 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/1/c/i/solution

Suppose

g (y) = g (y^{'})

. Applying

f

and using the right inverse identity gives

y = (f \circ g) (y) = (f \circ g) (y^{'}) = y^{'} .

This is exactly injectivity, so every right inverse is injective.

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