Passing to the quotient (topology)
From Maths
 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 passingtothequotient 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 passingtothequotient 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
