Difference between revisions of "Surjection"

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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>
 
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>
  
{{Definition}}
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==Definition==
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Given a [[Function|function]] {{M|f:X\rightarrow Y}}, we say {{M|f}} is ''surjective'' if:
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* <math>\forall y\in Y\exists x\in X[f(x)=y]</math>
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* Equivalently <math>\forall y\in Y</math> the set <math>f^{-1}(y)</math> is non-empty. That is <math>f^{-1}(y)\ne\emptyset</math>
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==Theorems==
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===Obvious results===
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{{Begin Theorem}}
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====The composition of surjective functions is surjective====
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{{Begin Proof}}
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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 wish to show that <math>\forall z\in Z\exists x\in X[h(x)=z]</math>
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: Let {{M|z\in Z}} be given
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:: Then {{M|\exists y\in Y}} such that {{M|1=g(y)=z}}
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:: Of course also {{M|\exists x\in X}} such that {{M|1=f(x)=y}}
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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}}
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: We have shown {{M|1=\forall z\in Z\exists x\in X[h(x)=z]}} as required.<ref>Alec Teal's (own) work</ref>
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{{End Proof}}
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{{End Theorem}}
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==See also==
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* [[Injection]]
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* [[Bijection]]
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* [[Function]]
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==References==
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<references/>
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{{Definition|Set Theory}}

Revision as of 17:07, 10 May 2015

Surjective is onto - for f:AB

every element of B
is mapped onto from at least one thing in A

Definition

Given a function f:XY, we say f is surjective if:

  • yYxX[f(x)=y]
  • Equivalently yY
    the set f1(y)
    is non-empty. That is f1(y)

Theorems

Obvious results

[Expand]

The composition of surjective functions is surjective


See also

References

  1. Jump up Alec Teal's (own) work