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I am trying to show that if X is a topological space, ~ an equivalence relation on X and q:X-->X/~ the quotient map (i.e. q(x)=[x]), then the quotient topology on X/~ (U in X/~ open iff q^{-1}(U) open in X) is characterized by the following universal property:
"If f:X-->Y is continuous and constant on each equivalence class of ~, then there exists a unique continuous map f':X/~-->Y such that f=f' o q"
That the quotient topology satisfies this is easy to show: just define f'([x]):=f(x). Then f' is continuous since for V open in Y, [itex]q^{-1}(f'^{-1}(V)) = (f' \circ q)^{-1}(V)=f^{-1}(V)[/itex], which is open in X.
But, for the other direction, we must show that if some topology T on X/~ satisfies the above universal property, then T is the quotient topology. I.e., we must show that [itex]q^{-1}(U)[/itex] is open in X <==> U is open in T.
I don't see how.
"If f:X-->Y is continuous and constant on each equivalence class of ~, then there exists a unique continuous map f':X/~-->Y such that f=f' o q"
That the quotient topology satisfies this is easy to show: just define f'([x]):=f(x). Then f' is continuous since for V open in Y, [itex]q^{-1}(f'^{-1}(V)) = (f' \circ q)^{-1}(V)=f^{-1}(V)[/itex], which is open in X.
But, for the other direction, we must show that if some topology T on X/~ satisfies the above universal property, then T is the quotient topology. I.e., we must show that [itex]q^{-1}(U)[/itex] is open in X <==> U is open in T.
I don't see how.