Every surjective map gives rise to an equivalence relation

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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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Why was it even created?! Fix the surjective problem

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Flesh out, check existing content then demote to grade F
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Something in Topological Manifolds (Lee) should cover this. Under Quotient topology


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)


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.


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This is easy. Truly

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