Difference between revisions of "Surjection"
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m (Linking to surjection's problem, making note to apply to bijection - finishing proof that really should be in its own page.) |
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| − | + | {{Requires work|grade=A*|msg=See [[Injection]]'s requires-work box [https://wiki.unifiedmathematics.com/index.php?title=Injection&oldid=9337 permalink] [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 21:56, 8 May 2018 (UTC) | |
| + | Also: | ||
| + | * Factor out composition theorem into own page. [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 21:56, 8 May 2018 (UTC) | ||
| + | * Apply this to the [[Bijection]] page too [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 21:56, 8 May 2018 (UTC)}} | ||
| + | |||
| + | :: Surjective is onto - for <math>f:A\rightarrow B</math> every element of <math>B</math> is mapped onto from at least one thing in <math>A</math> | ||
| + | __TOC__ | ||
==Definition== | ==Definition== | ||
{{:Surjection/Definition}} | {{:Surjection/Definition}} | ||
==Theorems== | ==Theorems== | ||
| − | {{Begin Theorem}} | + | {{Begin Inline Theorem}}<!-- TODO: MOVE TO OWN SUBPAGE ! --> |
The composition of surjective functions is surjective | The composition of surjective functions is surjective | ||
| − | {{Begin Proof}} | + | {{Begin Inline Proof}} |
Let {{M|f:X\rightarrow Y}} and {{M|g:Y\rightarrow Z}} be surjective maps, then their composition, {{M|1=g\circ f=h:X\rightarrow Z}} is surjective. | Let {{M|f:X\rightarrow Y}} and {{M|g:Y\rightarrow Z}} be surjective maps, then their composition, {{M|1=g\circ f=h:X\rightarrow Z}} is surjective. | ||
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:: We now know {{M|\exists x\in X}} with {{M|1=f(x)=y}} and {{M|1=g(y)=g(f(x))=h(x)=z}} | :: We now know {{M|\exists x\in X}} with {{M|1=f(x)=y}} and {{M|1=g(y)=g(f(x))=h(x)=z}} | ||
| − | : | + | : Thus it is shown that: |
| + | :* {{M|1=\forall z\in Z\exists x\in X[h(x)=z]}} | ||
| + | : as required.<ref>[[User:Alec|Alec's]] work - the proof speaks for itself</ref> | ||
{{End Proof}} | {{End Proof}} | ||
{{End Theorem}} | {{End Theorem}} | ||
Latest revision as of 21:56, 8 May 2018
Grade: A*
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- Surjective is onto - for f:A→B every element of B is mapped onto from at least one thing in A
Contents
[hide]Definition
Given a function f:X→Y, we say f is surjective if:
- ∀y∈Y∃x∈X[f(x)=y]
- Equivalently ∀y∈Y the set f−1(y) is non-empty. That is f−1(y)≠∅
Theorems
[Expand]
The composition of surjective functions is surjective
