Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 1 / f / ii / Solution

The inverse factors are generally in the wrong order. For a counterexample, take the bijections

f (x) = x + 1

and

g (x) = 2 x

, with inverse functions

f^{- 1} (x) = x - 1

and

g^{- 1} (x) = x /2

. Then

((f \circ g) \circ (f^{- 1} \circ g^{- 1})) (x) = 2 (x /2 - 1) + 1 = x - 1.

Thus

False in general; the displayed composition need not be id .

For particular bijections that commute the equality can hold; the claim fails as a universal assertion.

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