View source for Every surjective map gives rise to an equivalence relation

==Warning:==
{{Warning|There is something wrong here, as the surjective property is never used! It is true though that every map, {{M|f:X\rightarrow Y}} gives rise to an equivalence relation, where {{M|x_1\sim x_2}} if {{M|1=f(x_2)=f(x_2)}} but this doesn't use the surjective part!}}
* Notice also that in such a case we can {{link|factor|function}} {{M|f}} through {{M|\pi:X\rightarrow X/\sim}} always to yield {{M|\bar{f} }}, and "distil" the information of {{M|f}} into this new map, {{M|\bar{f} }}.
{{Dire page|1=Why was it even created?! Fix the surjective problem}}


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{{Stub page|grade=C|msg=Flesh out, check existing content then demote to grade {{C|F}}}}
{{Requires references|grade=D|msg=Something in Topological Manifolds (Lee) should cover this. Under Quotient topology}}
==Statement==
Let {{M|X}} and {{M|Y}} be [[sets]], and let {{M|f:X\rightarrow Y}} be a ''[[surjective]]'' [[mapping]]. Then we can define an [[equivalence relation]] on {{M|X}}, {{M|\sim\subseteq X\times X}}, by:
* For {{M|x_1,x_2\in X}}, we say {{M|x_1\sim x_2}} if {{M|1=f(x_1)=f(x_2)}} 
So we have {{M|1=\forall x_1,x_2\in X[x_1\sim x_2\iff f(x_1)=f(x_2)]}} (for the "and only if" part, see [[definitions and iff]])
* '''See also: ''' [[every equivalence relation gives rise to a surjective map]], usually written {{M|\pi:X\rightarrow X/\sim}} given by {{M|\pi:x\mapsto[x]}} where {{M|1=[x]:=\{y\in X\ \vert x\sim y\} }}, 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==
{{Requires proof|grade=C|easy=True|This is easy. Truly}}
==References==
<references/>
{{Theorem Of|Elementary Set Theory|Set Theory}}