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

The subspace topology isThese sets contain the empty set and $A$, are closed under arbitrary unions, and are closed under finite intersections because intersections distribute over the common factor $A$. Thus they form a topology on $A$.

For $U\in \mathcal{T}$, the inverse image under inclusion is $i^{-1}(U)=A\cap U$, which is open in the subspace topology by definition. The inverse-image criterion for continuity therefore proves the inclusion is continuous.

Every subspace-open set has the form $A\cap U=i^{-1}(U)$ for some $U\in \mathcal{T}$. Continuity of $i:(A,{\mathcal{T}}^{\prime })\to (X,\mathcal{T})$ makes every such inverse image open in ${\mathcal{T}}^{\prime }$. Hence $\boxed{{\mathcal{T}}_A\subseteq {\mathcal{T}}^{\prime }}$.

If $A\cap U$ is subspace-open, thenThe right side is open in $Z$ by the assumed continuity of the composition. This holds for every subspace-open set, proving $f$ is continuous into the subspace topology.

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 $f:(A,{\mathcal{T}}_A)\to (A,{\mathcal{T}}^{\prime })$ continuous. Its inverse image of any ${\mathcal{T}}^{\prime }$-open set is that set itself, so it must lie in ${\mathcal{T}}_A$. Thus $\boxed{{\mathcal{T}}^{\prime }\subseteq {\mathcal{T}}_A}$.