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

Latest revision as of 20:23, 11 October 2016

Statement

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 $\pi$^{[Note 1]} 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

Notes

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

References