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

From Maths
Jump to: navigation, search
m (Statement)
m (See also)
Line 11: Line 11:
 
{{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==
 
==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]]
+
* [[Factoring a function through the projection of an equivalence relation induced by that function yields an injection]]
{{Todo|Link to continuous version}}
+
** [[If a surjective function is factored through the canonical projection of the equivalence relation induced by that function then the yielded function is a bijection]]
 +
{{Todo|Link to continuous version ([[File:MondTop2016ex1.pdf]] - Q5)}}
 +
 
 
==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 22:28, 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:
Homework assignment isn't sufficient

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 (File:MondTop2016ex1.pdf - Q5)



References

  1. File:MondTop2016ex1.pdf