Quotient Space

Definition

Let $X$ be a space, $Y$ a set and $q:X\to Y$ a surjective function (the quotient map). Define the quotient topology on $Y$ by setting $U\subseteq Y$ open $⟺$ $q^{-1}(U)$ open in $X$.

so if a map $q:X\to Y$ is continuous, surjective, and open or closed, then it is a quotient map.

The function $q$ is of course always a continuous map. The quotient topology is also clearly the largest topology we can put on $Y$ such that $q$ is going to be continuous. Since all open sets on $X$ is mapped to the open sets on $Y$, we cannot have an extra set on the quotient topology, otherwise this extra set’s preimage will not be open and $q$ will not be continuous.

We can prove that the quotient topology is indeed a topology: for “Closed under unions”: Let $U_i\subseteq Y$ open, $q^{-1}\left({\bigcup }_i{u}_i\right)={\bigcup }_i{q}^{-1}(u_i)$ by property of preimage. Thus arbitrary union of $U_i$ is open. analogously, $q^{-1}\left({\bigcap }_i{u}_i\right)={\bigcap }_i{q}^{-1}(u_i)$

Different way to think of a quotient space

Similar to factor groups or factor rings we divide space $X$ into partitions (fibers or formally $\{q^{-1}(\{y\})|y\in Y\}$. $X$ can be viewed as disjoint union of non-empty sets ($X={\coprod }_{y\in Y}{S}_y$, with $S_y$ being a indexed partition, see Disjoint Union Space).

We can define such a function $q$ and get the partitions OR We can define the partitions $X$ indexed by $Y$ and construct such a function $q$ OR We can define a equivalence relation and view the equivalence classes as partitions

View from universal property

Lemma

Let $q:X\to Y$ be a quotient map. For any space $Z$, a map $f:Y\to Z$ is continuous if and only if $f\circ q$ is continuous.

Proof $⟹$ is obvious Proof $⟸$ : assume $f\circ q$ is continuous Let $S\subseteq Z$ open $⟹$ $(f\circ q{)}^{-1}(S)$ is open $⟹$ $q^{-1}(f^{-1}(S))$ is open $⟹$ $f^{-1}(S)$ is open ($q$ is continuous) $⟹$ $f$ is continuous

Important

The quotient topology is unique with this property

Universal Property:

Let $q:X\to Y$ be a quotient map, $f:X\to Z$ continuous such that $q(x)=q(x^{\prime })⟹f(x)=f(x^{\prime })$ Then there exists a unique continuous map $\tilde{f}:Y\to Z$ satisfying $f=\tilde{f}\circ q$ (makes graph commute)

Uniqueness of Quotients

Suppose $q_1:X\to Y_1$ and $q_2:X\to Y_2$ are quotient maps satisfying $q_1(x)=q_1(x^{\prime })⟺q_2(x)=q_2(x^{\prime })$ Then we must have $Y_1\cong Y_2$ Then we must have unique ${\tilde{q}}_1$ and ${\tilde{q}}_2$ that makes the graph commute. $⟹id\circ q_2={\tilde{q}}_2\circ q_1={\tilde{q}}_2\circ {\tilde{q}}_1\circ q_2$ $⟹{\tilde{q}}_2\circ {\tilde{q}}_1={\tilde{q}}_1\circ {\tilde{q}}_2=id$ $⟹Y_1\cong Y_2$

This relates to the universal property of quotient groups by replacing continuous map by group homomorphism and replacing homeomorphism by group isomorphism.

Examples

See Examples of Quotient Space

Property

Warning

Quotient space doe NOT preserve the properties like other spaces do ($e$.$g$. product space or subspace)

Let $q:X\to Y$ be an open quotient map (For every open set $U\subseteq X$, the image $q(U)\subseteq Y$ is open, see open maps), then $Y$ is hausdorff $⟺$ $R=\{(x_1,x_2)|q(x_1)=q(x_2)\}$ is closed in $X\times X$ Proof: $⟹$ Take $(x_1,x_2)\notin R$ $⟹$ $q(x_1),q(x_2)$ are distinct $Y$ is hausdorff, by definition $⟹\exists \text{ nbhds}{v}_1\ni q(x_1),v_2\ni q(x_2),v_1\cap v_2=\varnothing$ since $q^{-1}(v_1)$ and $q^{-1}(v_2)$ are both open in $X$, $q^{-1}(v_1)\times q^{-1}(v_2)$ is neighborhood of $(x_1,x_2)$ This shows the complement of $R$ is open, so $R$ is closed.

It is usually easier to check if $R$ is closed in the product space $X\times X$ than to directly check whether $Y$ is hausdorff

Other Properties

• Any composition of quotient maps is a quotient map
• An injective quotient map is a homeomorphism
  • an injective quotient map is a bijection
• A subset $K\subseteq Y$ is closed $⟺$ $q^{-1}(K)$ is closed
• If $U\subseteq X$ is saturated open or closed subset, then $q_{|U}:U\to q(U)$ is a quotient map
• If $\{q_{\alpha }:X_{\alpha }\to Y_{\alpha }\}$ are a family of quotient maps, then $\tilde{q}:{\coprod }_{\alpha }{X}_{\alpha }\to {\coprod }_{\alpha }{Y}_{\alpha }$ is a quotient map
• Any quotient of a connected space is connected, also works for Path Connectedness