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

Table of contents
- Solution 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.

 Ancestors (12)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version