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> | ||
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==Theorems== | ==Theorems== | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
Revision as of 17:37, 10 May 2015
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]
Given a function [ilmath]f:X\rightarrow Y[/ilmath], we say [ilmath]f[/ilmath] 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
The composition of surjective functions is surjective
Let [ilmath]f:X\rightarrow Y[/ilmath] and [ilmath]g:Y\rightarrow Z[/ilmath] be surjective maps, then their composition, [ilmath]g\circ f=h:X\rightarrow Z[/ilmath] is surjective.
- We wish to show that [math]\forall z\in Z\exists x\in X[h(x)=z][/math]
- Let [ilmath]z\in Z[/ilmath] be given
- Then [ilmath]\exists y\in Y[/ilmath] such that [ilmath]g(y)=z[/ilmath]
- Of course also [ilmath]\exists x\in X[/ilmath] such that [ilmath]f(x)=y[/ilmath]
- We now know [ilmath]\exists x\in X[/ilmath] with [ilmath]f(x)=y[/ilmath] and [ilmath]g(y)=g(f(x))=h(x)=z[/ilmath]
- We have shown [ilmath]\forall z\in Z\exists x\in X[h(x)=z][/ilmath] as required.[1]
See also
References
- ↑ Alec Teal's (own) work
