# Definition Given a [[Topological Space|topological space]] $X$ equipped with an [[Equivalence Relation]] $\sim$, the *quotient space* $X/\sim$ is the set of all [[Equivalence Class|equivalence classes]]. # Examples ## Example 1 Let $X = \mathbb{R}$ and $x, y \in \mathbb{R}$. Introduce an [[Equivalence Relation]] $\sim$ such that $x \sim y$ if there exists $n \in \mathbb{Z}$ such that $x = y + 2 \pi n$. The [[Equivalence Class]] $[x] = \{\ldots, x - 2\pi, x, x + 2\pi, \ldots\}$ and any number $x \in [0, 2\pi)$ can serve as a representative of a an [[Equivalence Class]] $[x]$. Note that $0 \sim 2\pi$ even though they're different points in $\mathbb{R}$, similarly with $\pi \sim -\pi$, etc. However, these points are looked upon as the same point in $\mathbb{R}/\sim$, the quotient space, since they belong to the same equivalence class. We can conclude that the quotient space $R/\sim \cong S^1$, i.e. the quotient space is [[Isomorphism|isomorphic]] to the [[circle|circle]] $S^1 = \{e^{i\theta}|0\leq \theta < 2\pi\}$. This is shown in the figure below from Nakahara. ![[Pasted image 20210630104545.png]] ## Example 2 Let $X$ be a square disc, i.e., $X = \{(x,y) \in \mathbb{R}\times{R} | \,\,|x| \geq 1, |y| \geq 1\}$. Now identify the points on a pair of facing edges together, i.e., $(-1, y) \sim (1, y), \,\, \forall y$. The quotient space is then the [[cylinder|cylinder]]. If we we have equivalence relation $(-1, -y) \sim (1, y), \,\, \forall y$, we get a [[Möbius Strip|Möbius strip]]. See the figure below. ![[Pasted image 20210630105030.png]] ![[Pasted image 20210630105045.png]] Note that the cylinder is an [[Orientability|orientable]] surface while the Möbius strip is a [[Orientability|non-orientable]] one. ## Example 3 Let $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ be two points in $\mathbb{R}\times \mathbb{R}$ and introduce an equivalence relation $\sim$ by: $\left(x_{1}, y_{1}\right) \sim\left(x_{2}, y_{2}\right)$ if $x_{2}=x_{1}+2 \pi n_{x}$ and $y_{2}=y_{1}+2 \pi n_{y}$, $n_{x}, n_{y} \in \mathbb{Z} .$ Then $\sim$ is an [[Equivalence Relation]]. The quotient space $(\mathbb{R}\times\mathbb{R}) / \sim$ is the [[torus]] $T^{2}$. ![[Pasted image 20210630110456.png]] ## Example 4 If we identify the edges of the rectangle in different ways we get the [[Klein Bottle|Klein bottle]] and the [[Projective Plane|projective plane]] ($RP^2$). ![[Pasted image 20210630110549.png]] ![[Pasted image 20210630110605.png]] These surfaces cannot be [[Embedding|embedded]] in $\mathbb{R}^3$ without self-intersecting. ## Example 5 We can get a [[genus|genus]] 2 [[torus]] from an octagon as follows ![[Pasted image 20210630110825.png]] ## Example 6 Let $D^{2}=\left\{(x, y) \in \mathbb{R}\times\mathbb{R} \mid x^{2}+y^{2} \leq 1\right\}$ be a closed disc. Identify the points on the boundary $S^1 = \left\{(x, y) \in \mathbb{R}\times\mathbb{R} \mid x^{2}+y^{2}=1\right\} ;\left(x_{1}, y_{1}\right) \sim\left(x_{2}, y_{2}\right)$ if $x_{1}^{2}+y_{1}^{2}=x_{2}^{2}+y_{2}^{2}=1.$, i.e., points that lie on the boundary are equivalent. By gluing the boundary together, we obtain the [[2-sphere|2-sphere]] $S^{2}$ as the quotient space $D^{2} / \sim$, also written as $D^{2} / S^{1}$, If we take an $n$ -dimensional disc $D^{n}=\left\{\left(x_{0}, \ldots, x^{n}\right) \in \mathbb{R}^{n+1} \mid\left(x_{0}\right)^{2}+\cdots+\left(x^{n}\right)^{2} \leq 1\right\}$ and identify the points on the surface $S^{n-1}$, we obtain the $n$ -sphere $S^{n}$, namely $D^{n} / S^{n-1}=S^{n}$. ![[Pasted image 20210630110909.png]] ![[Disk_to_Sphere_using_Quotient_Space.gif]]