Solution: Let $\mathcal{T} _{cc}$ and $\mathcal{T} _{cf}$ denote the co-countable and co-finite topology on $X$ respectively. We need to show that the identity map \[ f : \left( X, \mathcal{T} _{cc} \right) \to \left( X, \mathcal{T} _{cf} \right) \] is continuous. Let $U \in \mathcal{T} _{cf}$. This means that $U^c$ is either $X, \emptyset$ or finite. If $U^c$ is either $X$ or $\emptyset$, then $f^{-1} (U) $ will also be either $X$ or $\emptyset$ which belongs to $\mathcal{T} _{cc}$. So let us assume that $U^c$ is neither the full set nor the empty set. Consider $f^{-1} (U) = U$. Since, $U^c$ is finite and hence countable, $U \in \mathcal{T} _{cc}$. Thus, $I$ is continuous.

Now, let \[ g: \left( X, \mathcal{T} _{cf} \right) \to \left( X, \mathcal{T} _{cc} \right) \] is the identity map. Let $g$ be continuous. We need to show that $X$ is finite. If $X$ is not finite, choose a set $U$ such that $U^c$ is not finite but countable. Thus, $U \in \mathcal{T} _{cc}$. Note that $g^{-1} (U) (=U)$ is not finite and hence $U \notin \mathcal{T} _{cf}$. Thus, $X$ has to be finite. On the other hand if $X$ is finite, then we note that $\mathcal{T} _{cc} = \mathcal{T} _{cf}$ and hence the identity map is continuous.