Well, consider you are in a metric space. The open sets of the space are sets $A$ such that each point of $A$ has a ball around it still inside of $A$. We will say any open set around a point is a neighbourhood of the point. Open sets have two interesting properties:1 - Arbitrary union of open sets is still open.

2- Finite intersection of open sets is still open.

Turns out these two properties are enough to demonstrate a good amount of results. Just like when we leave $\mathbb{R}^n$ and go to metric spaces, having realized that (for a lot of purposes) all we need is the triangle inequality (plus some good conditions) to talk meaningfully of

**distance**, we will abandon metric spaces, desiring to talk meaningfully of**neighbourhoods**(open sets). This generalization is, in my point of view, a good motivation by itself. But the fact is that topology has far-reaching uses in a lot of fields of mathematics, and helps us understand and formalize things like "a circle is a closed interval with its endpoints glued".So, what is a Topological Space?

**Definition:**A*Topological Space*is a set $X$, together with a set $\tau$ of subsets of $X$, such that:i) $\emptyset \in \tau$; $X \in \tau$

ii) Given a family $A_{\alpha}$ in $\tau$, $\displaystyle \bigcup A_{\alpha} \in \tau$

iii) $A_1, A_2 \in \tau \Rightarrow A_1 \cap A_2 \in \tau$

The elements of $\tau$ are called

*open sets*of the topology.Examples:

i) Metric Space - A metric space is a topological space, where the open sets are precisely its open sets (note that the first "open sets" refers to "the open sets of the topology", whereas the second "open sets" refers to the "open set as defined in a metric space).

ii) Given a set $X$, the topology $\tau=\{\emptyset, X\}$ is called the

*trivial topology*. It is easily verified that it is a topology.iii) Given a set $X$, the topology $\tau=\mathcal{P} (X)$ is called

*discrete topology*. It is also easily verified that it is a topology.For an example that is new to first-readers of topology:

iv) Given a set $X$, consider $\tau=\{\text{subsets of } X \text{ that have finite complement, and the empty set}\} $. This is a topology, called the

*finite complement topology*, and a simple-to-describe one. Moreover, it has interesting features. First, let's define a property that is kind of reasonable to expect:**Definition:**We say the topological space $(X,\tau)$ is*Hausdorff*if, for every two points $p_1, p_2$ of $X$, there exists a neighbourhood $U_1$ of $p_1$ and a neighbourhood $U_2$ of $p_2$ such that $U_1 \cap U_2=\emptyset$.Essentially, this says we can "properly" separate points. It is obvious a metric space is a Hausdorff topological space. In fact, every topological space should be Hausdorff... or not?

Take our example (iv): Consider the set $\mathbb{Z}$ with the finite complement topology. Take the points $1$ and $2$. A neighbourhood $U_1$ of $1$ is, for example, $\{1,2,6,7,8,9,10,...\}$ and a neighbourhood $U_2$ of $2$ is, for example, $\{2,3,4,1210,1430,1431,1432,1433,1434,...\}$. But it doesn't matter what neighbourhoods of $1$ and $2$ you use, since they always have finite complement, there is a number $N_1$ such that any $n>N_1$ will be in $U_1$. Likewise, there is a number $N_2$ such that any $n>N_2$ will be in $U_2$. So, the intersection will always be non-empty ($\max\{N_1, N_2\}+1$ will be in the intersection, for example). Therefore, $\mathbb{Z}$ with the finite complement topology is

**not Hausdorff**.Well, now that we know what is a topological space, let's introduce one of the most fundamental concepts of Topology: Continuous Functions. As we will see, continuous functions with a continuous inverse will behave quite analogous to what a "isomorphism" is in linear algebra.

So, let's begin with the analysis definition of continuity:

**Definition 1 of continuity:**$f:X\rightarrow Y$ is said to be continuous at $x_0$ if $\forall \epsilon > 0 \exists \delta > 0 : d(x,x_0)<\delta \Rightarrow d(f(x),f(x_0))<\epsilon$.There is, in this definition, an explicit use of distance. Here comes one that is easy to see to be equivalent, but quite more appealing:

**Definition 2 of continuity:**$f:X\rightarrow Y$ is said to be continuous at $x_0$ if for all neighbourhood $V$ of $f(x_0)$, there exists a neighbourhood $U$ of $x_0$ that is taken inside $V$ by $f$.In fact, this definition is automatically generalized to topological spaces, where there is no distance. Now, we have the following proposition:

**Proposition:**A function $f:X \rightarrow Y$ is continuous if and only if, for every open set $V$ of $Y$, $f^{-1}(V)$ is open.Note that this can be rephrased more compactly with:

*Inverse image of open is open*.Demonstration: Let's prove $\Rightarrow$:

Take an open set $V$ of $Y$. We have to prove $f^{-1}(V)$ is open. So, take a point $p$ in $f^{-1}(V)$. By continuity, there is a neighbourhood $U$ of $p$ that goes into $V$. So, this implies $U \subset f^{-1}(V)$. Therefore, $p$ is an interior point. Since $p$ is arbitrary, $f^{-1}(V)$ is open.

For $\Leftarrow$,

let $p$ be a point of X. We will prove $f$ is continuous at $p$. Let $V$ be a neighbourhood of $f(p)$. By hypothesis, $f^{-1}(V)$ is open. Hence, there is a neighbourhood $U$ of $p$ in $f^{-1}(V)$. So, this neighbourhood will be taken to $V$. Therefore, $f$ is continuous at $p$. $\square$

Turns out that, for topological purposes, this property of continuity is so important that it is common to

*define*"continuous" as "pre-image of open is open". In fact, if a continuous function is bijective and has a continuous inverse (a homeomorphism), it is easy to see that it will take open sets from one topology to open sets of another topology. Hence, homeomorphisms "preserve the topology" in some sense. If there is a homeomorphism between two topological spaces, we can say they are*topologically equivalent*.This "Pt.1" serves as an introduction to "Pt.2", where I'll talk about quotient spaces, and adjunction spaces.