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See Injection's requires-work box permalink Alec (talk) 21:56, 8 May 2018 (UTC)


  • Factor out composition theorem into own page. Alec (talk) 21:56, 8 May 2018 (UTC)
  • Apply this to the Bijection page too 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]


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]


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]
Thus it is shown that:
  • [ilmath]\forall z\in Z\exists x\in X[h(x)=z][/ilmath]
as required.[1]

See also


  1. Alec's work - the proof speaks for itself