Difference between revisions of "Equivalence relation induced by a function"

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Let {{M|X}} and {{M|Y}} be [[sets]] and let {{M|f:X\rightarrow Y}} be any [[mapping]] between them. Then {{M|f}} induces an [[equivalence relation]], {{M|\sim\subseteq X\times X}} where<ref>[[File:MondTop2016ex1.pdf]]</ref>:
 
Let {{M|X}} and {{M|Y}} be [[sets]] and let {{M|f:X\rightarrow Y}} be any [[mapping]] between them. Then {{M|f}} induces an [[equivalence relation]], {{M|\sim\subseteq X\times X}} where<ref>[[File:MondTop2016ex1.pdf]]</ref>:
 
* 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)}}
 
* 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)}}
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Note that {{M|f}} may be [[factor (function)|factored]] through the [[canonical projection of an equivalence relation]] to yield an injection. Furthermore if {{M|f}} is [[surjective]], then so is the induced map, and then the induced map is a bijection.
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* See: [[factoring through the projection of an equivalence relation generated by a map yields an injection and if the map is surjective then the factored map is surjective also and is thus a bijection]]
 
==Proof==
 
==Proof==
 
{{Requires proof|grade=D|easy=true|msg=Easy proof, marked as such. Just gotta show it's an [[equivalence relation]]}}
 
{{Requires proof|grade=D|easy=true|msg=Easy proof, marked as such. Just gotta show it's an [[equivalence relation]]}}
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==See also==
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* [[factoring through the projection of an equivalence relation generated by a map yields an injection and if the map is surjective then the factored map is surjective also and is thus a bijection]]
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{{Todo|Link to continuous version}}
 
==References==
 
==References==
 
<references/>
 
<references/>
 
{{Relations navbox|plain}}
 
{{Relations navbox|plain}}
 
{{Definition|Elementary Set Theory|Set Theory}}
 
{{Definition|Elementary Set Theory|Set Theory}}

Revision as of 14:14, 8 October 2016

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Something weird happened with every surjective map gives rise to an equivalence relation this page is what it SHOULD be. I also have a reference, granted not that strong of one
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Statement

Let [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] be sets and let [ilmath]f:X\rightarrow Y[/ilmath] be any mapping between them. Then [ilmath]f[/ilmath] induces an equivalence relation, [ilmath]\sim\subseteq X\times X[/ilmath] where[1]:

  • 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]

Note that [ilmath]f[/ilmath] may be factored through the canonical projection of an equivalence relation to yield an injection. Furthermore if [ilmath]f[/ilmath] is surjective, then so is the induced map, and then the induced map is a bijection.

Proof

Grade: D
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
Easy proof, marked as such. Just gotta show it's an equivalence relation

This proof has been marked as an page requiring an easy proof

See also


TODO: Link to continuous version


References

  1. File:MondTop2016ex1.pdf