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}} | + | ==Definition== |
+ | Given a [[Function|function]] {{M|f:X\rightarrow Y}}, we say {{M|f}} is ''surjective'' if: | ||
+ | * <math>\forall y\in Y\exists x\in X[f(x)=y]</math> | ||
+ | * 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> | ||
+ | ==Theorems== | ||
+ | ===Obvious results=== | ||
+ | {{Begin Theorem}} | ||
+ | ====The composition of surjective functions is surjective==== | ||
+ | {{Begin 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. | ||
+ | |||
+ | : We wish to show that <math>\forall z\in Z\exists x\in X[h(x)=z]</math> | ||
+ | |||
+ | |||
+ | : Let {{M|z\in Z}} be given | ||
+ | :: Then {{M|\exists y\in Y}} such that {{M|1=g(y)=z}} | ||
+ | :: Of course also {{M|\exists x\in X}} such that {{M|1=f(x)=y}} | ||
+ | :: 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 have shown {{M|1=\forall z\in Z\exists x\in X[h(x)=z]}} as required.<ref>Alec Teal's (own) work</ref> | ||
+ | {{End Proof}} | ||
+ | {{End Theorem}} | ||
+ | |||
+ | ==See also== | ||
+ | * [[Injection]] | ||
+ | * [[Bijection]] | ||
+ | * [[Function]] | ||
+ | |||
+ | ==References== | ||
+ | <references/> | ||
+ | |||
+ | {{Definition|Set Theory}} |
Revision as of 17:07, 10 May 2015
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∈Ythe set f−1(y)is non-empty. That is f−1(y)≠∅
Theorems
Obvious results
[Expand]
The composition of surjective functions is surjective
See also
References
- Jump up ↑ Alec Teal's (own) work