 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/1/c/ii/solution

f (x) = f (y)

, applying the function

f

gives

f (f (x)) = f (f (y))

. Since

f \circ f = id_{A}

, this implies

x = y

, proving injectivity. Also every

y \in A

is the image of

f (y)

, because

f (f (y)) = y

Thus

f

f^{- 1} = f .

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