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*Proposition:*
The composition of two bijective functions is bijective:

forallA,B,C,f:A->^{bijective}B,g:B->^{bijective}C:fog:A->^{bijective}C.

*Proof:*
Take arbitrary `f`: `A` ->^{bijective} `B`, `g`: `B`
->^{bijective} `C`.

- We show
`f`o`g`is injective. Take arbitrary`x`_{0}in`A`and`x`_{1}in`A`with (`f`o`g`)(`x`_{0}) = (`f`o`g`)(`x`_{1}). We have to show`x`_{0}=`x`_{1}.We know, by definition of o , that

`g`(`f`(`x`_{0})) =`g`(`f`(`x`_{1})) and thus, because`g`is injective,`f`(`x`_{0}) =`f`(`x`_{1}). Since`f`is injective, we then have`x`_{0}=`x`_{1}. - We show
`f`o`g`is surjective. Take arbitrary`z`in`C`; we have to find some`x`such that (`f`o`g`)(`x`) =`z`.Since

`g`is surjective, we have some`y`in`B`such that`g`(`y`) =`z`. Since`f`is surjective, we have some`x`in`A`such that`f`(`x`) =`y`. Thus (`f`o`g`)(`x`) =`g`(`f`(`x`)) =`g`(`y`) =`z`.

Author: Wolfgang Schreiner

Last Modification: December 7, 1999