 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability-resit/1/d/i/solution

g \circ f : A \to C

is defined by

(g \circ f) (a) = g (f (a))

. Suppose

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

. Applying

g

gives

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

. Since the composite is an injective function,

a = a^{'}

. Thus $f$ is an injective function.

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