So, what is a quotient topology? Well, the idea is that we will "identify points". More colourfully, we will shrink a whole subset (or whole subsets) to a point (or points). For example: when you take the disk in $\mathbb{R}^2$ and identify all the points of the boundary (the 1-dimensional sphere $S^1$), you get the 2-dimensional sphere $S^2$:

When you take a square, and identify a point of each side with its "opposite", you get a torus.

When you take a square, and identify a point of each side with its "opposite", you get a torus.

Ok, so now we go to the theory. (First, let me remark the following: I will assume that the equivalence between a "partition of a set" and "equivalence classes" of some equivalence relation is well-understood. To every partition of a set, we can define the equivalence class: "$x \sim x'$ if $x$ is in the same subset of $x'$". And every equivalence class determines a partition of a set.)

With that in mind, let's begin:

Equivalently, if $X$ is a topological space and we have an equivalence relation $\sim$, call $X/ \sim$ the quotient space (meaning, the set of equivalence classes). We then have the map: $\pi :X \rightarrow X / \sim$ that takes $x$ to its equivalence class $\bar{x}$. Now, give $X/ \sim$ the following topology: $U'$ is open in $X/ \sim$ if and only if $\pi^{-1}(U')$ is open in $X$. This is the

It may not be easy to see that they are equivalent at a first glance if you are not familiar with equivalence relations, or with topology, but a bit of thought will make it clear. I will adopt the latter definition in the calculations (but will write $X/ \sim $ as $X^*$ for convenience) since it is more algebraic, hence, easier to handle.

Now, before introducing the next useful theorem, let me give some acquaintance to

What this theorem says can be given more precision, in the following language:

Take a minute to understand this and then we can proceed:

With that in mind, let's begin:

**Definition:**If $X$ is a topological space and we have a partition of $X$, call $X^*$ the set of those subsets that form the partition. Give $X^*$ the following topology: A subset of $X^*$ (which is a collection of subsets of X) is open if and only if the union of that collection is open in $X$. This is the*quotient space.*Equivalently, if $X$ is a topological space and we have an equivalence relation $\sim$, call $X/ \sim$ the quotient space (meaning, the set of equivalence classes). We then have the map: $\pi :X \rightarrow X / \sim$ that takes $x$ to its equivalence class $\bar{x}$. Now, give $X/ \sim$ the following topology: $U'$ is open in $X/ \sim$ if and only if $\pi^{-1}(U')$ is open in $X$. This is the

*quotient space*.It may not be easy to see that they are equivalent at a first glance if you are not familiar with equivalence relations, or with topology, but a bit of thought will make it clear. I will adopt the latter definition in the calculations (but will write $X/ \sim $ as $X^*$ for convenience) since it is more algebraic, hence, easier to handle.

Now, before introducing the next useful theorem, let me give some acquaintance to

*commutative diagrams*. Consider the isomorphism theorem:**Isomorphism Theorem:**Given an homomorphism $f: G \rightarrow H$ between two groups, we have $\displaystyle G/\ker(f) \simeq Im(\phi)$.What this theorem says can be given more precision, in the following language:

**Isomorphism Theorem:**Every homomorphism $f: G \rightarrow H$ between two groups induces an injective homomorphism $\bar{f}: G / \ker(\phi) \rightarrow H$ making the following diagram commute (meaning: $f= \bar{f} o \pi$)Take a minute to understand this and then we can proceed:

**Proposition:**Given $X, Y$ topological spaces, a continuous $f:X\rightarrow Y$ which is constant on every equivalence class induces a continuous map $f^*: X^* \rightarrow Y$ making the following diagram commute:Demonstration: Well, recall the proof of the isomorphism theorem: what happens is that, when you define the map, you must show that it independs of the representative of the class for it to be well-defined. In that case, that happens because the quotient is on the kernel. In this case, it is even simpler: $f$ is constant in the equivalence class!

Define $f^*(\bar{x})=f(x)$. It is well-defined by the previous observation, and obviously $f=f^*o\pi$. It remains to show that $f^*$ is continuous. Lets show pre-image of open is open! But, a set $U$ is open in $X^*$ if and only if $\pi^{-1} U$ is open in $X$. So, we have to show that, for every open $V$ in $Y$, $\pi^{-1} o f^{*-1} (V)$ is open. But this set is exactly $f^{-1}(V)$. Since $f$ is continuous, we have proved $f^*$ is continuous.

Let's see an application of this and prove... that the circle is a segment where you identify the endpoints!

Consider $I:=[0,2\pi]$, and $S^1:=(\cos(t),\sin(t))$, for $t \in [0,2\pi)$. Now, make the following equivalence relation in $I$: every point is equivalent only to itself, except $0$ and $2\pi$, which are equivalent (they are in the same equivalence class). Therefore, we have $I^*$

Take the map $f: I \rightarrow S^1$ that takes $x$ to $(\cos(x), \sin(x))$. Note that $f(0)=f(1)$. So, $f$ satisfies the hypothesis of our preceding proposition (namely, it is constant in every class!). Then, we have an induced continuous map $f^*: I^* \rightarrow S^1$. (Note that $f^*$ is bijective). Now, take the following map: $g: S^1 \rightarrow I^* $ that takes $(\cos(x), \sin(x))$ to $\bar{x}$). It is obviously continuous on all points, except possibly at $f(1,0)$ (since all other classes are points). But the neighbourhoods of $f^*(1,0)=\bar{1}=\bar{0}$, now that we are at the quotient space, must "contain" a neighbourhood of $1$ and a neighbourhood of $0$. Why? Suppose you have a neighbourhood $\bar{U}$ of $\bar{1}$. So, $\pi^{-1}(\bar{U})$ must be an open set containing $0$ and $1$. Therefore, there are two small intervals around $0$ and $1$ in $\pi^{-1}(\bar{U})$, which will be taken to the corresponding classes. So, this $g$ is continuous, since we can take a small enough piece of the circle around $(1,0)$ for which the function will fall inside those intervals. But this $g$ is the inverse of $f^*$. So, we found a homeomorphism. We could also do something more elegant (but would need more knowledge about topology):

In Pt.3, we shall talk about another style of glueing things.

Define $f^*(\bar{x})=f(x)$. It is well-defined by the previous observation, and obviously $f=f^*o\pi$. It remains to show that $f^*$ is continuous. Lets show pre-image of open is open! But, a set $U$ is open in $X^*$ if and only if $\pi^{-1} U$ is open in $X$. So, we have to show that, for every open $V$ in $Y$, $\pi^{-1} o f^{*-1} (V)$ is open. But this set is exactly $f^{-1}(V)$. Since $f$ is continuous, we have proved $f^*$ is continuous.

Let's see an application of this and prove... that the circle is a segment where you identify the endpoints!

Consider $I:=[0,2\pi]$, and $S^1:=(\cos(t),\sin(t))$, for $t \in [0,2\pi)$. Now, make the following equivalence relation in $I$: every point is equivalent only to itself, except $0$ and $2\pi$, which are equivalent (they are in the same equivalence class). Therefore, we have $I^*$

Take the map $f: I \rightarrow S^1$ that takes $x$ to $(\cos(x), \sin(x))$. Note that $f(0)=f(1)$. So, $f$ satisfies the hypothesis of our preceding proposition (namely, it is constant in every class!). Then, we have an induced continuous map $f^*: I^* \rightarrow S^1$. (Note that $f^*$ is bijective). Now, take the following map: $g: S^1 \rightarrow I^* $ that takes $(\cos(x), \sin(x))$ to $\bar{x}$). It is obviously continuous on all points, except possibly at $f(1,0)$ (since all other classes are points). But the neighbourhoods of $f^*(1,0)=\bar{1}=\bar{0}$, now that we are at the quotient space, must "contain" a neighbourhood of $1$ and a neighbourhood of $0$. Why? Suppose you have a neighbourhood $\bar{U}$ of $\bar{1}$. So, $\pi^{-1}(\bar{U})$ must be an open set containing $0$ and $1$. Therefore, there are two small intervals around $0$ and $1$ in $\pi^{-1}(\bar{U})$, which will be taken to the corresponding classes. So, this $g$ is continuous, since we can take a small enough piece of the circle around $(1,0)$ for which the function will fall inside those intervals. But this $g$ is the inverse of $f^*$. So, we found a homeomorphism. We could also do something more elegant (but would need more knowledge about topology):

*Since $I$ is compact, so is $I^*$. But $f^*$**is bijective, and $S^1$ hausdorff, so $f^*$**is in fact a homeomorphism.*In Pt.3, we shall talk about another style of glueing things.