Every surjective map gives rise to an equivalence relation
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Warning:
Warning:There is something wrong here, as the surjective property is never used! It is true though that every map, [ilmath]f:X\rightarrow Y[/ilmath] gives rise to an equivalence relation, where [ilmath]x_1\sim x_2[/ilmath] if [ilmath]f(x_2)=f(x_2)[/ilmath] but this doesn't use the surjective part!
- Notice also that in such a case we can factor [ilmath]f[/ilmath] through [ilmath]\pi:X\rightarrow X/\sim[/ilmath] always to yield [ilmath]\bar{f} [/ilmath], and "distil" the information of [ilmath]f[/ilmath] into this new map, [ilmath]\bar{f} [/ilmath].
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Statement
Let [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] be sets, and let [ilmath]f:X\rightarrow Y[/ilmath] be a surjective mapping. Then we can define an equivalence relation on [ilmath]X[/ilmath], [ilmath]\sim\subseteq X\times X[/ilmath], by:
- For [ilmath]x_1,x_2\in X[/ilmath], we say [ilmath]x_1\sim x_2[/ilmath] if [ilmath]f(x_1)=f(x_2)[/ilmath]
So we have [ilmath]\forall x_1,x_2\in X[x_1\sim x_2\iff f(x_1)=f(x_2)][/ilmath] (for the "and only if" part, see definitions and iff)
- See also: every equivalence relation gives rise to a surjective map, usually written [ilmath]\pi:X\rightarrow X/\sim[/ilmath] given by [ilmath]\pi:x\mapsto[x][/ilmath] where [ilmath][x]:=\{y\in X\ \vert x\sim y\}[/ilmath], an equivalence class
Purpose
The quotient topology can be defined using a map to identify parts together, but also an equivalence relation can do the same job, this page is half way to showing them to be the same.
Proof
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This proof has been marked as an page requiring an easy proof
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