\b[True.] If open sets cover $Y$, their <inverse images> under the continuous <surjection> $f$ cover $X$. <Compactness> of $X$ gives a finite subcover; <surjectivity> then makes the corresponding finite family cover $Y$. This proves the <compactness> of a <continuous image of a compact space>.

Solution

Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / E / 3 / a / Solution

 Solution (source code)