 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-4/7e/a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

f

x \in f^{- 1} (f (A))

, then

f (x) = f (a)

for some

a \in A

, so injectivity gives

x = a \in A

. The reverse inclusion always holds; hence

f^{- 1} (f (A)) = A

Conversely, if (ii) holds and

f (x) = f (y)

, then

y \in f^{- 1} (f ({x})) = {x}

, so

x = y

. Thus (i) and (ii) are equivalent.

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