\b[False:] the target need not equal the image. Take the one-point <compact space> $M=\{0\}$, $N=\mathbb R$, and the <continuous map> $f(0)=0$. The target is not <compact>, for example because it is unbounded.

The valid theorem is that a <continuous image of a compact space> is <compact>. Thus $f(M)$ is <compact>; <compactness> of $N$ follows if <surjectivity> is added.

Solution

Past exam of the metric spaces and complex analysis course of the University of Oxford / 2014 / 1 / b / ii / Solution

 Solution (source code)