Quotient topology

Note: Motivation for quotient topology may be useful

Definition of Quotient topology

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

The quotient topology is actually a topology

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TODO: Easy enough

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Quotient map

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

[ilmath]p[/ilmath] is a quotient map if we have [math]U\in\mathcal{K}\iff p^{-1}(U)\in\mathcal{J}[/math]

That is to say [math]\mathcal{K}=\{V\in\mathcal{P}(Y)|p^{-1}(V)\in\mathcal{J}\}[/math]

Also known as:

• Identification map

Stronger than continuity

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

Theorems

Theorem: The quotient topology, [ilmath]\mathcal{Q} [/ilmath] is the largest topology such that the quotient map, [ilmath]p[/ilmath], is continuous

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

• [math]\mathcal{Q}=\{V\in\mathcal{P}|p^{-1}(V)\in\mathcal{J}\}[/math]

is the largest topology on [ilmath]Y[/ilmath] we can have such that [ilmath]p[/ilmath] is continuous

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

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

Suppose that [ilmath]\mathcal{K}\ne\mathcal{Q} [/ilmath]

Let [ilmath]V\in\mathcal{K} [/ilmath] such that [ilmath]V\notin \mathcal{Q} [/ilmath]

By continuity of [ilmath]p[/ilmath], [ilmath]p^{-1}(V)\in\mathcal{J} [/ilmath]

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

Thus any topology on [ilmath]Y[/ilmath] where [ilmath]p[/ilmath] 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 [ilmath](X,\mathcal{J})[/ilmath] and an Equivalence relation [ilmath]\sim[/ilmath], then the map: [math]q:(X,\mathcal{J})\rightarrow(\tfrac{X}{\sim},\mathcal{Q})[/math] with [math]q:p\mapsto[p][/math] (which is a quotient map) is continuous (as above)

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

Also known as:

• Identification space