 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2004/d/d4/c/solution

True in both cases. For every open

U

f

makes

f^{- 1} (U)

open. Then

(f \circ g)^{- 1} (U) = g^{- 1} (f^{- 1} (U))

g

. The value

f (0)

has no bearing on this preimage argument.

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