Examples of Quotient Space

Example: Gluing a segment

By defining the endpoints as being in the same equivalence class, we basically connect the endpoints and the resulting quotient space is homeomorphic to a circle.

Consider this orange subset, it is open (see Subspace) and includes $[0]$ and $[1]$, so on our quotient space, the corresponding set must also be open. A better way to draw this quotient space is of course to draw a circle

Here we have the quotient map $w:I\to S^1$ $s\mapsto e^{2\pi i\cdot s}$

we check that $q(0)=e^{2\pi i\cdot 0}=1=e^{2\pi i}=q(1)$

Example: glue ball into sphere

Consider a closed disk (2-dimensional ball) ${\bar{\mathbb{B}}}^2\subseteq {\mathbb{R}}^2$, and an equivalence relation $\sim :=((x,y)\sim (x,-y))$ We get ${\bar{\mathbb{B}}}^2/\sim \cong {\mathbb{S}}^2$

Example: gluing square edges

Cone on a space X

Let $X$ be a space Consider the product space $X\times I$ we define $\sim :=((x,0)\sim (y,0)\forall x,y\in X)$

We denote $X\times I/\sim =CX$

NOTE

we also have $C{\mathbb{S}}^2\cong {\mathbb{B}}^3$ Whether the “vertex” is inside of the sphere or outside of the sphere does not matter

We have here a quotient map $q:X\times I\to CX$

Wedge sum

$\sim :=((x_i\sim x_j)\forall i,j)$ We then define the wedge sum as

$$\underset{i}{\bigvee }{X}_i:=\coprod _i{X}_i/\sim$$

In the case of having only two spaces we have: $X\vee Y=(X\bigsqcup Y)/\sim$ This can also be viewed as a Adjunction Space

Example: Normalization

$q:{\mathbb{R}}^{n+1}\setminus \{0\}\to {\mathbb{S}}^n$ $x\mapsto \frac{x}{|x|}$

Example: R/Z

$\mathbb{R}/\mathbb{Z}=\mathbb{R}/(x\sim y⟺x-y\in \mathbb{Z})$ Let $q:\mathbb{R}\to \mathbb{R}/\mathbb{Z}$

To define $\tilde{f}$ we only need to have $q(x)=q(x^{\prime })⟹f(x)=f(x^{\prime })$ using universal property. This is indeed true (easily checked) So there is an unique continuous $\tilde{f}$ to make the graph commute

To find for example $\tilde{f}([0])$ we can just find an element $x$ of the preimage of $[0]$ in $\mathbb{R}$, and find $f(x)$. In this case let’s choose $x=1$, and $f(1)$ gives $\sin (2\pi )=0$. When we choose $x=2$, then $f(2)$ still gives 0.

Actually $\mathbb{R}/\mathbb{Z}$ is homeomorphic to ${\mathbb{S}}^1$

To show $\mathbb{R}/\mathbb{Z}$ is homeomorphic to ${\mathbb{S}}^1$: Note that we have $x-x^{\prime }\in \mathbb{Z}⟺e^{2\pi i\cdot x}=e^{2\pi i\cdot x^{\prime }}$ (Example Gluing a segment) By using uniqueness of quotient maps we can clearly see that $\mathbb{R}/\mathbb{Z}\cong {\mathbb{S}}^1$