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*Proposition:*
If a function is injective, its inverse is also a function:

forallA,B,f:A->^{injective}B: :f^{-1}:B->A.

*Proof:*
Take arbitrary `f`: `A` ->^{injective} `B`.
We have to show
`f`^{-1}: `B` -> `A`.

We have `f`^{-1} subset `B` x `A`. Thus it remains to be shown

(Take arbitraryforallx,y_{0},y_{1}: (<x,y_{0}> inf^{-1}/\ <x,y_{1}> inf^{-1}) =>y_{0}=y_{1}).

(1) <We have to show (2)x,y_{0}> inf^{-1}/\ <x,y_{1}> inf^{-1}.

(3) <i.e.,y_{0},x> inf/\ <y_{1},x> inf,

Author: Wolfgang Schreiner

Last Modification: December 7, 1999