Exercise 22.5

Consider any open set $U$ in the subspace $A$ so that $U=A\cap V$ for some open set $V$ in $X$ by the definition of a subspace. Since $A$ is also open in $X$ we have that $A\cap V=U$ is also open in $X$ by the definition of a topology. Then $q(U)=p(U)$ is open in $Y$ since $p$ is an open map. Since $U\subset A$, it follows that $p(U)\subset p(A)$ by Exercise 2.2 part (e) so that $p(U)\cap p(A)=p(U)=q(U)$. This shows that $q(U)$ is open in the subspace $p(A)$ since we have shown that $p(U)$ is open in $Y$. Therefore $q$ is an open map since $U$ was an arbitrary open set of $A$.