Quotient topology

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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 $X$

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 1: these definitions are equivalent

Universal property of the quotient topology

$\xymatrix{ X \ar[d]_{q} \ar[dr]^{f\circ q} & \\ Y \ar[r]_f & Z }$
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

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 $$p$$

[Expand]

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

[Expand]

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