 Solution

ID: 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/v/solution

The proposed inverse function cancels on both sides:

(f \circ g) \circ (g^{- 1} \circ f^{- 1}) (g^{- 1} \circ f^{- 1}) \circ (f \circ g) = id, = g^{- 1} \circ (f^{- 1} \circ f) \circ g = id .

The uniqueness of an inverse function therefore gives

True: (f \circ g)^{- 1} = g^{- 1} \circ f^{- 1} .

The operation applied last must be undone first.

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