Quotients

Question

• Does the diagram in Munkres 22.2 commute?

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Overview §

The word “quotient” refers to dividing a space into disjoint pieces. One way to formalize this idea is by constructing quotient spaces: given an equivalence relation $\sim$ on a topological space $X$, the set of equivalence classes

$$X/\sim =\{C_x|x\in X\}$$

can be thought as what’s left after “crushing” all points that are similar to each other down to a single element. The quotient topology is the topology that makes the canonical projection, or quotient map $p:X\to X/\sim$ defined by $x\mapsto [x]$ continuous.

Quotient are generally thought of in terms of their universal property: if $f$ is a function that is constant on equivalence classes, then $f$ is still well-defined after “crushing” the equivalence class to a single point. In fact, the universal property implies that the quotient topology is chosen precisely so that such a map $f$ is still continuous.

Notation

Given $A\subseteq X$, let ${\sim }_A$ be the equivalence relation given by $x=y$ or $x,y\in A$ for all $x,y\in X$. Then we write $X/A:=X/{\sim }_A$ for the quotient where all of $A$ is “crushed” to a point.

Related: Disjoint unions

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Quotient spaces §

Definition: Quotient topology

Let $\sim$ be an equivalence relation on a topological space $X$, and let

$$C_x=\{z\in X|z\sim x\}\subset X$$

be an equivalence class. The set of equivalence classes

$$X/\sim =\{C_x|x\in X\},$$

where $C_x=C_y$ if and only if $x\sim y$, has a canonical projection or quotient map $p:X\to X/\sim$ sending point $x$ to its equivalence class $[x]$; the notation $[x]$ indicates an element of $X/\sim$, rather than a subset

$$p^{-1}([x])=\{y\in X|y\sim x\}=C_x\subset X.$$

The quotient topology on $X/\sim$ is the topology such that $p$ is continuous:

$$U\subset X/\sim \text{ open}⟺p^{-1}(U)\subset X\text{ open}.$$

Theorem: Universal property of quotient spaces

If $X$ is a topological space and $\sim$ is an equivalence relation on $X$, the projection map $p:X\to X/\sim$ is continuous.

Moreover, if $f:X\to Y$ is a continuous map that is constant on equivalence classes, meaning that if $x\sim z$, then $f(x)=f(z)$, then $f$ induces a continuous map $\bar{f}:X/\sim \to Y$ defined by

$$\bar{f}([x])=f(x),$$

and $\bar{f}\circ p=f$. We say that $f$ “factors through” the quotient $X/\sim$.

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#wip “Gluing”

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Quotient maps §

Definition: Quotient map, relative quotient topology

A continuous function $p:X\to Y$ is a quotient map if $p$ is surjective and

$$U\subset Y\text{ open}⟺p^{-1}(U)\text{ open}.$$

The quotient topology on $Y$ (relative to $p$) is the unique topology such that $p$ is a quotient map.

Lemma:

Let $p:X\to Y$ be continuous and surjective. Then $p$ is a quotient map if either of the following hold:

• (i) $p$ is an open or closed map (sends open sets to open sets, or closed sets to closed sets);
• (ii) $X$ is compact and $Y$ is Hausdorff (❓ compare to compactness & homeomorphisms?)

Theorem (Munkres 22.2): Characterizing continuous functions out of quotient spaces

Let $p:X\to Y$ be a quotient map. Let $Z$ be a space and let $f:X\to Z$ be a function which is constant on $p^{-1}(z)$ for all $z\in Z$. Then there exists a unique $\overline{f}:Y\to Z$ such that $\overline{f}\circ p=f$.

Moreover:

• (i) The universal property of quotients. $\overline{f}$ is continuous if and only if $f$ is continuous;
• (ii) $\overline{f}$ is a quotient map if and only if $f$ is a quotient map.

#concept-question Does this diagram commute?

Theorem: The relationship between quotient maps and quotient spaces

Let $p:X\to Y$ be a quotient map. Define and equivalence relation $\sim$ on $X$ by setting

$$x_1\sim x_2⟺p(x_1)=p(x_2).$$

Then there exists a homeomorphism $\overline{p}:X/\sim \to Y$, and the diagram

commutes. That is, every quotient map is of the form $X\to X/\sim$.

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Examples §

Interval with endpoints glued together §

Consider the interval $[0,1]$ with the equivalence relation $t\sim s$ if either: $t=s$; $t=0$ and $s=1$; or $t=1$ and $s=0$. This is the simplest equivalence relation under which $0\sim 1$. Note the following:

• There are two types of points in $[0,1]$: if $t\ne 0,1$, then the point $[t]\in [0,1]/\sim$ corresponds to the equivalence class $\{t\}$. If $t=0,1$, then the point $[t]\in [0,1]/\sim$ corresponds to the equivalence class $\{0,1\}$.
• Every open interval $(a,b)$ is saturated, meaning it is a union of equivalence classes $C_t=\{t\}$ for all $t\in (a,b)$, which project to (i.e., their images under $p:[0,1]\to [0,1]/\sim$ are) open sets in the quotient.
• The half-open interval $[0,b)$ for $b<1$ does not project to an open set $p([0,b))$. (❓ why would $1\in [0,b)$ if saturated?).

TBD §

$p:[0,1]\to S^1$, $p(t)=(\cos (2\pi t),\sin (2\pi t))$.

Making the torus from $X=[0,1{]}^2$, define $\sim$ by $(t,0)\sim (t,1)$ and $(0,s)\sim (1,s)$ for all $t,s\in [0,1]$.

• Take the minimal equivalence relation generated by these conditions (a relation) — taking the intersection of all relations $R\subseteq X\times X$.
• Technically $(0,0)\sim (0,1)$, $(0,0)\sim (1,0)\sim (1,1)$, have declared symmetry and transitivity by doing so? (the four points in the corners are equivalent)
  • ❓ so classes are explicitly what?

Quotients with $S^n$ and $D^n$ §

• $D^n\cong (S^{n-1}\times [0,1])/(S^{n-1}\times \{0\})$
• $D^n/S^{n-1}\cong S^n$.

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Review §

• 🔺 Show that if $U\subset X/\sim$ is open, then $p^{-1}(U)\subset X$ is an open subset of $X$ which is a union of equivalence classes (i.e., saturated); and that if $V\subset X$ is saturated, then $p(V)$ is open (note that $p^{-1}(p(V)))=V$.
• 🔺 Check that the quotient topology is indeed a topology.

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Proof appendix §

Lemma:

Let $p:X\to Y$ be continuous and surjective. Then $p$ is a quotient map if either of the following hold:

• (i) $p$ is an open or closed map (sends open sets to open sets, or closed sets to closed sets);
• (ii) $X$ is compact and $Y$ is Hausdorff (❓ compare to compactness & homeomorphisms?)

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Proof from Topology.

• (i) Suppose $p^{-1}(U)$ is open. If $p$ is an open map, then $p(p^{-1}(U))$ is an open set, and this is equivalent to $U$ by surjectivity. If $p$ is a closed map, $p(X\backslash p^{-1}(U))=Y\backslash U$ is closed, implying $U\subseteq Y$ is open.
• (ii) If $C\subseteq X$ is a closed set, then $C$ is also compact. Since $p$ is continuous, $p(C)$ is also compact, means $p(C)\subseteq Y$ closed subset of Hausdorff space. Then $p$ is a closed map and we are back in (i).#wip

Theorem (Munkres 22.2): Characterizing continuous functions out of quotient spaces

Let $p:X\to Y$ be a quotient map. Let $Z$ be a space and let $f:X\to Z$ be a function which is constant on $p^{-1}(z)$ for all $z\in Z$. Then there exists a unique $\overline{f}:Y\to Z$ such that $\overline{f}\circ p=f$.

Moreover:

• (i) The universal property of quotients. $\overline{f}$ is continuous if and only if $f$ is continuous;
• (ii) $\overline{f}$ is a quotient map if and only if $f$ is a quotient map.

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Proof from Topology.

• Existence of $\overline{f}$. Let $y\in Y$. Since $p$ is surjective, we may pick $x\in X$ such that $p(x)=y$, and set $\overline{f}(y)=f(x)$. The condition on $f$ guarantees $\overline{f}$ is well-defined ❓, uniqueness is clear.
• (i) $\overline{f}$ continuous $⟹f$ continuous. Let $U\subseteq Z$ be open. Then $f^{-1}(U)=p^{-1}({\overline{f}}^{-1}(U))$ is open in $X$. Since $p$ is a quotient map, this implies $f^{-1}(U)$ is open.
• (i) $f$ continuous $⟹\overline{f}$ continuous.
  1. 🔺 Exercise: check composition of quotient maps is a quotient map.

Proof of (ii) from MME Topology notes.

We restate (ii) as the following:

Proposition: Surjections from compact spaces to Hausdorff spaces are quotients

Let $f:X\to Y$ be a continuous surjection from a compact space to a Hausdorff space. Then if $\sim$ is the equivalence relation

$$x\sim z⟺f(x)=f(z),$$

then the induced map $g:X/\sim Y\to Y$ is a homeomorphism; equivalently, $f$ is a quotient map.

Theorem: The relationship between quotient maps and quotient spaces

Let $p:X\to Y$ be a quotient map. Define and equivalence relation $\sim$ on $X$ by setting

$$x_1\sim x_2⟺p(x_1)=p(x_2).$$

Then there exists a homeomorphism $\overline{p}:X/\sim \to Y$, and the diagram

commutes. That is, every quotient map is of the form $X\to X/\sim$.

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