Difference between revisions of "Passing to the quotient (topology)/Statement"

Latest revision as of 20:23, 11 October 2016

Statement

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

• Let [ilmath](Y,\mathcal{ K })[/ilmath] be any topological space and let [ilmath]f:X\rightarrow Y[/ilmath] be a continuous map that is constant on the fibres of [ilmath]\pi[/ilmath]^{[Note 1]} then:
• there exists a unique continuous map, [ilmath]\bar{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] such that [ilmath]f=\overline{f}\circ\pi[/ilmath]

We may then say [ilmath]f[/ilmath] descends to the quotient or passes to the quotient

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

Notes

1. ↑ That means that:
   • [ilmath]\forall x,y\in X[\pi(x)=\pi(y)\implies f(x)=f(y)][/ilmath] - as mentioned in passing-to-the-quotient for functions, or
   • [ilmath]\forall x,y\in X[f(x)\ne f(y)\implies \pi(x)\ne\pi(y)][/ilmath], also mentioned
   • See :- Equivalent conditions to being constant on the fibres of a map for details

References