The forgetful functor $U:\mathsf{Top}\to\mathsf{Set}$ sends a topological space to its underlying set of points. We have the following adjunction : where $D:\mathsf{Set}\to\mathsf{Top}$ sends a set to the discrete topology on that set, i.e. the one in which all sets are open, and $I:\mathsf{Set}\to\mathsf{Top}$ sends a set to the indiscrete topology on that set, i.e. the one in which the empty set and the entire set are the only open sets.
For each set $A$, we have a universal arrow $\eta_A:A\to UDA$, where $UDA$ is simply the set $A$ itself.
Universality states that for each function $f:A\to UDB$, or $f:A\to B$, there is a unique $f':DA\to DB$ such that the above diagram commutes. This is clear, because every function originating from a topological space endowed with the discrete topology is continuous, meaning that any set function $f:A\to B$ extends to a continuous function $f':DA\to DB$. In the adjunction $D\dashv U$, we have:
The unit $\eta:1_{\mathsf{Set}}\Rightarrow UD$ is the identity natural transformation.
The counit $\epsilon:DU\Rightarrow 1_{\mathsf{Top}}$ shows how for each topological space $X$, an "exploded" version of that space in which all points are separated, given by $DUX$, can be continuously embedded in the original space.
For each set $A$, we have a universal arrow $\epsilon_A:UIA\to A$, where $UIA$ is simply the set $A$ itself.
Universality states that for each function $f:UIB\to A$, or $f:B\to A$, there is a unique $f':IB\to IA$ making this diagram commute. Again, this is clear because any function into a topological space endowed with the indiscrete topology is continuous, and hence any set function $f:B\to A$ extends to a continuous function $f':IB\to IA$. In this adjunction $U\dashv I$, we have:
The unit $\eta:1_{\mathsf{Top}}\Rightarrow IU$ shows how each topological space can be continuously embedded in a "smashed together" version of that space in which no two points are separated.
The counit $\epsilon:UI\Rightarrow 1_{\mathsf{Set}}$ is simply the identity natural transformation.
The discrete topology functor $D$ does not admit a left adjoint. To see why, we shall prove that there exists no universal arrow from the topological space $X=\mathbb Q$ with the usual topology to the functor $D$. Let us suppose that such a universal arrow were to exist: In order for $u$ to be continuous, it would have to be locally constant (i.e. each point would have to be contained in some neighborhood on which $u$ is constant) because $DY$ has the discrete topology. In particular, $N=u^{-1}(u(0))$ must be a neighborhood of $0\in\mathbb Q$ in which $u$ is constant. We may choose an open ball $B_\epsilon(0)\subset N$ on which $u$ is constant as well.
Now, consider $D2$, the space consisting of the discrete topology on $2$ points. We may define a function $f:\mathbb Q\to D2$ that sends $x\mapsto 0$ if $x\in B_{\epsilon/2}(0)$, and $x\mapsto 1$ otherwise, and this function is continuous because it is locally constant and $D2$ has the discrete topology. However, there cannot exist $f'$ such that $f=(Df')\circ u$, for $u$ is constant on $B_\epsilon(0)$ but $f$ is nonconstant on this set. Hence, we cannot have such a universal arrow $u$ at all, and the desired adjunction is impossible.
The indiscrete topology functor $I$ does not admit a right adjoint. Notice that the only continuous functions originating from a topological space with the indiscrete topology are ones that send all points of the domain to points of the codomain which cannot be separated by any neighborhood. That is, for any continuous function $f:IA\to X$, there is a subset $S\subset X$ such that for any $x_1,x_2\in S$ and for all open sets $\mathcal O\subset X$, we have $x_1\in \mathcal O \iff x_2\in \mathcal O$. If $X$ is Hausdorff like $\mathbb R$, this means that $S$ must be a singleton. Hence, if there were to exist a universal arrow then it would have to be a constant map, mapping all elements of $IA$ to some real number $r\in\mathbb R$. However, we may also define a continuous (constant) map $f:IB\to\mathbb R$ sending $b\mapsto r+1$ for all $b\in IB$. Clearly there cannot exist $f'$ such that $f=u\circ (If')$, since $u$ only takes on the value $r$ but $f$ only takes on the value $r+1$. Hence, the universal arrow cannot exist, nor can the desired adjunction.
The adjoint pair $D\dashv U$ gives rise to a monad , but not a very interesting one! The composite $UD$ is simply equal to the identity functor $1_\mathsf{Set}$, and the unit and multiplication of the monad are just identity natural transformations. So much for that monad.
The adjoint pair $U\dashv I$ also gives rise to a monad, which, although its behavior is not so complex, is at least nontrivial. This monad we will call the indiscrete topology monad, and its data is given by $\langle T,\eta,\mu\rangle=\langle IU,\eta,I\epsilon U\rangle$. Here's an interpretation for what each of these pieces do:
The endofunctor $T=IU:\mathsf{Top}\to\mathsf{Top}$ sends each topological space $X$ to indiscrete topological space with the same set of points. Intuitively, we may think of $T$ as "gluing together" all of the points of $X$ into a single homogenous mass.
The unit $\eta:1_{\mathsf{Top}}\Rightarrow T$ lifts the identity map between the point-sets of $X$ and $IUX$ - which is a mere set function - to a morphism of $\mathsf{Top}$, i.e. a continuous function from $X$ into $TX$. This is possible because any function into an indiscrete topological space is continuous.
The multiplication $\mu:T^2\Rightarrow T$ is just the identity natural transformation, because the functor $T$ is idempotent by its definition.