Difference between revisions of "Quotient topology"

Revision as of 19:53, 10 June 2015

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 $$p$$

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^[1] 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

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^[2] where $\mathcal{Q}$ is the topology induced by the quotient

Also known as:

• Identification space

See also

• Open and closed maps

References

1. Jump up ↑ Topology - Second Edition - James R Munkres
2. Jump up ↑ Introduction to topological manifolds - John M Lee - Second edition