Difference between revisions of "Passing to the quotient (topology)"
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Latest revision as of 17:43, 8 October 2016
- See passing to the quotient for other forms of this theorem
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Contents
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
This is an instance of passing to the quotient (function) with functions between sets. In this proof we apply this theorem and show the result yields a continuous mapping (by assuming both [ilmath]f[/ilmath] and [ilmath]\pi[/ilmath] are continuous)
Proof
Notes
- ↑
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
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