Choose $Z=(A,\mathcal T_A)$ and let $f$ be the identity function on $A$. The composite $i\circ f$ is continuous by part (i). The stipulated universal implication therefore makes $f:(A,\mathcal T_A)\to(A,\mathcal T')$ continuous. Its inverse image of any $\mathcal T'$-<open set> is that set itself, so it must lie in $\mathcal T_A$. Thus $\boxed{\mathcal T'\subseteq\mathcal T_A}$.

Solution

Past exam of the options I course of the University of Oxford / 2007 / E / E2 / iv / Solution

 Solution (source code)