Difference between revisions of "Equivalence relation induced by a function"
From Maths
(Created page with "{{Stub page|grade=B|msg=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, g...") |
m |
||
Line 5: | Line 5: | ||
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)}} | ||
+ | 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. | ||
+ | * 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]]}} | ||
+ | ==See also== | ||
+ | * [[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]] | ||
+ | {{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
Stub grade: B
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
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
Grade: A
This page requires references, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:
The message provided is:
Homework assignment isn't sufficient
Contents
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:
This proof has been marked as an page requiring an easy proof
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
|