Passing to the quotient (topology)
From Maths
- See passing to the quotient for other forms of this theorem
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[hide]Statement
Suppose that (X,J) is a topological space and ∼ is an equivalence relation, let (X∼,Q) be the resulting quotient topology and π:X→X∼ the resulting quotient map, then:
- Let (Y,K) be any topological space and let f:X→Y be a continuous map that is constant on the fibres of π[Note 1] then:
- there exists a unique continuous map, ˉf:X∼→Y such that f=¯f∘π
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
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 f and π are continuous)
Proof
Notes
- Jump up ↑
That means that:
- ∀x,y∈X[π(x)=π(y)⟹f(x)=f(y)] - as mentioned in passing-to-the-quotient for functions, or
- ∀x,y∈X[f(x)≠f(y)⟹π(x)≠π(y)], also mentioned
- See :- Equivalent conditions to being constant on the fibres of a map for details
References
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