\b[False.] Give $X=\mathbb R$ its usual <topology>, let $Y$ be a one-point space, and take the constant <continuous map> $f:X\to Y$. The set $K=Y$ is <compact>, but $f^{-1}(K)=\mathbb R$ is not <compact>. For example, the <open cover> $\{(-n,n):n\ge1\}$ has no finite subcover. Thus continuity alone does not preserve <compactness> under inverse images.

Solution

Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 2 / 4A / ii / Solution

 Solution (source code)