previous up next
Go backward to Function Composition
Go up to Top
Go forward to Inverse Function Properties
RISC-Linz logo

Function Inversion

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

forall A, 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

(forall x,y0, y1: (<x, y0> in f-1 /\  <x, y1> in f-1) => y0=y1).
Take arbitrary x, y0, and y1 and assume
(1) <x, y0> in f-1 /\  <x, y1> in f-1.
We have to show (2) y0=y1. From (1) and the definition of inverse, we know
(3) <y0, x> in f /\  <y1, x> in f,
i.e., f(y0) = x and f(y1) = x. Since f is injective, we thus know (2).
Author: Wolfgang Schreiner
Last Modification: December 7, 1999

previous up next