 Solution

ID: past-exam-of-the-topology-course-of-the-university-of-oxford/2014/1/b/ii/solution

Because

f

f^{- 1} (B)

is infinite whenever

B

is infinite; a finite preimage could have only finitely many image points. By part (i), its closure is

X

. Apply part (a)(ii):

Y = f (X) = f (\overline{f^{- 1} (B)}) \subseteq \overline{f (f^{- 1} (B))} = \overline{B} .

The other inclusion is automatic, so

\overline{B} = Y for every infinite B \subseteq Y .

Both continuity and surjectivity enter the argument.

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