Given a topological space $X$, let $\mathbf{Bundle}(X)=\mathbf{Top}/X$ be the category of bundles over $X$, and let $\mathbf{Sh}(X)$ be the category of sheaves over $X$. Then there's a sheafification functor $\mathbf{Bundle}(X)\to\mathbf{Sh}(X)$ that takes a bundle to its sheaf of sections, and this functor has a left adjoint that sends a sheaf to its corresponding étalé bundle.
In the case where $X$ is the one-point space these functors are just the forgetful functor $U:\mathbf{Top}\to\mathbf{Set}$ and the functor $\mathbf{Set}\to\mathbf{Top}$ that equips a set with its discrete topology. In this case $U$ also has a right adjoint that equips a set with its codiscrete (a.k.a. indiscrete) topology.
Does the sheafification functor have a right adjoint in the case of general $X$? Presumably this would be a functor that sent each sheaf to a bundle whose fibres had codiscrete topology.
Best Answer
No: sheafification does not preserve colimits. For example, let $X=\mathbb{R}$. Then in $\mathbf{Bundle}(\mathbb{R})$, the identity map $\mathbb{R}\to\mathbb{R}$ can be written as the filtered colimit of the inclusion maps $A\to\mathbb{R}$ where $A$ ranges over all countable subspaces of $\mathbb{R}$ (this is because a set in $\mathbb{R}$ is closed iff it is closed under convergence of sequences). No such inclusion map $A\to\mathbb{R}$ has a section over any nonempty open set, so the sheafification of each of them is the empty sheaf, and the colimit of the sheafifications is again the empty sheaf. But the identity map $\mathbb{R}\to\mathbb{R}$ has sections over every open set, so its sheafification is not the empty sheaf.