Exercise 18.3

Proof. First note that clearly, the inverse of the identity function is itself with the domain and image reversed, and that for any subset $A\subset X=X^{\prime }$ we have $i(A)=i^{-1}(A)=A$.

$\text{(⇒)}$ Suppose that $i$ is continuous and consider any open set $U\in T$. Then we have that $i^{-1}(U)=U$ is open in $T^{\prime }$ since $i$ is continuous. Since $U$ was arbitrary, this shows that $T\subset T^{\prime }$ so that $T^{\prime }$ is finer.

$\text{(⇐)}$ Now suppose that $T^{\prime }$ is finer so that $T\subset T^{\prime }$. Consider any open set $U\in T$ so that also clearly $U\in T^{\prime }$, i.e. $U$ is also open in $T^{\prime }$. Since $i^{-1}(U)=U$, this shows that $i$ is continuous by the definition of continuity. □

Proof. Clearly, $i$ is a bijection since its domain and image are the same set, and $i^{-1}=i$. We then have that

as desired. □