 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/2/d/iv/solution

Suppose

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

. Since

f

is surjective, choose

a, a^{'} \in A

with

f (a) = b

and

f (a^{'}) = b^{'}

. Then

(g \circ f) (a) = (g \circ f) (a^{'})

, so injectivity gives

a = a^{'}

and therefore

b = b^{'}

. Thus

g

is injective. The use of surjectivity is what makes every point of

B

accessible to this argument.

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