The following is taken from "An introduction to Category Theory" by Harold Simmons
$\color{Green}{Background:}$
$\textbf{Example (Galois connection):}$ We modify the category of $\textbf{Pos}$(Poset) to produce a new category $\textbf{Pos}^{\vdash}.$ As with $\textbf{Pos},$ the objects of $\textbf{Pos}^{\vdash}.$ are posets, but the arrows are different.
Given a pair $S,T$ of posets, an $\mathrm{adjunction}$ from $S$ to $T$ is a pair of monotone maps as on the left such that the equivalence on the right
$$S\substack{ \xrightarrow{f}\\ \xleftarrow[g]\\}T\quad\quad f(a)\leq b \text{ iff } a\leq g(b)$$
holds for all $a\in S$ and $b\in T.$ We call
$$f\text{ the left adjoint }\quad\quad g\text{ the right adjoint }$$
of the pair, and sometimes write
$$f\dashv g$$
to indicate an adjunction.
$\textbf{(1) Exercise:}$ Each set $S$ can be converted into a preset (pre-order set) in two extreme ways. The discrete version uses equality as the comparison. For the indiscrete version any two elements are comparable. Show that the forgetful functor
$$\textbf{Set}\leftarrow \textbf{Pre}$$
has both a left and a right adjoint, and these are different.
$\textit{Solution:}$ Consider the forgetful functor
$\textbf{Set}\xleftarrow{U} \textbf{Pre}$
We produce a left adjoint and a right adjoint to this functor:
$$D\vdash U\quad\quad \small\textbf{Set}\substack{ \xrightarrow{D}\\ \xleftarrow{U}\\ \xrightarrow[I]\\}\small\textbf{Pre}\quad\quad U\vdash I$$
Each set $X$ $\color{blue}{can}$ $\color{blue}{be}$ $\color{blue}{converted}$ $\color{blue}{into}$ a preset $\color{blue}{in}$ $\color{blue}{extreme}$ $\color{blue}{ways}$
$$DX=(X,=)\quad\quad IX=(X,\parallel)$$
On the left we use equality as the comparison. This gives a poset. On the right any two elements are comparable. This is not a poset if $X$ has least two elements. We call these the |
$$\mathrm{discrete}\quad\quad\mathrm{indiscrete}$$
presets....
$\textbf{(2) Exercise:}$ Show that the forgetful functor
$$\textbf{Set}\leftarrow \textbf{Top}$$
has both a left and a right adjoint, and these are different.
$\textit{Solution:}$ This is more or less the same as solution of the previous exercise except we now use topologies rather than pre-orders.
Let
$$\textbf{Set}\leftarrow \textbf{Top}$$
to be the forgetful functor. We product a left adjoint and a right adjoint to this functor.
$$\small\textbf{Set}\substack{ \xrightarrow{D}\\ \xleftarrow{U}\\ \xrightarrow[I]\\}\small\textbf{Top}$$
Each set $X$ $\color{blue}{can}$ $\color{blue}{be}$ $\color{blue}{converted}$ $\color{blue}{into}$ a topological space $\color{blue}{in}$ $\color{blue}{extreme}$ $\color{blue}{ways}.$
$$DX=(X,PX)\quad\quad IX=(X,\{\emptyset, X\})$$
On the left we use discrete topology in which each subset is open. On the right we use the indiscrete topology in which only the two extreme subsets are open. Naturally we call these the
$$\mathrm{discrete}\quad\quad\mathrm{indiscrete}$$
space...
$\color{Red}{Questions:}$
In the solutions for the above two questions for finding the left and right adjoints, Simmons stated that "Each set $X$ $\color{blue}{can}$ $\color{blue}{be}$ $\color{blue}{converted}$ $\color{blue}{into}$ $...$ $\color{blue}{in}$ $\color{blue}{extreme}$ $\color{blue}{ways}.$" In the case of question 1, it is for pre-order sets, and for question 2, it is for topological spaces. Am I right to assume that for any objects in a category where there is order set kind of structure, in the sense of objects that can be compared like in a pre-order or partial order, and that can also be formed into a lattice/complete lattice, (As a result, the morphisms in the category have to be monotone. This is only a guess on my part) then I can take the discrete and indiscrete version of those objects for left and right adjoints respectively? I know for the case of the forgetful functor sending posets to sets, the notion of galois connections ($\textbf{Example}$ above) come into it for adjunction. How does that related to the solutions of the two questions above.
Thank you in advance.