= Solution
{c}
Choose $Z=(A,\mathcal T_A)$ and let $f$ be the identity set map $A\to A$. The composition $i\circ f$ is the inclusion and is continuous by part (i). The hypothesized universal implication therefore says that the identity map
$$
\operatorname{id}:(A,\mathcal T_A)\longrightarrow(A,\mathcal T')
$$
is continuous. Its <preimage> of a set $V\in\mathcal T'$ is $V$ itself, so every such $V$ lies in $\mathcal T_A$. Therefore
$$
\boxed{\mathcal T'\subseteq\mathcal T_A.}
$$
Combining this with part (ii), when both properties hold, characterizes the <subspace topology> uniquely.
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