 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-4/1d/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

True. For each

a \in A

a \in f^{- 1} (Z \cap W) ⟺ f (a) \in Z \cap W ⟺ f (a) \in Z and f (a) \in W ⟺ a \in f^{- 1} (Z) \cap f^{- 1} (W) .

Thus

f^{- 1} (Z \cap W) = f^{- 1} (Z) \cap f^{- 1} (W)

. Unlike the corresponding statement for images of a function, this intersection identity requires no injectivity.

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