Quotient topology

• Need to add Equivalent conditions to a map being a quotient map

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Definition

There are a few definitions of the quotient topology however they do not conflict. This page might change shape while things are put in place

Quotient topology via an equivalence-relation definition

Given a topological space, $(X,\mathcal{J})$ and an equivalence relation on $X$, $\sim$, the quotient topology on $\frac{X}{\sim}$, $\mathcal{K}$ is defined as:

• The set $\mathcal{K}\subseteq\mathcal{P}(\frac{X}{\sim})$ such that:
  • $\forall U\in\mathcal{P}(\frac{X}{\sim})[U\in\mathcal{K}\iff \pi^{-1}(U)\in\mathcal{J}]$ or equivalently
• $\mathcal{K}=\{U\in\mathcal{P}(\frac{X}{\sim})\ \vert\ \pi^{-1}(U)\in\mathcal{J}\}$

In words:

• The topology on $\frac{X}{\sim}$ consists of all those sets whose pre-image (under $\pi$) are open in $X$

Claim 1: $\mathcal{K}$ is indeed a topology on $\frac{X}{\sim}$

Quotient topology via a mapping to a set definition

Let $(X,\mathcal{J})$ be a topological space and let $h:X\rightarrow Y$ be a surjective map onto a set $Y$, then the quotient topology, $\mathcal{K}\subseteq\mathcal{P}(Y)$ is a topology we define on $Y$ as follows:

• $\forall U\in\mathcal{P}(Y)[Y\in\mathcal{K}\iff h^{-1}(U)\in\mathcal{J}]$ or equivalently:
• $\mathcal{K}=\{U\in\mathcal{P}(Y)\ \vert\ h^{-1}(U)\in\mathcal{J}\}$

The quotient topology on $Y$ consists of all those subsets of $Y$ whose pre-image (under $h$) is open in $X$

Claim 2: these definitions are equivalent

Immediate theorems

The next two theorems demonstrate the purpose, the job if you will, of the quotient topology. The second (passing to the quotient) is the most important.

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Universal property of the quotient topology

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$\xymatrix{ X \ar[d]_{q} \ar[dr]^{f\circ q} & \\ Y \ar[r]_f & Z }$
In this commutative diagram $f$ is continuous $\iff$ $f\circ q$ is continuous

Let $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$ be topological spaces and let $q:X\rightarrow Y$ be a quotient map. Then:

• For any topological space, $(Z,\mathcal{ H })$ a map, $f:Y\rightarrow Z$ is continuous if and only if the composite map, $f\circ q$, is continuous

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Passing to the quotient

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$\xymatrix{ X \ar[d]_\pi \ar[dr]^f & \\ \frac{X}{\sim} \ar@{.>}[r]^{\overline{f} }& Y}$
$f$ descends to the quotient

Suppose that $(X,\mathcal{ J })$ is a topological space and $\sim$ is an equivalence relation, let $(\frac{X}{\sim},\mathcal{ Q })$ be the resulting quotient topology and $\pi:X\rightarrow\frac{X}{\sim}$ the resulting quotient map, then:

• Let $(Y,\mathcal{ K })$ be any topological space and let $f:X\rightarrow Y$ be a continuous map that is constant on the fibres of $\pi$ then:
• there exists a unique continuous map, $\bar{f}:\frac{X}{\sim}\rightarrow Y$ such that $f=\overline{f}\circ\pi$

We may then say $f$ descends to the quotient or passes to the quotient

Note: this is an instance of passing-to-the-quotient for functions

OLD PAGE

Note: Motivation for quotient topology may be useful

Definition of the Quotient topology

$$\begin{xy} \xymatrix{ X \ar[r]^p \ar[dr]_f & Q \ar@{.>}[d]^{\tilde{f}}\\ & Y } \end{xy}$$

(OLD)Definition of Quotient topology

If

$$(X,\mathcal{J})$$ is a topological space, $$A$$ is a set, and $$p:(X,\mathcal{J})\rightarrow A$$ is a surjective map then there exists exactly one topology $$\mathcal{J}_Q$$ relative to which $$p$$ is a quotient map. This is the quotient topology induced by $$p$$

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The quotient topology is actually a topology

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TODO: Easy enough

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Quotient map

Let $(X,\mathcal{J})$ and $(Y,\mathcal{K})$ be topological spaces and let $p:X\rightarrow Y$ be a surjective map.

$p$ is a quotient map if we have

$$U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}$$

That is to say

$$\mathcal{K}=\{V\in\mathcal{P}(Y)|p^{-1}(V)\in\mathcal{J}\}$$

Also known as:

• Identification map

Stronger than continuity

If we had $\mathcal{K}=\{\emptyset,Y\}$ then $p$ is automatically continuous (as it is surjective), the point is that $\mathcal{K}$ is the largest topology we can define on $Y$ such that $p$ is continuous

Theorems

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Theorem: The quotient topology, $\mathcal{Q}$ is the largest topology such that the quotient map, $p$, is continuous. That is to say any other topology such on $Y$ such that $p$ is continuous is contained in the quotient topology

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For a map $p:X\rightarrow Y$ where $(X,\mathcal{J})$ is a Topological space we will show that the topology on $Y$ given by:

• $$\mathcal{Q}=\{V\in\mathcal{P}|p^{-1}(V)\in\mathcal{J}\}$$

is the largest topology on $Y$ we can have such that $p$ is continuous

Proof method: suppose there's a larger topology, reach a contradiction.

Suppose that $\mathcal{K}$ is any topology on $Y$ and that $p:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})$ is continuous.

Suppose that $\mathcal{K}\ne\mathcal{Q}$

Let $V\in\mathcal{K}$ such that $V\notin \mathcal{Q}$

By continuity of $p$, $p^{-1}(V)\in\mathcal{J}$

This contradicts that $V\notin\mathcal{Q}$ as $\mathcal{Q}$ contains all subsets of $Y$ whose inverse image (preimage) is open in $X$

Thus any topology on $Y$ where $p$ is continuous is contained in the quotient topology

This theorem hints at the Characteristic property of the quotient topology

Quotient space

Given a Topological space $(X,\mathcal{J})$ and an Equivalence relation $\sim$, then the map:

$$q:(X,\mathcal{J})\rightarrow(\tfrac{X}{\sim},\mathcal{Q})$$ with $$q:p\mapsto[p]$$ (which is a quotient map) is continuous (as above)

The topological space $(\tfrac{X}{\sim},\mathcal{Q})$ is the quotient space where $\mathcal{Q}$ is the topology induced by the quotient

Also known as:

• Identification space

See also

• Open and closed maps